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Received: 26 November 2019 Revised: 10 January 2020 Accepted: 19 January 2020 DOI: 10.1111/obr.13005 PEDIATRICS/PHYSIOLOGY Adipokines: A gear shift in puberty Desirée Nieuwenhuis | Natàlia Pujol-Gualdo Amanda J. Kiliaan Department of Anatomy, Radboud university medical center, Donders Institute for Brain, Cognition and Behaviour, Preclinical Imaging Center PRIME, Nijmegen, The Netherlands Correspondence Amanda J. Kiliaan, PhD, Associate Professor, Department of Anatomy, Donders Institute for Brain, Cognition, and Behaviour, Preclinical Imaging Center PRIME, Radboud university medical center, 6500 HB Nijmegen, Geert Grooteplein 21N 6525 EZ Nijmegen, The Netherlands. Email: amanda.kiliaan@radboudumc.nl Funding information Europees Fonds voor Regionale Ontwikkeling (EFRO), Grant/Award Number: BriteN 2016 1 | INTRODUCTION The prevalence of obesity in adolescents and children is increasing in | Ilse A.C. Arnoldussen | Summary In this review, we discuss the role of adipokines in the onset of puberty in children with obesity during adrenarche and gonadarche and provide a clear and detailed overview of the biological processes of two major players, leptin and adiponectin. Adipokines, especially leptin and adiponectin, seem to induce an early onset of puberty in girls and boys with obesity by affecting the hypothalamic-pituitary- gonadal (HPG) axis. Moreover, adipokines and their receptors are expressed in the gonads, suggesting a role in sexual maturation and reproduction. All in all, adipokines may be a clue in understanding mechanisms underlying the onset of puberty in child- hood obesity and puberty onset variability. KEYWORDS adipokines, obesity, puberty 1,2 the age of 5 years were overweight or were with obesity in 2016, and 3 Obesity is defined by an excessive accumulation of white adipose tissue (WAT), and it is often indicated by a body mass index (BMI) 4 above 30. Two main types of adipose tissue were described: WAT and brown adipose tissue (BAT), which differ in morphology and func- 5-7 Ilse A.C. Arnoldussen and Amanda J. Kiliaan contributed equally to this work. This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. © 2020 The Authors. Obesity Reviews published by John Wiley & Sons Ltd on behalf of World Obesity Federation Obesity Reviews. 2020;21:e13005. wileyonlinelibrary.com/journal/obr 1 of 10 https://doi.org/10.1111/obr.13005 alarming rates. Specifically, worldwide, 41 million children below this number is expected to increase to 70 million in 2025. obesity is associated with various severe health complications, includ- ing increased risk of diabetes mellitus type 2, hypertension, heart dis- eases, and disturbances in sex hormone levels. 5,6 and mitochondria and plays a role in thermogenesis. Adipocytes in tion. BAT consists of adipocytes containing multiple lipid droplets WAT contain only a few mitochondria and a single lipid droplet. Adipose tissue has several functions including the storage of energy, thermogenesis, and the production and secretion of adipokines Generally, two physiological processes, adrenarche and gonadarche, 11,24 Childhood 5,7,8 a key role in puberty onset. Puberty is known as a period through which the body changes physically, being a physiological process resulting in the maturation of children, i.e. they develop sexual characteristics and obtain reproduc- 9,11 Adipokines are involved in a number of physiological processes including blood pressure, metabo- lism, glucose, and vascular homeostasis and may play amongst others 8-10 (hormones, cytokines, and peptides). tive functions. between obesity and puberty,2,12-23 the biological mechanisms under- lying obesity and puberty onset remain unclear. Hereafter, we review in detail the role of adipokines in the onset of puberty in childhood obesity. Although many studies have shown associations 2 | INITIATION OF PUBERTY PHYSIOLOGICAL PROCESSES IN THE interact to regulate the onset of puberty. During adrenarche, the adrenal cortex secretes steroid hormones (including 2 of 10 NIEUWENHUIS ET AL. androstenedione, dehydroepiandrosterone, dehydroepiandrosterone sulfate (DHEAS), androstenedione, and cortisol), insulin-like growth factor, and growth hormone, which contribute to the pubertal insights on new genetic loci (e.g. melanocortin-4 receptor, mitochon- drial carrier 2, and mitogen-activated protein kinase 13) and on sev- eral pathways that regulate the timing of puberty; however, it partly 34 9,24,25 Both adrenarche and gonadarche are involved in the development growth spurt, body odor, skin oiliness, and skeletal maturation. explains puberty timing variation. Thereby, defining the role of 25 adipokines is of importance in elucidating the variability in puberty as the expression of adipokines is sex-specific and is altered with body composition, adiposity, and during growth spurts. Moreover, adipokines and their receptors are expressed in gonads and several brain regions suggesting involvement in the onset of puberty and sex- ual maturation. Lastly, adipokines interfere in processes regulating timing and duration of puberty, for instance in the HPA and HPG axes which are both key players during adrenarche and gonadarche. Involvement of adipokines in the onset of puberty and specifically in individuals with obesity will be further reviewed in the next 2,24 3 | Puberty onset in girls is assessed using different markers, such as thelarche (breast development), menarche (the start of of pubic hair. pituitary-gonadal (HPG) axis is activated,2,26 and several hormones have been identified to participate in the activation of the HPG axis During gonadarche (Figure 1), the hypothalamic- 2,27 Kisspeptin, neurokinin B, and dynorphin are released by specialized including kisspeptin, neurokinin B, dynorphin, leptin, and ghrelin. 28 key regulator of the pulsatile secretion of gonadotropin releasing neurons, the KNDy neurons in the hypothalamus. Kisspeptin is a 29,30 B stimulates, and dynorphin inhibits the release of kisspeptin, which hormone (GnRH) from the hypothalamus. In addition, neurokinin implies that both coordinate a pulsatile release of kisspeptin. 31 Sub- sections. sequently, the activated HPG axis induces the pituitary gland to secrete luteinising hormone (LH) and follicle stimulating hormone (FSH). As a result, gametogenesis occurs, and the gonads will release sex hormones. Consequently, secondary sex characteristics develop including breast development in girls and an increased testicular vol- 2,26,32 is possibly due to differences in levels of body fat, hypothalamic-pitui- THE ONSET OF PUBERTY IN GIRLS ume in boys. The age at puberty onset varies greatly among individuals, which 19 35 menstruation), and pubic hair development. 33 genome-wide association studies have provided important new tary-adrenal (HPA) axis activity, and genetic background. Recent The average age of However, this age differs between cultures and ethnicities, and since 1980, age at menarche is girls at start of menarche is 12.4 years. 36 significantly decreasing. 36-39 F I G U R E 1 Hormonal regulation in the initiation of puberty in boys and girls. The secretion of kisspeptin, neurokinin B, and dynorphin from KNDy neurons initiate the release of gonadotropin releasing hormone (GnRH) from the hypothalamus. This activates the pituitary gland to produce and secrete luteinising hormone (LH) and follicle stimulating hormone (FSH), which in turn stimulate the gonads to produce estrogen and testosterone in girls and boys, respectively 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License NIEUWENHUIS ET AL. 3 of 10 T A B L E 1 Summary of included studies Authors Year Country Study Design Primary Outcome Sex Sample Size (n) Age (y) Data Collection Lian et al21 2019 China Cross-sectional Puberty starts earlier in Chinese Han girls with obesity compared with Chinese Han girls with normal weight. Girls 2996 9-19 2012 and 2013 Biro et al12 Lazzeri et al20 2018 USA 2018 Italy Longitudinal Cross-sectional Body mass index had a greater effect on age at menarche than did race and ethnicity. Girls 946 6-16 2004-2014 Li et al23 2018 China Longitudinal For both, boys and girls, a higher BMI (ie, overweight and obese) is associated with earlier onset of puberty Girls Girls Boys Girls 542 Deng et al22 Flom et al15 2017 China Cross-sectional Increased BMI is associated with early timing spermarche and menarche. Boys Girls Girls 1278258 9-15 2005-2012 He et al24 Holmgren et al17 2017 China 2017 Sweden Cross-sectional Longitudinal Onset of puberty is not related to obesity in boys. Boys Boys Girls Girls 782 7-17 972 929 5839 Kelly et al19 2017 UK 2016 Brazil 2016 USA Longitudinal prospective cohort Higher BMI in girls is associated with the onset of menstruation at an earlier age. 11 10-18 11-17 Barcellos Gemelli et al25 Cross-sectional Longitudinal Excess weight is associated with early age of menarche. Girls 727 2014 2003-2009 Glass et al16 Lee et al26 In girls, but not in boys, greater adiposity is associated with the earlier onset of puberty. Boys Girls 135 Cabrera et al27 Leonibus et al14 2014 USA 2013 Italy Cross-sectional Longitudinal Thelarche occurred earlier than recently reported, while age of menarche remained unchanged. Girls 610 3-17.9 2007 2005-2012 Currie et al13 2012 Europe, USA, Canada Cross-sectional Overweight/obesity during childhood predicts the early onset of puberty in girls. Girls 20410 11, 13, 15 2005-2006 2017 USA Prospective birth cohort Overweight/obese status at the age of 7 ye was associated with increased risk of early menarche 788 From birth to menarche occurred Pregnancies 1959-1966 2016 USA Cross-sectional Boys with overweight enter puberty earlier compared with boys with normal weight or obesity, while puberty starts later in boys with obesity compared with boys with normal weight and overweight. Boys 3872 6-16 2005-2010 Overweight during childhood shows a relation with the early onset of puberty in girls. 6535 4259 695 11 15 5.8-12.2 2009/2010 2013/2014 2014-2017 Higher BMI during childhood is associated with early puberty. 2008 and 2009 2000-2002 Obesity during childhood is related to the earlier onset of puberty. Boys Girls 84 123 71 (Continues) 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 4 of 10 NIEUWENHUIS ET AL. 3.1 | Fat storage For the initiation of puberty, the timing of stimulation and/or inhibi- tion of different hormones is important, and additionally, a certain amount and distribution of body fat is needed in order to start menar- che, which emphasizes the importance of body fat. From an evolution- ary point of view, body fat increases in mammalian females during puberty onset, and it highlights the need to guarantee a healthy preg- 40 women with anorexia nervosa. particularly body fat localized predominantly on the gluteofemoral fat depots, is profoundly associated with start of menarche, more than nancy, offspring, and maternal survival. fat, sex-hormones, and neuroendocrine alterations can evolve in men- strual dysfunction, for instance, in women with severe obesity or in 41-43 44-46 to gluteofemoral fat depots suggesting that leptin may convey infor- amount of total body fat. mation on body fat distribution to the hypothalamus during puberty. An improper level of body Importantly, body fat distribution, Blood leptin levels are strongly related 45 3.2 | HPG axis The HPG axis is activated by the release of kisspeptin resulting in the release of GnRH from the hypothalamus, and LH and FSH from the pituitary gland. In girls, FSH is involved in the development of the folli- cles in the ovaries, and it promotes the secretion of estrogen. LH stim- ulates the production of androgen hormones and induces ovulation 48 9,47 the release of kisspeptin and neurokinin B, and kisspeptin thereby (Figure 1). The secretion of estrogen has an inhibitory effect on inhibits the GnRH release from the hypothalamus. pattern of GnRH is important for the regulation of the menstrual cycle. This roughly 28-day-cycle comprises several phases, including the follicular phase and luteal phase. During the follicular phase, increasing levels of FSH stimulate the maturation of follicles and the production of estrogen from the ovaries. This in turn inhibits the release of FSH from the pituitary gland. A high level of estrogen will induce the production of LH by the pituitary gland, resulting in ovula- tion. The matured follicle secretes progesterone thereby inhibiting the release of GnRH. When the corpus luteum is demolished, there is less 48 3.3 | Adipokines According to results from studies reported in Table 1, girls with obe- sity enter puberty earlier compared with girls with normal higher leptin concentrations inhibit the intake of food and increases inhibition of GnRH. As a consequence, the cycle will start again. whole process, starting from the activated HPG axis, results in the development of the secondary sex characteristics in girls including 9,47 thelarche and menarche. 13,14,16-23,49-51 weight. these girls might be found in the secretion of adipokines. For instance, leptin is positively associated with the amount of body fat. Generally, energy expenditure. 9,52-54 An explanation for the early onset of puberty in The expression This TABLE 1 (Continued) Authors Year Country Study Design Primary Outcome Sample Sex Size (n) Age (y) Data Collection Herman-Giddens et al28 2012 USA Cross-sectional Observed mean ages of beginning genital and pubic hair growth and early testicular volumes were earlier than in past studies, depending on the characteristic and race/ethnicity. Boys 4131 6-16 2005-2010 Sorensen et al29 Aksglaede et al30 2010 2009 Denmark Denmark Cross-sectional/longitudinal Longitudinal Puberty onset at earlier ages was associated with an increased BMI in boys. Boys 1528 5.8-19.9 1991-1993/2006-2008 1930-1969 Juul et al31 Ribeiro et al32 2007 2006 Denmark Portugal Retrospective cohort Cross-sectional Higher BMI is associated with early voice break. 11-15 10-15 1990-1999 Kaplowitz et al18 Abbreviation: BMI, body mass USA Cross-sectional The early onset of puberty in Caucasian girls is likely related to an increased BMI. 5-12 1992-1993 2001 index. The higher BMI in boys and girls at 7 y of age, the earlier they enter puberty. Boys 21 612 Girls 135 223 Boys 463 Boys 382 Girls 437 Girls 10 750 Early sexual maturation in boys and girls is associated with overweight. 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License NIEUWENHUIS ET AL. 5 of 10 Leptin may possibly play a role in adrenarche as its plasma level increases with higher levels of body fat and as it can modulate both girls. 33 ing adrenarche. In coherence, in children with obesity, the androgen These findings suggested that lower reproductive status was associated with higher total adiponectin concentrations and that a higher reproductive status was related to higher HMW adiponectin the HPA and HPG axes. These axes are functionally integrated dur- DHEAS was positively associated with leptin levels. Nevertheless, concentrations in girls. In addition, individuals with obesity often another study showed that enhanced adrenal androgen secretion in girls with premature adrenarche was not explained by leptin or BMI 55 ated with androgen levels in girls ; however, it was not related to levels. and IL-6. TNF-α alters, and IL-6 inhibits the expression of 56 8 In addition, the adipokine adiponectin was negatively associ- 57 differences of adiponectin seem to develop during the progression of 56 adiponectin (Figure 2). Thereby, a low level of total adiponectin and/or high levels of inflammatory cytokines in individuals with obe- sity can promote the onset of puberty. Many more adipokines are secreted by WAT including omentin, 52,65-67 9,36,62,68 adrenarche in girls with Prader-Willi syndrome. Interestingly, sex puberty. adrenarche; however, both are not required factors. Thus, leptin and adiponectin might be able to influence In gonadarche, leptin can stimulate the secretion of kisspeptin, and subsequently activation of the HPG axis, which eventually increases the expression of estrogen and androstenedione in the ova- 58 2,60 65-67 The expression of these ries (Figure 2). Ob gene in WAT, resulting in the synthesis and secretion of leptin. Thus, high levels of leptin promote onset of puberty in girls via secre- tion of kisspeptin, and estrogen stimulates leptin secretion addition- ally. Moreover, adiponectin can affect the HPG axis due to the expression of adiponectin receptors in the hypothalamus, pituitary In return, estrogen stimulates the expression of the 59 gland, and gonads. onset as it inhibits the secretion of kisspeptin and GnRH in the hypo- thalamus and the release of GH and LH in the pituitary gland, and 2,60-62 52,60 63 girls with central precocious puberty (CPP). Moreover, total adiponectin had negative correlations with progression of puberty in girls (defined by Tanner stages), whereas HMW adiponectin had FIGURE 2 Adipokinesaffectingthe initiation of puberty in girls. Leptin stimulates the release of kisspeptin in KNDy neurons, which activates the hypothalamus to produce gonadotropin releasing hormone (GnRH). In response to the release of GnRH, the pituitary gland secretes follicle stimulating hormone (FSH) and luteinising hormone (LH), which stimulates the ovaries to release estrogen resulting in the formation of secondary sex characteristics in girls. Estrogen stimulates the production of leptin. Adiponectin inhibits GnRH release resulting in reduced levels of GnRH and thereby a delayed onset of puberty. TNF- α and IL-6 inhibit the production of adiponectin and therefore stimulate the onset of puberty In detail, adiponectin is a regulator of puberty thereby inhibiting the onset of puberty (Figure 2). with obesity often have low levels of adiponectin. et al. showed that total adiponectin was significantly lower, whereas high molecular weight (HMW) adiponectin was significantly higher in ment. 55 63 develop a chronic low-grade inflammatory state, which can be indi- cated by a high level of circulating inflammatory cytokines like TNF-α 64 Individuals Sitticharoon positive associations with LH levels and the progression of puberty in 63 visfatin, resistin, and chemerin. and visfatin are expressed in the ovaries. adipokines in the ovaries suggests a role within the reproductive sys- tem; however, the exact biological processes have to be examined. Thus, specifically leptin, adiponectin, and inflammatory cytokines pro- duced by WAT could be permissive key players during an early onset of puberty in girls with obesity. As an exception, HMW adiponectin seems to have a stimulatory effect on peripheral repro- ductive function as HMW is not able to cross the blood brain 63 barrier. 4 | Markers that are used to assess puberty onset in boys are THE ONSET OF PUBERTY IN BOYS spermarche, voice break, testicular volume, and pubic hair develop- 35 spermarche develop in the early stages of puberty onset, voice In women, omentin, chemerin, While pubic hair development, larger testicular volume, and 69 testicular volume increases, which occurs at an average age of break usually appears in later stages of puberty. Generally, first 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 6 of 10 NIEUWENHUIS ET AL. 11.9 years, followed by the development of pubic hair at 12.2 years of average, and lastly, boys experience spermarche around an aver- 55 related with leptin levels. Thereby, leptin plausibly has a minor impact in adrenarche in boys. Since leptin receptors are found in the hypothalamus, pituitary gland, and testes, they might be involved in the onset of puberty by affecting the HPG axis during gonadarche. Leptin stimulates the release of kisspeptin and GnRH, and as a consequence, it accelerates the onset of puberty (Table 1, Figure 3). In contrast, adiponectin inhibits the secretion of GnRH, GH, LH, and FSH therewith delaying the onset of puberty. However, adiponectin levels are generally lower in men compared with women and even lower in men with obe- age age of 13.4 years. 70 4.1 | Fat storage Many aspects of the reproductive physiology are energetically demanding,71 and therefore, an adequate energy level is necessary. In boys, a dynamic change in body composition occurs around the age of 10 to 13 years, in which they gain approximately 40% of sity. culating inflammatory cytokines. levels can stimulate the HPG axis and therewith an early onset of puberty in boys. Nevertheless, leptin can inhibit the production of tes- 72 mostly consisting of lean mass, which causes exhaustion of most of fat. Subsequently, a growth spurt follows in which they gain tissue 72 in boys, an adequate amount of body fat is important in the onset of their body fat. These alterations in amount of body fat indicate that 4.2 | Puberty in boys is initiated by the release of kisspeptin. As mentioned before, this activates the HPG axis, resulting in the release of GnRH from the hypothalamus, and consequently the release of LH and FSH 9,74 puberty. tosterone from the testes, to estrogen (Figure 3). of the development of secondary sex characteristics in boys. Additionally, leptin can affect fertility in men as it can modulate the nutritional support of spermatogenesis, and moreover, dysfunction of spermatogenesis is associated with an increased leptin level and 73 58 2,60-62 HPG axis from the pituitary gland (Figure 1). and LH stimulates the secretion of testosterone from the testes, which inhibits the release of kisspeptin from the KNDy neurons and 9,48 in men, the release of kisspeptin is more consistent, causing a con- 29,48 subsequently GnRH from the hypothalamus. receptors expressed on KNDy neurons. In humans, KNDy neurons Contrarily to women, LH-induced testosterone levels lead to the stant release of LH. development of secondary sex characteristics in boys. differences between sexes in kisspeptin release are related to a sex- specific and sex steroid-dependent kisspeptin system as estrogen and progesterone modulate kisspeptin activity through the sex-steroid 48 in the infundibular nucleus are involved in negative and positive sex- 48 tal exposure to sex steroids and result in sex-specific differences in steroid feedbacks. kisspeptin release. These sexual dimorphisms are induced by perina- 75,76 4.3 | Adipokines The association between obesity and puberty onset in boys is rather controversial compared with findings in girls. Most studies reported an early onset of puberty in boys associated with increased ate adipose tissue from actual breast tissue. stages are more difficult to assess than female stages as boys lack a more determined marker such as menarche. Thirdly, puberty onset can be indicated by the activation of the HPG axis, and the presence of these secondary sex characteristics is the result of hormonal 2 14,17,22,23,50,51,77,78 BMI, 20,49 all while others reported no associations at Current markers used 79 16,80 or a delayed onset of puberty (Table 1). The presence of excessive adipose tissue can be involved in puberty onset in boys as the secretion of adipokines can modulate both adrenarche and gonadarche. Leptin can affect adrenarche by modulating both the HPG and HPA axes,33 and moreover, androgen levels were positively 55 nal androgen secretion in boys with premature adrenarche was not associated with plasma leptin levels. Nevertheless, enhanced adre- 9 In more detail, 61,62 adiponectin, and individuals with obesity often have high levels of cir- Moreover, inflammatory cytokines, TNF-α, and IL-6, inhibit expression of the leptin receptor in the testis. FSH induces spermatogenesis, too. function and role still have to be examined. 64 High leptin and low adiponectin and fat tissue can convert testosterone Both processes might result in the delay 29,61,79 81,82 In men, other adipokines like chemerin are found in the gonads 65 Thus, particularly high leptin and low adiponectin levels stimulate the HPG axis and thereby accelerate the onset of puberty in boys. Additionally, leptin can dysregulate the development of secondary sex characteristics and spermatogenesis by affecting testosterone levels and nutritional sup- port of spermatogenesis. 5 | LIMITATIONS AND FUTURE RESEARCH DIRECTIONS Even though multiple epidemiological studies have shown the link between puberty onset and obesity, there are some important limita- tions. Firstly, determining both the onset and stage of puberty is rather difficult. For instance, assessing the stage of breast develop- ment in girls with obesity is complicated as clinicians should differenti- 2 changes in response to the activated HPG axis. to determine the onset of puberty refer to secondary sex characteris- tics, such as testicular volume in boys and breast development in girls. A more accurate measurement of puberty onset would be to combine secondary sex characteristics with plasma or serum hormone level measurements such as LH, FSH, adipokines, e.g. leptin. Thereby, differences in puberty measurements could explain variations in the age of puberty onset between boys and girls within different Thereby, resistin is expressed in the testes of rats, but its exact 83 Secondly, male pubertal 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License NIEUWENHUIS ET AL. 7 of 10 FIGURE 3 Adipokines affecting the initiation of puberty in boys. Leptin activates kisspeptin secretion in KNDy neurons, this activates the production of gonadotropin releasing hormone (GnRH) from the hypothalamus. GnRH stimulates the pituitary gland to secrete follicle stimulating hormone (FSH) and luteinising hormone (LH), activating the production of testosterone from the testes allowing the development of secondary sex characteristics. Leptin also inhibits the production of testosterone, which may cause a delayed onset of puberty. Adiponectin inhibits GnRH release. Low levels of adiponectin, as a result of TNF-α and IL-6 expression, lead to a reduced inhibition of GnRH. In response to GnRH release, the pituitary gland will secrete FSH and LH, and the testes will produce testosterone resulting in the development of secondary sex characteristics in boys countries, and In addition, the inclusion of a of puberty. ferent time points is complicated, as subjects examined several decades ago presented pronounced differences concerning lifestyle patterns such as nutrition and exercise habits. Lastly, obesity or over- weight is often determined by BMI, a classification based on weight and height measurements. Additionally, it is important that all studies studies or across continents, ethnicities proper age range (8-16 years) is important when assessing the onset (Figure 4). 12-15,17,20-23,49,77-79,84,85 30,47 Furthermore, comparison between studies from dif- 86 Specifically in children, BMI is often dependent on age and growth use the same anthropometric standards and sex-specific cut-offs. 13,14,16-23,49-51,77-80 fat and would represent a more accurate measurement in its regard. Based on this review, several suggestions can be made for further research. Firstly, the roles of adipokines like resistin, chemerin, visfatin, and omentin in puberty onset, fertility, and sexual maturation should be examined in detail. Secondly, future research examining the onset of puberty should combine indicators of puberty onset (e.g. breast development or testicular volume) with plasma or serum hor- mone measurements such as LH, FSH, sex-steroids, adipokines (e.g. spurts. ment in case of growth spurts. distribution of body fat should be taken into account in determining puberty and obesity in children. For instance, the body adiposity index (BAI), which was introduced in 2011 by Bergman et al.,87 uses hip cir- cumference and height in order to estimate the percentage of body 87 Thereby, BMI is a less accurate measure- F I G U R E 4 87,88 Therefore, both percentage and Average age of puberty onset in Europe, China, and the United States according to several studies from Table 1. Age of puberty onset ranges from 8.47 to 13.33 years in girls and from 8.63 leptin), and body fat distribution (e.g. BAI,87 waist-hip ratio's and/or dual-energy X-ray absorptiometry (DXA)2). Additionally, defining con- sistent and general measurements of puberty in both boys and girls, combined with a proper age range (8-16 years), would facilitate the comparisons between different studies and their results. 12-15, 17, 20-23, 25-29, 31 to 13.7 years in boys. included if average age of markers used to assess puberty was not reported. Pink: girls. Blue: boys Studies (Table 1) were not 39, 56 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 8 of 10 NIEUWENHUIS ET AL. 6 | CONCLUSION In conclusion, epidemiological data regarding obesity and puberty onset in girls show similar outcomes as adiposity results in the early onset of puberty in girls. 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Ann Epidemiol. 2017;27(3):187-93.e2. 16. Glass NA, Torner JC, Letuchy EM, et al. The relationship between greater prepubertal adiposity, subsequent age of maturation, and bone strength during adolescence. Journal of bone and mineral research: the official journal of the American Society for Bone and Min- eral Research. 2016;31(7):1455-1465. 17. Holmgren A, Niklasson A, Nierop AF, et al. Pubertal height gain is inversely related to peak BMI in childhood. Pediatr Res. 2017;81(3): 448-454. 18. Kaplowitz PB, Slora EJ, Wasserman RC, Pedlow SE, Herman- Giddens ME. Earlier onset of puberty in girls: relation to increased body mass index and race. Pediatrics. 2001;108(2):347-353. 19. Kelly Y, Zilanawala A, Sacker A, Hiatt R, Viner R. Early puberty in 11-year-old girls: Millennium Cohort Study findings. Arch Dis Child. 2017;102(3):232-237. 20. Lazzeri G, Tosti C, Pammolli A, et al. Overweight and lower age at menarche: evidence from the Italian HBSC cross-sectional survey. BMC Womens Health. 2018;18(1):168-174. thereby an early onset of obesity. leptin can inhibit the production of testosterone in boys and subse- quently inhibit the development of secondary sex characteristics affecting spermatogenesis. for other adipokines, like resistin and omentin, are present in the testes and ovaries suggesting a role in puberty or reproduction; 58, 71 however, their plausible function is still unknown. that adipokines may be key regulators in an early onset of puberty in both girls and boys with obesity, specifically by affecting the HPG axis during gonadarche. Future research should focus on assessing puberty onset by measuring consistent puberty markers and determine the percentage of body fat and its distribution and adipokines and hormone serum levels particularly involved in the HPG axis. CONFLICTS OF INTEREST The authors declare no conflict of interest. FUNDING INFORMATION This research was funded by Europees Fonds voor Regionale Ontwikkeling (EFRO), project BriteN 2016. ORCID Ilse A.C. Arnoldussen Amanda J. Kiliaan https://orcid.org/0000-0002-7395-5284 https://orcid.org/0000-0002-2158-6210 13, 14, 16-26, 29-32 Furthermore, several receptors Nevertheless, We conclude Search strategy We searched PubMed for articles published before Novem- ber 15th, 2019 using relevant keywords, including ‘onset of puberty and adiposity/obesity’, ‘onset of puberty’, ‘children with obesity’, ‘adipose tissue’, ‘childhood obesity’, ‘adiposity’, ‘obesity’, ‘adipokine(s)’, ‘HPG axis’, ‘adipokines ovary/ova- ries’, or ‘adipokines testes’, either alone or in combination. Selection criteria used were English language, longitudinal or cross-sectional studies assessing the onset of puberty, including menarche, thelarche, spermarche, or voice break, combined with high BMI or obesity/adiposity, and articles assessing or reviewing adipokines and its effects on the reproductive system. 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License NIEUWENHUIS ET AL. 9 of 10 21. Lian Q, Mao Y, Luo S, et al. Puberty timing associated with obesity and central obesity in Chinese Han girls. BMC Pediatr. 2019; 19(1):1-7. 22. Deng Y, Liang J, Zong Y, et al. Timing of spermarche and menarche among urban students in Guangzhou, China: trends from 2005 to 2012 and association with Obesity. 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Circulating plasma leptin and IGF-1 levels in girls with premature adrenarche: potential implications of a preliminary study. Horm Metab Res. 2001;33(3):138-143. 34. Cousminer DL, Widén E, Palmert MR. The genetics of pubertal timing in the general population: recent advances and evidence for sex-spec- ificity. Curr Opin Endocrinol Diabetes Obes. 2016;23(1):57-65. 35. Marshall WA, Tanner JM. Variations in pattern of pubertal changes in girls. Arch Dis Child. 1969;44(235):291-303. 36. Lacroix AE, Whitten R. Physiology. Treasure Island (FL): Menarche. StatPearls. StatPearls Publishing; 2018. 37. McDowell MA, Brody DJ, Hughes JP. Has Age at Menarche Chan- ged? Results from the National Health and Nutrition Examination Sur- vey (NHANES) 1999–2004. J Adolesc Health. 2007;40(3):227-231. 38. de Muinich Keizer SM, Mul D. Trends in pubertal development in Europe. Hum Reprod Update. 2001;7(3):287-291. 39. Talma H, Schönbeck Y, van Dommelen P, Bakker B, van Buuren S, Hirasing RA. Trends in menarcheal age between 1955 and 2009 in the Netherlands. PLoS ONE. 2013;8:e60056-e60056. 40. Kaplan HS, Lancaster JB. An evolutionary and ecological analysis of human fertility, mating patterns, and parental investment. Off- spring: Human fertility behavior in biodemographic perspective. 2003;1: 170-223. 41. Mitan LA. Menstrual dysfunction in anorexia nervosa. J Pediatr Adolesc Gynecol. 2004;17(2):81-85. 42. Xu H, Li P-H, Barrow TM, et al. Obesity as an effect modifier of the association between menstrual abnormalities and hypertension in young adult women: Results from Project ELEFANT. PLoS ONE. 2018; 13(11):e0207929-e0207929. 43. Tauqeer Z, Gomez G, Stanford FC. Obesity in women: insights for the clinician. J Womens Health (Larchmt). 2018;27(4):444-457. 44. de Ridder CM, Thijssen JH, Bruning PF, Van den Brande JL, Zonderland ML, Erich WB. 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Obesity Reviews. 2020;21:e13005. https://doi.org/ 10.1111/obr.13005 1467789x, 2020, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1111/obr.13005, Wiley Online Library on [10/03/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are goverBased on the PowerPoint you shared, here is a simple quiz focusing on vocabulary, exponent laws, and identifying function types. --- Quiz: Exponent Laws & Rational Exponents (3.1–3.3) Multiple Choice (5 questions) Choose the correct answer. 1. In the expression 5^3, the number 3 is called the a) base b) power c) exponent d) coefficient 2. Which law of exponents would you use to simplify (x^2)^3? a) Product rule b) Quotient rule c) Power of a power rule d) Zero exponent rule 3. According to the zero exponent law, 7^0 = a) 0 b) 1 c) 7 d) undefined 4. If the first differences in a table of values are constant, the function is a) linear b) quadratic c) exponential d) not a function 5. Which expression is equivalent to \frac{2^5}{2^3}? a) 2^2 b) 2^8 c) 2^{15} d) 2^{-2} True or False (5 questions) Write T for true or F for false. 1. When multiplying powers with the same base, you add the exponents. 2. A negative exponent means the answer will always be negative. 3. For an exponential function, the ratios of consecutive y-values are constant. 4. The power 16^{\frac{1}{2}} is equal to 8. 5. The quotient rule for exponents says \frac{x^a}{x^b} = x^{a-b}. Completion (2 questions) Fill in the blank with the correct term. 1. The _____________ rule states that when raising a power to another power, you multiply the exponents. 2. If the second differences in a table of values are constant, the function is ______________. --- Answer Key Multiple Choice 1. c) exponent 2. c) Power of a power rule 3. b) 1 4. a) linear 5. a) 2^2 True or False 6. T 7. F (a negative exponent indicates a reciprocal, not a negative value) 8. T 9. F (16^{\frac{1}{2}} = \sqrt{16} = 4) 10. T Completion 11. power of a power (or power rule) 12. quadraticEff..rs of ott.-PoFllat i What woLrld hoppen ro our colnrry i, it is ovetsp.pulored? When our counrry is ov€.-populdted, re @ €xp€ri.nce rh€ foll.wirg: Food is our bdsic h@d. Wh€n th€.Cs an ih.re.se ir populdtion it neans thar hore ,@d is iealed. It rheds ho .naJgh food, rrtrple irll srruggle wirh eddr oth€r in ordeLro €!'r- As o l!fllr, lhde rill be o f@d -- , ond ou, now]nert of on ihdiyiduol fron d c..tair - the move$eni o, on individudl our of o cerrain pla.e which help r€duce ihe populotion of th6t fr Arcih€. b.sic ned is w.ra. Wde. shorroge ocu.s when there is on ircreare of hu,nber of p@ple ro be $pptied. rn owr-popur.t d ore.s, woler is rdior€d, Ir rEB rhoi supplies like ti,tWSS ond ,IWSI can'i $pply enoish worer. Do you hdve enough supply of sai.. in your oreo? Aside f.om food alld worer, shelier is olso ohe o, our inportant heeds. As the populoiion ihcre.!e!, building n.w hoLr!€s or rhelt€r is limit.i. To find solulion to this prcbl€n, some goverihent og.ncies dnd orhs non{ov€Ihrehl offi.iofs (N6O) .onvefied sot@ ti.elields, du,np site. dnd nountcirlr inlo flbdivisions dnd relidentiols. Sut whot uould be ths effect o{ coMrtiig .i@fields to .6id€nri6l uits in our food supply? z , 2 Z Z :'", becouse there ore no enough space for prcpex garbage dkposol. ^s o r€sulr, sore peoPle lend to ihrow'their gorbdge onywh.f€. oorbdge baones brc{niry ond rursing ground of iEecrs and onidols ihot @se horm ro pe.ple. Dec.yiry garboge olso produces r,hpleaiant odor ard ehen burn if pmduces pois.nour qds @lled nelhohe As ihe populdtion incr€a3*, the 9d6.9e dso incraes. nris is T't ,,8 T H Wha you de living in on oa-populdi.d pla@, you moy oqaiae halrh prcblerns. Ir is be@@. the woi.r srpply is limit.d ihct will l..d you to poor hygi.ni. hobirs. In plo.4 like rhis, the surrouhdiigs naybe uniidy. o focrorthoi @uld oko cfFe.t your h4l'th. The common oilments rhot yd @uld oc$rire in ovesfDpllar€d ploces ore bEnchil is, o5l hnq. diqrrha and rube.culosis. 7,\ ,\\ \1" 6. Lnck of Herlrh sarvice llosi Pelple in 6n oM-populci€d 6ra 90 ro rubli. heilrh @trtas ond governhent hospirols be6u.e ii prcvides fr@ @Eulrorion oid los @sr rEdicdrions. A3 a ..suli, lh€s€ gow.nnenr dg€rciB b.@ne itud.4$re in mcetiig ihe n eds b..ou!€ df ihsrffici€nr funds. Lock of medicol personnel ,o odmaiisi€I is also s problen in mosr hosptols ev€n rhere or. od.audtc supply of hedicire!. 7_ Do you how wlry rhe crim€ roi€ hexs ih becdur€ fiDre pe.ple o.e fnJrrct€d d@ ro sLffici€.i naE io supp.rr their forniliG. ouf country inclY{ses? If is uh.mploym€ni dnd hdve no arinet .re u$dv gr€{rer ia dn dq-popltdled ra whq. tl, , a, v, tlr I E. Air ard Wat€r Pollutioh How dir be.o'nes pollut€d? I11€ dir b@'n€s p.llurn be.4ne of rhe hormfolgoees thot ser. produ.4 by the fdciori€s and vehicles. Itete {octories ond whi.l6 @ fuel ro run nochiB ond .JBin6. In ,h€ prc.ess, they give our Cdrboh Dioxide ond other ho.6ful gars.r such 6 Nittugei Oxide, Corbon l oioxide dnd Le.d iiio the oir. Do you know whot .ontdbute io ihe incr€asing number of whides qnd foctories? It is ihe inc.6e o, populdtion. As whdt I hove dis.!sse!, wirh a lihired sra.e 9@bd9e disposalie one of the problens thot .o!ld ise i, dh o!er-pop!,.t€d ploce. exn,jple ot thie orc rhos€ pelpl€ livi,rg oh the raverside teid 'ro ,hrou, lheir gEr&ge Hde you seen 'th. P6si9 river or the Tulyahan river? Did you {ind it Whdr do you think i! ihe eff€.t of ihis ih the.re4iures sho lives ih Ahothd f6do.s thal could.on rlbule to wdtd pollutioh dre oil s?ills, gorbqg€ fro,n boa, or ships ahd som€ ihdust.iol wosre. 9. Ite l@96f p4.enroge group. Individuols who orc this grclp. of olr popllarioh is compos.n of the working @pobla of s'rpporting ,heir fomilies nok !-up Though rhas group hol& the lojgeei percenroge of d. populaiion, rhis olso becomer one o{ oveFpopulored probl€]ns b€4use there ore rc jobs awildble fo. oll of iha10. Erergy Shortdge ltere will be on energy shortdge iJ ihe populdtion incre63"l be.dise rhe d.,nand i. €le.iriciry is high. Why is thai wh.n th. PoPqldion inclE.g, rhe d4ord in el4tricity is high? Ir B be.ouse there $,ould be 8to.e hdsat dnd blildirys to lighr ond nore el?riric oPPliohces ro run. rt.6rcznho!3.Ef+ed Whor is rhe grernho@ eff€.r? In whoi say il c.uld offect c2 6re.hhG. effed is rhe wdrniltg of rhe drltlosphee. lvhen the 5un worft rhe.nrrh s1jrf.@, sone of rhe h@r go€J bo.k ro rhe ornos?herc. Air an the dtnDsphere which is C@boi Dioxid. ,rops ihe heot 6hd it mok6 the a.th very worm. As ihe populdtion coniinuou!|,l gtol4 , the gt@rl$use etfe.t b@res no.e visible. Ir is becaosu ,hera ore mo.e focrories snd whicl.s iha, produce wdst€s ond fuma5 which cduses more C{.bo. Diodde ir the ormosphere. As a rcsutt, ,herc eiould be nore h4, ,rop in the ornosphere uhich osk6 th. @ih nuch wornerIf this will hoppen continuously, ,h€ fish ih th€ ocah *ill di€, ricerields/f@mlands will dry too due to lh€ wcm clitnole 12. Destruction of rhe Ozone Loyer A5 whot you hove l@med lrheh you de in v5-6, rhot the qzore ldver is 'the proiecrive loy€. of the olnosPhd€. ft protects us {rom the homful effects of ultrdviolei rays of the su. Do you khow ,hot our Ozore lol€t q4. dQ4tt\!ci.d? Il olreadY hod holes lhai dllow the ulrroviolet rdys to .4dt ihe @rrh. How do6 this hdpPei? Does th. in rc$e of poPuldioh h@€ sonething 'to do tr,lh ir? Yes, rhe I6i grov/irts PoPuldiion .odribuied o lot be@use 6 th' populotion incre3es, rhe u5e of refrigerd'tors, d€rosol lProvs 6nd pl4srics 6bo ihcre&s6. The sid producls coiiojn chemicol called Chlorofluorocdrbons (CFCS) which is mix wafh ihe dir in ihe ormosPher€. As o resulr. ihe hcrmfirl chernicol r€oches the Prolectiw ldver dnd lhrowh. hole in {hid ult@iolzi cahders aid cai4.ct3 ,F.*Y.iis hi!586$q€9.7,- Ho$ doas dcid rdin form? Is cid roii hdmful ro rEn? In the prcvious dis.ussions, yodt€ t.on€d rhd more vel .1e3 dnd fdctori€s or€ necded fo het the iii:.e.siry number o{ P@Pla. Lefs now fihd af hd f@tot.i€s dnd vehicle! .ontribure in the forrEtion of ocid When foctories 6nd whi.ler give off woste gd..3 ,hot will ,nix on lhe noisture i. rhe oir, it will ihen Produ.e sulPhu.i. ocld dnd Nitri. o.id. 'Ihe clol,Jd folb will ,h€h obsorb rhese ccids ond ehei ihe clold f.lls os .oin, ih. ccid is ahady Pdrr of itU/ha d.id ftin falls oh lok"!, ,46 or ocan ih€ fish sill die d.d if h fdlls oh fopnlonds,lhe pldni. together oith the soil B desrroyed. When you inhole dir with Niiric acid, your blood will los. irs @pobilily io fonspori Oxyg€h to your diff€.ai bodY Po.i3. ScieniisB include other rorns oJ dcidic pr€cipiigrion. Thes€ drc nisi, Do you krcw ihot Nuclerr power slotionr Use .adiodctive ,ndie.ials in producirE fuels, yet, rhey do and those .odioactiw rndlqlotE gi\e otf radio'ting en.rgy thoi is harmrul 'to livirq thilEs. wlren rodiotion enlert ihe body ot living things it {ill srq rhere for o lorg ,eriod of ri'ne. Exonple fhe rodiqtion vG srilled to the c.m. Then rhe @rn will be aie by rhe chicken, the .odiotion o the c.rn 'rill also 'tronsf€r to the chi.k€n. Wha on individuol als ihe nat of the chickeh sith mdiarion, helshe rill .ko oblorb ihe rodi@.tirc mtaid that will destrcy hB/her .€lls ond ruket hnn/hd si.r. Over-populoiion .on leld to food shoridg€, wdter shorroqe, housiB probl€ms, qdrbog€ probl€rs, lock of halrh sdi.e. tisa ol clit@ rote, oir ond woi€r pollution, uhanpl6ynat, eiergy 5horr69e, grenhoq3€ efreci, desrruction o( th. ozo@ lat/e?, rci.l roi. olld e.l€d. wattaEarth's History. All the processes that have been discussed require long periods of time to create a noticeable change on Earth's surface. You can just imagine how long it would take to create an oceanS as vast as the Pacific Ocean if the ocean floor moves only at about 10 cm/year. It is then important to know the history of Earth to learn the complexities of its past and be able to use it to understand the present. Just like learning the history of a country that requires one to read a lot of books, learning the history of Earth involves studying a lot of rocks. Rocks, especially sedimentary rocks, contain a lot of information about Earth's past. It holds the key to most of the geologic processes that happened on Earth and the key to uncovering how life on Earth evolved. But these discoveries are worthless if there is no time perspective. Thus, one of the most important contributions of geologists to mankind is the geologic time scale, which holds a history that is exceedingly long.The geologic time scale divides the history of Earth into different blocks of time by using relative dating. Since geologists use rocks to understand Earth's history, dating does not give accurate numerical dates, it only tells that an event preceded the relative dating places these rocks in their proper sequence of formation. But relative other. This method is still widely used today, alongside a more accurate method called absolute dating, which uses radioactive elements. With relative and absolute dating. geologists can trace the history of Earth. Relative Dating. Relative dating requires one to know the basic principles such as law of super-position, principle of original horizontality, principle of cross-cutting relationships, and unconformities.Law of Superposition The law of superposition is the most basic principle in relative dating. It states that in an undeformed sequence of sedimentary rock, the layers found at the top are the youngest rocks and the layers at the bottom are the oldest. It may seem too obvious, but this principle has only been clearly stated in 1669 by the Danish anatomist, geologist, and priest, Nicolaus Steno. Principle of Original Horizontality Along with the law of superposition, Steno stated that an undeformed sequence is the one where the layers are still in a horizontal position. This follows the principle of original horizontality, which states that sediments are deposited horizontally. Principle of Cross-Cutting Relationships The principle of cross-cutting relationships determines which events occurred first depending on which rocks are affected. The geologic layer that cuts another is younger than the layer it cuts across.Unconformities Rock layers that have not been interrupted are considered conformable. These sites represent spans of geologic time. But there is no place on Earth that has a complete conformable stratum since external and internal processes have always interrupted the deposition of the sediments. These breaks in the record of the rock strata are called unconformities. Using unconformities, geologic events are determined. There are three basic types of unconformities angular unconformity, disconformity, and nonconformity. Angular unconformity is characterized by having tilted or folded sedimentary rocks below younger, horizontal layers of rock. Disconformity is determined where there are missing parallel rock layers. Erosion takes place and removes the younger top layers and then deposition would once again happen. Nonconformity is characterized by an igneous or metamorphic rock found below a sedimentary rock. Figure 3-13. Three basic types of unconformities Using these principles for relative dating, one can determine the order of events However, relative dating does not give a time element as to when they happened. Absolute Dating For a much more accurate method of determining the history of Earth, geologists make use of absolute dating. This method uses unstable elements to determine the exact age of rocks. Isotopes are elements that have the same number of protons but different number of neutrons. Most isotopes are stable but some may be unstable. This is because the forces that bind the protons and neutrons in the nucleus of the isotope are not strong enough to hold them together, resulting in a radioactive decay, The unstable isotopes are called radioactive isotopes or parent isotopes. When these parent isotopes undergo radioactive decay, new isotopes, known as daughter products, are formed. The time it takes for one-half of the nuclei in the sample to decay is called half-life. This amount of time is fixed for each kind of radioactive isotope no matter what physical conditions it is subjected to. The ratio of parent daughter isotope determines how many half-lives have passed. If it is 1:1, then one half-life has passed; if it is 1:3, then two half-lives have passed; and if 1:7, then three half-lives have passed, and so on. Therefore, using the concept of half-life and parent-daughter ratio, geologists can determine the exact age of the sample. This method is called radiometric dating. It uses five radioactive isotopes to determine the age of rocks. For dating rocks that are about a million years old, rubidium-87, thorium-232, and the two isotopes of uranium (U-238 and U-235) are used. The fifth radioactive isotope is potassium-40, which has a half-life of 1.3 billion years. With these radioactive elements, determining the accurate age of rocks becomes easier. For dating events that are more recent, radiocarbon dating is used. This method uses carbon-14. Carbon-14 has a half-life of 5730 years and can be used to date back events up to 75000 years. All organisms contain a small amount of carbon-14, which is proportional with the amount of carbon-12. When an organism dies, the carbon-14 decays and is no longer replaced. The amount of carbon-14 left in the sample is then compared to the amounts of carbon-12 present, and radiocarbon dates can then be determined. This method has been particularly useful for anthropologists, archeologists, historians, and geologists for events that are much more recent.Fossils Aside from rocks, geologists also use the remains of living organisms in understanding Earth's history. Some fossils are formed from parts of an organism (body fossil), while some provide signs or clues as to which life-forms were present at that time (Frace fossils). Fossils contain a lot of information about the past the kind of organisms that have lived, the environment where organisms lived, and the evolution organisms underwent as their environment changed. However, not all organisms turned into fossils, therefore, scientists cannot learn everything about the past using fossils alone. There are also fossils that are used to determine the age of a rock. These are index fossils and these are only found in rocks of a particular age. The organisms that turned into index fossils have a relatively short life-spanning from a few million years to a few hundred million years. Index fossils are also found in most of the common rocks around the world, which makes them easier to identify.The methods used for dating the age of rocks are also used for fossils. Absolute dating is more commonly used since it can give exact numerical dates for the age, but relative dating can also be used to determine which fossils are older.Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.Q1. A teacher designs a lesson where students compute real-life percentages such as discounts and savings. 👉 A student calculates 15% of 200 to determine savings in a purchase. What is the correct result? A. 20 B. 25 C. 30 D. 35 Q2. In a classroom activity, learners compare numbers to find the highest common factor for grouping materials evenly. 👉 What is the GCF of 24 and 36? A. 6 B. 8 C. 12 D. 18 📘 FRACTIONS, DECIMALS, AND POWERS Q3. A learner converts fractions into percentages for data interpretation. 👉 What is 3/4 expressed as a percentage? A. 50% B. 60% C. 75% D. 80% Q4. A student models exponential growth using repeated multiplication. 👉 What is the value of 252^525? A. 25 B. 30 C. 32 D. 64 📘 ALGEBRA (EQUATIONS AND EXPRESSIONS) Q5. A teacher guides students to solve equations that represent real-life situations. 👉 Solve: 2x+8=202x + 8 = 202x+8=20 A. x = 4 B. x = 6 C. x = 8 D. x = 10 Q6. Students simplify expressions to understand relationships between quantities. 👉 Simplify: 3(x+4)−2x3(x + 4) - 2x3(x+4)−2x A. x + 12 B. x + 4 C. 5x + 4 D. 5x + 12 📘 FUNCTIONS AND GRAPHING Q7. A student analyzes a linear equation to determine its rate of change. 👉 What is the slope of y=3x−5y = 3x - 5y=3x−5? A. -5 B. -3 C. 3 D. 5 Q8. A learner evaluates functions to predict outcomes. 👉 If f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, what is f(4)f(4)f(4)? A. 7 B. 9 C. 11 D. 14 📘 GEOMETRY Q9. Students explore geometric shapes and their properties through visual models. 👉 What is the sum of interior angles of a triangle? A. 90° B. 180° C. 270° D. 360° Q10. A student calculates the area of a classroom table with dimensions 8 cm by 5 cm. 👉 What is the area? A. 26 sq cm B. 30 sq cm C. 40 sq cm D. 48 sq cm 📘 MEASUREMENT AND FIGURES Q11. A learner determines the volume of a cube used in a science experiment. 👉 What is the volume of a cube with side 4 cm? A. 16 cubic cm B. 32 cubic cm C. 48 cubic cm D. 64 cubic cm Q12. Students identify shapes used in design projects. 👉 How many sides does a hexagon have? A. 5 B. 6 C. 7 D. 8 📘 STATISTICS AND PROBABILITY Q13. A teacher helps students interpret data sets using measures of central tendency. 👉 What is the mean of 4, 6, 8, 10, 12? A. 6 B. 8 C. 10 D. 12 Q14. A class experiment involves flipping a fair coin. 👉 What is the probability of getting heads? A. 1/4 B. 1/3 C. 1/2 D. 2/3 📘 WORD PROBLEMS (APPLICATION) Q15. A car travels 180 km in 3 hours during a learning task on speed. 👉 What is its average speed? A. 45 km/h B. 60 km/h C. 75 km/h D. 90 km/h Q16. Students analyze work efficiency in a project. 👉 If 5 workers complete a task in 12 days, how long will 10 workers take? A. 3 days B. 6 days C. 8 days D. 12 days Q17. A student solves a problem involving ratios in a classroom population. 👉 If the ratio of boys to girls is 3:2 and there are 30 students, how many boys are there? A. 12 B. 15 C. 18 D. 20 Q18. A learner determines the duration of a scheduled trip. 👉 A journey starts at 8:30 AM and ends at 11:15 AM. How long is the trip? A. 2 hrs 15 mins B. 2 hrs 30 mins C. 2 hrs 45 mins D. 3 hrs 15 mins Q19. A student computes simple interest for financial literacy. 👉 What is the simple interest on ₱1000 at 5% for 2 years? A. ₱50 B. ₱75 C. ₱100 D. ₱150 Q20. A learner solves a perimeter problem involving a rectangle. 👉 A rectangle has a length of 12 cm and perimeter of 34 cm. What is the width? A. 5 cm B. 7 cm C. 10 cm D. 11 cm ✅ ANSWER KEY (BASED ON YOUR REVIEWER) (All verified from your uploaded file) [ilide.info...002acd4e5a | PDF] QAnswer1C2C3C4C5B6A7C8C9B10C11D12B13B14C15B16B17C18C19C20ACovalent Molecules and Compounds Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in Figure 3.1.1). Similarly, a few pure elements exist as polyatomic (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in Figure 3.1.1). Each covalent compound is represented by a molecular formula, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H2O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO2. Covalent compounds that predominantly contain carbon and hydrogen are called organic compounds. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table, as in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H2SO4), are described elswhere. Note! For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up. Write the molecular formula of each compound. a. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule. b. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule. c. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule. Solution: a. • A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic. • B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first. • C Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula. b. • A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound. • B The formula for an organic compound is written with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO. • C Adding subscripts gives the molecular formula C2H6O. c. • A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions. • B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F. • C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl3F. Write the molecular formula for each compound. a. Nitrous oxide, also called “laughing gas,” has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream. b. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms. c. Sulfur hexafluoride, a gas used to pressurize “unpressurized” tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule. Answer: a. N2O b. C12H22O11 c. SF6. Ionic Compounds The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively-charged protons in the nucleus is equal to the number of negatively-charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge. Note! Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges. When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:Polyatomic Ions Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. P. The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3− ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42− form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3− form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43− ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43− ions.Summary • There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. • The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.Contributors The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration.CARBOHYDRATES Carbohydrates are organic compounds composed of carbon, hydrogen, and oxygen in a ratio of about one carbon atom to two hydrogen atoms to one oxygen atom. The number of carbon atoms in a carbohydrate varies. Some carbohydrates serve as a source of energy. Other carbohydrates are used as structural materials. Carbohydrates can exist as monosaccharides, disaccharides, or polysaccharides. Monosaccharides A monomer of a carbohydrate is called a monosaccharide (MAHN-oh-SAK-uh-RIED). A monosaccharide—or simple sugar— contains carbon, hydrogen, and oxygen in a ratio of 1:2:1. The gen- eral formula for a monosaccharide is written as (CH2O)n, where n is any whole number from 3 to 8. For example, a six-carbon mono- saccharide, (CH2O)6, would have the formula C6H12O6. The most common monosaccharides are glucose, fructose, and galactose, as shown in Figure 3-6. Glucose is a main source of energy for cells. Fructose is found in fruits and is the sweetest of the monosaccharides. Galactose is found in milk. Notice in Figure 3-6 that glucose, fructose, and galactose have the same molecular formula, C6H12O6, but differing structures. The different structures determine the slightly different properties of the three compounds. Compounds like these sugars, with a single chemical formula but different structural forms, are called isomers (IE-soh-muhrz). SECTION 2 OBJECTIVES ● Distinguish between monosaccharides, disaccharides, and polysaccharides. ● Explain the relationship between amino acids and protein structure. ● Describe the induced fit model of enzyme action. ● Compare the structure and function of each of the different types of lipids. ● Compare the nucleic acids DNA and RNA. VOCABULARY carbohydrate monosaccharide disaccharide polysaccharide protein amino acid peptide bond polypeptide enzyme substrate active site lipid fatty acid phospholipid wax steroid nucleic acid deoxyribonucleic acid (DNA) ribonucleic acid (RNA) nucleotide C HO H C H OH C OH H C CH2OH H C H OH O Glucose C OH C O H OH C OH H CH2OH C H CH2OH Fructose C H HO C OH H C OH H C CH2OH H C H OH O Galactose Glucose, fructose, and galactose have the same chemical formula, but their structural differences result in different properties among the three compounds. FIGURE 3-6 Copyright © by Holt, Rinehart and Winston. All rights reserved. 56 CHAPTER 3 Disaccharides and Polysaccharides In living things, two monosaccharides can combine in a condensa- tion reaction to form a double sugar, or disaccharide (die-SAK-e-RIED). For example in Figure 3-4, the monosaccharides fructose and glu- cose can combine to form the disaccharide sucrose. A polysaccharide is a complex molecule composed of three or more monosaccharides. Animals store glucose in the form of the polysaccharide glycogen. Glycogen consists of hundreds of glucose molecules strung together in a highly branched chain. Much of the glucose that comes from food is ultimately stored in your liver and muscles as glycogen and is ready to be used for quick energy. Plants store glucose molecules in the form of the polysaccha- ride starch. Starch molecules have two basic forms—highly branched chains that are similar to glycogen and long, coiled, unbranched chains. Plants also make a large polysaccharide called cellulose. Cellulose, which gives strength and rigidity to plant cells, makes up about 50 percent of wood. In a single cellu- lose molecule, thousands of glucose monomers are linked in long, straight chains. These chains tend to form hydrogen bonds with each other. The resulting structure is strong and can be broken down by hydrolysis only under certain conditions. PROTEINS Proteins are organic compounds composed mainly of carbon, hydrogen, oxygen, and nitrogen. Like most of the other biological macromolecules, proteins are formed from the linkage of monomers called amino acids. Hair and horns, as shown in Figure 3-7a, are made mostly of proteins, as are skin, muscles and many biological catalysts (enzymes). Amino Acids There are 20 different amino acids, and all share a basic structure. As Figure 3-7b shows, each amino acid contains a central carbon atom covalently bonded to four other atoms or functional groups. A single hydrogen atom, highlighted in blue in the illustration, bonds at one site. A carboxyl group, —COOH, highlighted in green, bonds at a second site. An amino group, —NH2, highlighted in yel- low, bonds at a third site. A side chain called the R group, high- lighted in red, bonds at the fourth site. The main difference among the different amino acids is in their R groups. The R group can be complex or it can be simple, such as the CH3 group shown in the amino acid alanine in Figure 3-7b. The differences among the amino acid R groups gives different proteins very different shapes. The different shapes allow pro- teins to carry out many different activities in living things. Amino acids are commonly shown in a simplified way such as balls, as shown in Figure 3-7c. (a) Many structures, such as hair and horns are made of proteins. (b) Proteins are made up of amino acids. Amino acids differ only in the type of R group (shown in red) they carry. Polar R groups can dissolve in water, but nonpolar R groups cannot. (c) Amino acids have complex structures, so, in this and other textbooks, they are often simplified into balls. FIGURE 3-7 (b) Alanine (an amino acid) (c) Simplified version of amino acid CH3 H N OH C C H O H (a) Copyright © by Holt, Rinehart and Winston. All rights reserved. BIOCHEMISTRY 57 H H N C C OH H O H CH3 H2O Glycine Alanine H N OH C C H O H H H N C C H O H CH3 N OH C C H O H (a) (b) (a) The peptide bond (shaded blue) that binds amino acids together to form a polypeptide results from a condensation reaction that produces water. (b) Poly- peptides are commonly shown as a string of balls in this textbook and elsewhere. Each ball represents an amino acid. FIGURE 3-8 Substrate Products Enzyme 1 2 3 In the induced fit model of enzyme action, the enzyme can attach only to a substrate (reactant) with a specific shape. The enzyme then changes and reduces the activation energy of the reaction so reactants can become products. The enzyme is unchanged and is available to be used again. 3 2 1 FIGURE 3-9 Dipeptides and Polypeptides Figure 3-8a shows how two amino acids bond to form a dipeptide (die-PEP-TIED). In this condensation reaction, the two amino acids form a covalent bond, called a peptide bond (shaded in blue in Figure 3-8a) and release a water molecule. Amino acids often form very long chains called polypeptides (PAHL-i-PEP-TIEDZ). Proteins are composed of one or more polypep- tides. Some proteins are very large molecules, containing hun- dreds of amino acids. Often, these long proteins are bent and folded upon themselves as a result of interactions—such as hydrogen bonding—between individual amino acids. Protein shape can also be influenced by conditions such as temperature and the type of solvent in which a protein is dissolved. For exam- ple, cooking an egg changes the shape of proteins in the egg white. The firm, opaque result is very different from the initial clear, runny material. Enzymes Enzymes—RNA or protein molecules that act as biological catalysts—are essential for the functioning of any cell. Many enzymes are proteins. Figure 3-9 shows an induced fit model of enzyme action. Enzyme reactions depend on a physical fit between the enzyme molecule and its specific substrate, the reactant being catalyzed. Notice that the enzyme has folds, or an active site, with a shape that allows the substrate to fit into the active site. An enzyme acts only on a specific substrate because only that substrate fits into its active site. The linkage of the enzyme and substrate causes a slight change in the enzyme’s shape. The change in the enzyme’s shape weakens some chemical bonds in the substrate, which is one way that enzymes reduce activation energy, the energy needed to start the reaction. After the reaction, the enzyme releases the products. Like any catalyst, the enzyme itself is unchanged, so it can be used many times. An enzyme may not work if its environment is changed. For example, change in temperature or pH can cause a change in the shape of the enzyme or the substrate. If such a change happens, the reaction that the enzyme would have catalyzed cannot occur.