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Ratio Introduction Lesson Mr. J
Quiz by Jaime Javetski
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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 â˘(-9.8m/s2) â˘d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) â˘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. 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 DesirĂŠe Nieuwenhuis | NatĂ lia 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. Š 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-Îą 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- Îą 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-Îą 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-Îą, 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-Îą 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. The majority of the studies examining boys with obesity indicate an early onset of puberty, while not all reported an earlier onset of puberty. In detail, high leptin, TNF-Îą, and IL-6 levels combined with low adiponectin levels stimulate the activation of the HPG axis in girls and boys with obesity, and 5, 45, 50, 51 REFERENCES 1. Kumar S, Kelly AS. Review of childhood obesity: from epidemiology, etiology, and comorbidities to clinical assessment and treatment. May- o Clin Proc. 2017;92(2):251-265. 2. Reinehr T, Roth CL. Is there a causal relationship between obesity and puberty? The Lancet Child & adolescent health. 2019;3(1):44-54. 3. WorldHealthOrganization. Facts and figures on childhood obesity. 2017. 4. Guglielmi V, Sbraccia P. Obesity phenotypes: depot-differences in adipose tissue and their clinical implications. Eat Weight Disord. 2018; 23(1):3-14. 5. Gomez-Hernandez A, Beneit N, Diaz-Castroverde S. Escribano O. 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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 âonset of puberty and adiposity/obesityâ, âonset of pubertyâ, âchildren with obesityâ, âadipose tissueâ, âchildhood obesityâ, âadiposityâ, âobesityâ, âadipokine(s)â, âHPG axisâ, âadipokines ovary/ova- riesâ, or âadipokines testesâ, either alone or in combination. 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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 goverIntroduction to ratiosIntroduction to Ratios - Starter QuizIntroduction to Ratios - Exit QuizIntroduction to Financial RatiosIntroduction to Trigonometric Ratios - Starter Quizratio