ENTEROS 2 (Nivel II)

Multiplicación de números enteros de hasta 3 por 2 dígitos

2 ovaries → follicles → oocyte Follicles: made of 1 oocyte + theca cells and granulosa cells Theca cells - make testosterone Granulosa cells - make estradiol Oocytes: fixed amount per lifetime (~400) Ovulation - when oocyte leaves the follicle Fertilization: Connects ovaries to uterus Open-system: fallopian tubes are open to peritoneal space → easy to get infection! NOTE: Endometriosis: when endometrial cells are found in places other than uterus Retrograde menstruation: shedding of endometrium → goes up to fallopian tube (instead of down to cervix → vagina) → enters peritoneal space → travels Endometrium: hormone-sensitive mucosal lining of the endometrium Non-pregnant menstrual cycle: bleeding = shedding of endometrium Pregnancy: endometrium doesn’t shed → implantation occurs here Myometrium: smooth muscle layer of uterus → contracts during period to shed endometrium → cramps Continuous with cervix Dilates during menses and during birth

7.2.1 Critical Angle 1. The critical angle is the angle of incidence at which the refracted ray: A. Bends toward the normal B. Bends away from the normal C. Travels along the boundary D. Is totally reflected Answer: C 7.2.2 Snell’s Law & Critical Angle 2. Which formula correctly represents the critical angle c when light travels from medium 1 to medium 2? A. n1cos⁡c=n2 B. n2sin⁡c=n1 C. n1sin⁡c=n2 D. n1sin⁡c=n2sin⁡90∘ Answer: D 7.2.3 Total Internal Reflection 3. Total internal reflection occurs only when: A. Light travels from air to glass B. Angle of incidence is less than the critical angle C. Light travels from a denser to a rarer medium D. Refractive index of the second medium is greater Answer: C 4. Which condition is not required for total internal reflection? A. Light must travel from a denser medium B. Angle of incidence must exceed the critical angle C. Refractive index of second medium must be lower D. Light must strike at 90° Answer: D 7.2.4 Ray Diagrams & Angle Calculations 5. A ray in water (n = 1.33) hits the surface at 40°. Critical angle = 48.8°. What happens? A. Refraction only B. Total internal reflection C. No refraction D. Light stops Answer: A 7.2.5 Snell’s Law in Glass Blocks & Prisms 6. A ray enters glass (n = 1.5) from air at 30°. Which statement is correct? A. It bends away from the normal B. It bends toward the normal C. It travels straight D. It undergoes total internal reflection Answer: B 7. In a prism, the deviation of light occurs mainly because: A. Light slows down in glass B. Light speeds up in glass C. Light reflects internally D. Light cannot pass through glass Answer: A 7.2.6 Mirages 8. A mirage appears on a hot road because: A. Light reflects off the sky B. Light refracts through layers of air with different densities C. Light undergoes dispersion D. Light travels in straight lines only Answer: B 7.2.7 Dispersion Through a Prism 9. Dispersion occurs because: A. All colors refract equally B. Different wavelengths refract differently C. The prism reflects light D. White light cannot be refracted Answer: B 7.2.8 Rainbow Formation 10. A rainbow is formed due to: A. Refraction only B. Total internal reflection only C. Dispersion only D. Refraction + TIR + dispersion Answer: D 7.2.9 Optical Fibers 11. Optical fibers work mainly due to: A. Refraction B. Diffraction C. Total internal reflection D. Dispersion Answer: C 12. Which is an advantage of optical fibers? A. High signal loss B. Immune to electromagnetic interference C. Very heavy D. Slow data transmission Answer: B

PHOTOSYNTHESIS LIGHT DEPENDENT REACTION 1. Photosystem II (PSII) – Light Absorption & Water Splitting • Light energy (photons) excites electrons in chlorophyll molecules. • These high-energy electrons leave PSII and are passed into the electron transport chain (ETC). • Meanwhile, water molecules are split (photolysis) into: o O₂ (released as a by-product into the atmosphere) o H⁺ ions (protons, which build up inside the thylakoid) o Electrons (e⁻), which replace the ones lost by PSII. 2. Electron Transport Chain (ETC) • Excited electrons move through protein carriers embedded in the thylakoid membrane. • As they move, their energy pumps H⁺ ions into the thylakoid space, creating a proton gradient (high H⁺ inside, low outside). 3. ATP Production (ATP Synthase) • The buildup of H⁺ ions acts like a “waterfall” of potential energy. • These protons flow back across the membrane through ATP synthase, a protein complex that acts like a turbine. • This flow drives the conversion of ADP + Pi → ATP, which provides energy for the Calvin cycle. 4. Photosystem I (PSI) • Electrons arriving from the ETC enter PSI. • Sunlight excites them again, boosting them to a higher energy level. 5. NADPH Production • The energized electrons are transferred to NADP⁺. • Along with a proton (H⁺), this forms NADPH, another energy carrier. • NADPH is then delivered to the Calvin cycle to help build glucose. End Products of Light-Dependent Reactions: • ATP (energy source for Calvin cycle) • NADPH (reducing power for glucose synthesis) • O₂ (released into the atmosphere as waste) Light-Independent Reactions (Calvin Cycle) • These reactions do not directly require sunlight. • They occur in the stroma of the chloroplast (the fluid-filled space surrounding the thylakoids). • The inputs are ATP and NADPH (from light-dependent reactions) and CO₂ (from the atmosphere). • The outputs are glucose (C₆H₁₂O₆) and other carbohydrates. Think of the Calvin cycle as a factory that uses the energy and “raw materials” made in Stage I (ATP & NADPH) to build sugars. The 3 Main Steps of the Calvin Cycle 1. Carbon Fixation • CO₂ from the atmosphere enters the chloroplast and diffuses into the stroma. • Each CO₂ molecule attaches to a 5-carbon sugar called RuBP (ribulose-1,5-bisphosphate). • This reaction is catalyzed by the enzyme RuBisCO (Ribulose-1,5-bisphosphate carboxylase/oxygenase — the most abundant enzyme on Earth!). • The result is a short-lived 6-carbon compound, which immediately splits into two 3-carbon molecules called 3-PGA (3-phosphoglycerate). Summary: CO₂ + RuBP → 2 × 3-PGA 2. Reduction Phase • The 3-PGA molecules are “energized” and converted into G3P (glyceraldehyde-3-phosphate), a more energy-rich 3-carbon sugar. • This transformation requires: o ATP (provides energy) o NADPH (provides high-energy electrons and hydrogen atoms). • Some of the G3P molecules will eventually be combined to form glucose and other sugars. Summary: 3-PGA + ATP + NADPH → G3P 3. Regeneration of RuBP • Not all G3P molecules leave the cycle. Most of them are used to regenerate RuBP so the cycle can continue. • This regeneration also requires ATP. • For every 3 turns of the cycle, 5 G3P molecules are recycled to regenerate 3 molecules of RuBP. Summary: G3P + ATP → RuBP The Full Cycle Balance • To make one G3P molecule that can exit the cycle (and later form glucose), the cycle must run 3 times, fixing 3 molecules of CO₂. • To make one glucose molecule (C₆H₁₂O₆), the cycle must run 6 times (since glucose needs 6 carbon atoms). Inputs (for 1 glucose): • 6 CO₂ • 18 ATP • 12 NADPH Outputs: • 1 glucose (C₆H₁₂O₆) • 18 ADP + 18 Pi • 12 NADP⁺ Day vs Night Clarification • The Calvin Cycle is called light-independent, but that doesn’t mean it only happens at night. • It usually happens during the day because it depends on ATP and NADPH, which are only produced in light-dependent reactions (when sunlight is available). Simplified Analogy • Carbon fixation = The factory brings in CO₂ as raw material. • Reduction = Workers use energy (ATP & NADPH) to shape the raw material into useful products (G3P). • Regeneration = Some products are recycled to keep the factory running (RuBP is re-formed). • Output = After enough cycles, the factory produces glucose, the “food” of the plant.

LESSON 4. Cellular Respiration • Define cellular respiration • Identify the stages of clan respiration You have just learned how the energy from the sun is captured, processed, and stored in the form of glucose. Cellular respiration, another important life process, is the means by which cells release the stored energy in glucose to make adenosine triphosphate (ATP). The primary goal of this life process is to convert stored energy into usable form, such as ATP, for the cells to carry out their functions. Cellular respiration involves several chemical reactions. The reactions can be summed up in the following equation: C6 H12 O6 + 602 ----- 6 CO₂ +6H₂O + ATP Glucose oxygen carbon dioxide water energy Aerobic respiration reactions, or cellular respiration that takes place in the presence of oxygen, can be grouped into three stages glycolysis, Krebs cycle, and electron transport chain (ETC). Stage 1: Glycolysis Glycolysis is the process that breaks down one molecule of 6-C glucose into 3-C pyruvates or pyruvic acids. It also releases four molecules of ATP. This process occurs in the cytoplasm of the cell. The following is the step-by-step process of glycolysis. Take note that several enzymes are involved in this process. 1. The first step of glycolysis requires energy. It can only proceed when the two ATP molecules donate energy to the glucose by transferring a phosphate group with the help of an enzyme, producing glucose 6-phosphate 2. Then, a specific enzyme promotes the rearrangement of the atoms, producing the fructose 6-phosphate. 3. The action of the enzyme in step 2 promotes the transfer of a phosphate group from another ATP molecule, forming fructose 1,6-bisphosphate. 4. The resulting fructose 1,6-bisphosphate molecules, with the help of another enzyme, splits into two molecules, each with three carbon backbones. These two sugars are dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. 5. Another important enzyme then rapidly interconverts the molecules of dihydro-xyacetone phosphate and glyceraldehyde 3-phosphate. This produces two molecules of glyceraldehyde 3-phosphate or 3-phosphoglyceraldehyde (PGAL) 6. The succeeding step involves another enzyme-mediated action. The hydrogen (H) from PGAL is transferred to the oxidizing agent, nicotinamide adenine dinucleotide (NAD), which forms NADH. A phosphate (P) is also added from the cytosol of the cell to oxidize the two molecules of PGAL, forming two 1.3-bisphosphoglycerate. 7. A phosphate (P) from 1,3-biphosphoglycerate is transferred to ADP to form ATP. This happens for each of the two 1,3-bisphosphoglycerate. resulting to a yield of two ATP and two 3-phosphoglycerate molecules. 8. A phosphate is transferred from 3-phosphoglycerate molecules from the third carbon to the second carbon, forming 2-phosphoglycerate molecules A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of 2- phosphoglycerate molecules. 9. A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of two phosphoenolpyruvic acid (PEP) 10. Phosphate (P) from PEP is transferred to ADP (and forms ATP) and the final product, pyruvic acid. This reaction yields two molecules of pyruvic acid and two ATP molecules In summary, a single glucose molecule that undergoes the process of glycolysis produces two molecules of pyruvic acid, four molecules of ATP, two molecules of NADEL and two molecules of H.O. However, only two molecules of ATP are counted as net products since two molecules of ATP are spent throughout the process. Stage II: Krebs Cycle The Krebs cycle, named after its proponent Sir Hans Adolf Krebs, is a cyclical series of enzyme-controlled reactions. This stage of cellular respiration occurs in the matrix of the mitochondria. It is sometimes. called the citric acid cycle (CAC) since it produces citric acid. Citric acid contains three carboxyl (COOH) groups; hence, it is also called the tricarboxylic acid cycle (TCA). This requires the pyruvic acids produced during glycolysis. The main function of this cycle is to produce high-energy-yielding molecules, namely, NADH and flavin adenine dinucleotide (FADH) that will later on be used in the electron transport chain reaction. Figure 6-7. Summary of glycolysis and corresponding products in each reaction presented (See Appendix F on page 285 for an enlarged and complete version of the image.) An initial process is needed for the Krebs cycle to begin. As a pyruvate molecule from glycolysis enters the mitochondrion, it undergoes an important preliminary ate to form acetyl-CoA reaction. Coenzyme-A (COA) combines with pyruvate help of an enzymatic complex. This conversion also produces CO, and NADH. The Krebs cycle is summarized as follows. Take note that several enzymes are involved in this process. 1. The Krebs cycle technically begins when the acetyl-CoA combines with oxaloacetic acid (OAA), a 4-C molecule, to produce citric acid, a 6-C molecule. 2. With the aid of an enzyme, the citric acid now goes through a series of reactions that releases energy. Water molecule is removed from the citric acid and is returned in a different location. The-OH group is repositioned, forming the molecule isocitrate. 3. Isocitrate is then oxidized, forming the a-ketoglutarate, a 5-C molecule. The byproducts of this reaction are NADH and CO, 4 The a-ketoglutarate loses its CO, and a coenzyme-A is added in its place. The decarboxylation occurs with the help of NAD, which then becomes NADH. The resulting molecule is called succinyl-CoA. 5. Succinyl-CoA is converted into succinate. Also in this reaction, a molecule of guanosine triphosphate (GTP) is synthesized. The GTP molecule has similar structure and energy properties to that of ATP and is used by cells the same way. The free phosphate group attacks the succinyl-CoA molecule, which detaches the COA. Then, phosphate is attached to GDP to come up with GTP, similar to the process that occur in ATP synthesis (from ADP to ATP). 6. Two hydrogens are removed from succinate, A molecule of flavin adenine dinucleotide (FAD), a coenzyme similar to NAD, is reduced to FADH, as it takes the hydrogens from the succinate. This reaction produces the fumarate. 7. Fumarate is then converted into malate as the addition of a water molecule is catalyzed. The final reaction is the regeneration of oxaloacetate. The resulting byproduct of this regeneration is NADH Recall that two pyruvate molecules were produced during glycolysis, causing the Krebs cycle to turn twice. Each tuts produces three molecules of NADH, single ATH one FADIH, and the by-product CO, which is exhaled. Stage III: Electron Transport Chain The electron transport chain (ETC) is a series of photon pumps on the inner membrane of the mitochondrion. Electron transport is the last stage of the cellular respiration. In this stage, the energy from NADH and FADH, from the Krebs cycle is transferred to ADP to produce ATP. This process is generally known as oxidative phosphorylation. This energy coupling mechanism in the cell was revealed by the work of Peter stored energy in the form of proton (1) gradient to phosphorylate (add phosphate) ADP and produce ATP. The pumping of hydrogen sons across the inner membrane creates higher concentration ions in the inner membrane than on the outside of the membrane. This chemiosmotic gradient causes the ions to flow back across the membrane where the concentration of ions is lower. ATP synthase lined in the matrix serve as a channel protein, helping the ions to move across the membrane. The chemiosmotic gradient powers the phosphorylation of ADP to ATP, which also occurs in the ATP synthase. After passing through the ETC, the oxygen, being the final hydrogen acceptor, combines with two electrons and two protons, forming a water molecule. Water is a by-product of cellular respiration and is excreted. MINI TEST 6-3 1. Which energy-releasing pathway yields the most ATF in each glucose molecule? 2. Briefly describe the two stages of aerobic respiration that follow glycolysis: (a) Krebs cycle (b) Electron transport chain Anaerobic Respiration Most cells carry out arrobic respiration when oxygen is present. Aerobic respiration is an efficient process that yields a lot of ATP. However, many organisms thrive in mud, marshes, animal gut, canned goods, sewage treatment pond, and deep oceans where oxygen is scarce. Organisms that can live without oxygen are called anaerobes. Cellular respiration that proceeds without the presence of oxygen is called anaerobic respiration. In the event that the oxygen supply becomes low, aerobic cells also perform fermentation and lactic acid fermentation anaerobic pathways. There are two common anaerobic pathways in these cells, alcoholic fermentation and lactic acid fermentation. In alcoholic fermentation, ethyl alcohol and carbon dioxide are produced by some cells using the pyruvate from glycolysis. Each pyruvate molecule is rearranged into acetaldehyde and carbon dioxide, which is eventually released. NADII gives up electrons to acetaldehyde to form ethanol Fermentation is widely used in the industry. Yeast, a fungus used in making bread. can undergo anaerobic respiration. Bakers aux sugar, flour, water, and yeast to form the bread dough. The dough rises due to the carbon dioxide and alcohol released by the yeast cells trapped in air bubbles. Beer and wine manufacturers, we yeast to ferment the sugars in wheat and grape juice, forming alcoholic beverages such as beer and wine. In some cells, glycolysis produces two pyruvates, two NADH molecules, and two ATP molecules. Pyruvate itself becomes the final acceptor of the electrons from the NADH that produces the final product: lactate. Oftentimes, this product is called lactic acid. Human skeletal muscles can carry out fermentation when the blood cannot supply the cells with adequate oxygen during strenuous activities. When lactic acid builds up in the muscles, fatigue, burning sensation, and cramps result. Lactic acid will continue to build up until there is adequate supply of oxygen. Lactic acid is then converted back into pyruvate in the liver. Muscles also restore normal functions. Have you ever wondered why milk or cream turns sour after some time? Bacterial cells that undergo fermentation are responsible in producing lactate that turns the milk sour. These bacteria are used in manufacturing yogurt and sour milk products. Fermentation pathways do not breakdown and utilize the glucose completely. ATP is no longer produced beyond the process of glycolysis. Thus, energy produced is just enough for some single-celled organisms, or the energy can only be used by multicellular organisms for a short period.

A solution is composed of a solute dissolved in a solvent. In the sugar water described in Figure 5-1, the solute was sugar and the solvent was water, and the solute molecules diffused through the solvent. It is also possible for solvent molecules to diffuse. In the case of cells, the solutes are organic and inorganic compounds, and the solvent is water. The process by which water molecules diffuse across a cell membrane from an area of higher concentration to an area of lower concentration is called osmosis (ahs-MOH-sis). Because water is moving from a higher to lower concentration, osmosis does not require cells to expend energy. Therefore, osmosis is the passive transport of water. Direction of Osmosis The net direction of osmosis depends on the relative concentra- tion of solutes on the two sides of the membrane. Examine Table 5-1. When the concentration of solute molecules outside the cell is lower than the concentration in the cytosol, the solution outside is hypotonic to the cytosol. In this situation, water diffuses into the cell until equilibrium is established. When the concentration of solute molecules outside the cell is higher than the concentration in the cytosol, the solution outside is hypertonic to the cytosol. In this situation, water diffuses out of the cell until equilibrium is established. Observing Diffusion Materials 600 mL beaker, 25 cm dialysis tubing, funnel, 15 mL starch solution (10 percent), 20 drops Lugol’s solution, 300 mL water, 100 mL graduated cylinder, 20 cm piece of string (2) Procedure 1. Put on your disposable gloves, lab apron, and safety goggles. 2. Pour 300 mL of water in the 600 mL beaker. 3. Add 20 drops of Lugol’s solution to the water. CAUTION: Lugol’s solution is a poison and eye and skin irritant. 4. Open the dialysis tubing, and tie one end tightly with a piece of string. 5. Using the funnel, pour 15 mL of 10 percent starch solution into the dialysis tubing. 6. Tie the other end of the dialysis tubing tightly with the second piece of string, forming a sealed bag around the starch solution. 7. Place the bag into the solution in the beaker, and observe the setup for a color change. Analysis What happened to the color in the bag? What happened to the color of the water around the bag? Explain your observations. Quick Lab www.scilinks.org Topic: Osmosis Keyword: HM61090 mb06se_homs01.qxd 11/27/07 8:52 AM Page 98 HOMEOSTASIS AND CELL TRANSPORT 99 When the concentrations of solutes outside and inside the cell are equal, the outside solution is said to be isotonic to the cytosol. Under these conditions, water diffuses into and out of the cell at equal rates, so there is no net movement of water. Notice that the prefixes hypo-, hyper-, and iso- refer to the relative solute concentrations of two solutions. Thus, if the solution outside the cell is hypotonic to the cytosol, then the cytosol must be hyper- tonic to that solution. Conversely, if the solution outside is hypertonic to the cytosol, then the cytosol must be hypotonic to the solution. Water tends to diffuse from hypo- tonic solutions to hypertonic solutions. How Cells Deal with Osmosis Cells that are exposed to an isotonic external environment usually have no difficulty keeping the movement of water across the cell membrane in balance. This is the case with the cells of ver- tebrate animals on land and of most other organ- isms living in the sea. In contrast, many cells function in a hypotonic environment. Such is the case for unicellular freshwater organisms. Water constantly diffuses into these organisms. Because they require a relatively lower concentration of water in the cytosol to function normally, unicel- lular organisms must rid themselves of the excess water that enters by osmosis. Some of them, such as the paramecia shown in Figure 5-2, do this with contractile vacuoles (kon-TRAK-til VAK-y ̄ ̄o ̄ ̄o-OL), which are organelles that remove water. Contractile vacuoles collect the excess water and then contract, pumping the water out of the cell. Unlike diffusion and osmosis, this pumping action is not a form of passive trans- port because it requires the cell to expend energy. Copyright © by Holt, Rinehart and Winston. All rights reserved. (a) (b) Vacuole filling with water Vacuole contracting TABLE 5-1 Direction of Osmosis Condition External solution is hypotonic to cytosol External solution is hypertonic to cytosol External solution is isotonic to cytosol Net movement of water into the cell out of the cell none H2O H2O H2O H2O H2O H2O The paramecia shown below live in fresh water, which is hypotonic to their cytosol. (a) Contractile vacuoles collect excess water that moves by osmosis into the cytosol. (b) The vacuoles then contract, returning the water to the outside of the cell. (LM 315) FIGURE 5-2 100 CHAPTER 5 (a) HYPOTONIC Cell walls (b) HYPERTONIC (a) ISOTONIC (b) HYPOTONIC (c) HYPERTONIC Other cells, including many of those in multicellular organisms, respond to hypotonic environments by pumping solutes out of the cytosol. This lowers the solute concentration in the cytosol, bring- ing it closer to the solute concentration in the environment. As a result, water molecules are less likely to diffuse into the cell. Most plant cells, like animal cells, live in a hypotonic environ- ment. In fact, the cells that make up plant roots may be surrounded by water. This water moves into plant cells by osmosis. These cells swell as they fill with water until the cell membrane is pressed against the inside of the cell wall, as Figure 5-3a shows. The cell wall is strong enough to resist the pressure exerted by the water inside the expanding cell. The pressure that water molecules exert against the cell wall is called turgor pressure (TER-GOR PRESH-er). In a hypertonic environment, water leaves the cells through osmosis. As shown in Figure 5-3b, the cells shrink away from the cell walls, and turgor pressure is lost. This condition is called plasmolysis (plaz-MAHL-uh-sis), and is the reason that plants wilt if they don’t receive enough water. Some cells cannot compensate for changes in the solute con-

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