biological sciences 8

BIOLOGICAL SCIENCES 8R 1. Which of the following conditions

could produce rickets?

I. Metabolic deficiency of parathyroid hormone

II. Impairment of conversion of vitamin D to its active form

III. Inability of the active form of vitamin D to act on its target tissue

A ) I onlyB ) I and II only C ) I and III onlyD ) II and III only

The passage states that rickets is caused by insufficient vitamin D activity. Insufficient vitamin D activity would reduce the ability of the body to absorb ingested calcium from the small intestine. To maintain calcium levels in the blood plasma, parathyroid hormone would promote the breakdown of bone tissue, causing the bones to become weak. If there were a metabolic deficiency of parathyroid hormone, the body would be unable to break down bone tissue (option I), causing a higher than normal ratio of mineral to organic matter in the bones instead of a lower than normal ratio. However, if the body were unable to convert vitamin D to its active form, or if vitamin D were unable to act on its target tissue, overall vitamin D activity would be impaired (options II and III), which can lead to rickets. Thus,

D is the best answer.

I missed this question because I...

Solution

Guess

2. Why do calcium supplements often include vitamin D?A ) Vitamin D is needed to prevent rickets.B ) The activated form of vitamin D stimulates the absorption of calcium into

the blood.

Activated vitamin D acts on the small intestine to stimulate the absorption of calcium into the bloodstream. The inclusion of vitamin D in calcium supplements would ensure that vitamin D is present in the body to help promote this absorption. Thus, B is the best answer.

C ) The activated form of vitamin D enhances the action of calcitonin.D ) The activated form of vitamin D enhances the uptake of calcium by bone

tissue. I missed this question because I...

Solution

Guess

3. A low level of calcium in the plasma will trigger an increase of:

I. osteoclast activity. II. parathyroid hormone.

III. vitamin C.

A ) I onlyB ) I and II only

When the level of calcium in the blood plasma is low, the body responds by mobilizing stores of calcium from the bones via the activity of parathyroid hormone. Parathyroid hormone will increase the number of osteoclasts, which break down bone cells. Therefore, one would expect an

increase in both parathyroid hormone and osteoclast activity in order to increase the level of calcium in the blood plasma (options I and II). However, vitamin C (option III) promotes bone formation, a process that would further lower the calcium level in the plasma. Thus, B is the best answer.

C ) I and III onlyD ) II and III only


Solution

Guess

4. Under what condition would the level of calcitonin tend to increase?A ) When there is a dietary deficiency of calciumB ) When there is a dietary deficiency of vitamin DC ) When the level of calcium in the plasma is high

Calcitonin reduces bone resorption. Bone resorption occurs when the level of calcium in the blood plasma is low, but resorption is not needed when the level of calcium is high. Therefore, resorption would be reduced by calcitonin under conditions in which the level of calcium in the plasma is high. Thus, C is the best answer.

D ) When the level of parathyroid hormone is too low I missed this question because I...

Solution

Guess

5. Which of the following persons would be most likely to have rickets? A ) A child with a dietary deficiency in fat-soluble vitamins living in a tropical

climate B ) A child with a dietary deficiency in fat-soluble vitamins living in a

northern climate

The passage states that vitamin D is nonpolar and that it can be obtained through the action of ultraviolet rays on the skin. Nonpolar molecules tend to be lipid-soluble rather than water-soluble, and exposure to ultraviolet

rays tends to be less in northern climates than in tropical climates. Of the choices given, a child who has a dietary deficiency of fat-soluble vitamins and lives in a northern climate would be most likely to develop rickets. Thus, B is the best answer.

C ) A child with a dietary deficiency in water-soluble vitamins living in a tropical climate

D ) A child with a dietary deficiency in water-soluble vitamins living in a northern climate


Solution

Guess

6. What would be the result of complete removal of the parathyroid glands?A ) Severe neural and muscular problems due to deficiency of calcium in the

plasma

Removal of the parathyroid gland would lead to hypocalcemia, a condition of low blood calcium, resulting from the lack of parathyroid hormone. This would cause increased neuromuscular excitability because of the change in membrane potential, which under normal physiological conditions, is partially kept in balance with extracellular calcium. Typically, the person would eventually die from severe respiratory muscle spasms. Thus, A is the best answer.

B ) An increase in calcitonin production to compensate for calcium deficiency in the plasma

C ) A drastic change in the ratio of mineral to matrix tissue in bones D ) Calcification of some organs due to accumulation of calcium in the plasma

7. Which of the following experiments

would provide the best supporting evidence that neutrophils are the cause of the reperfusion injury? A ) Performing the

ischemia/reperfusion experiment using B (antibody-producing) cell-depleted animals and

examining whether the degree of tissue damage is reduced

B ) Performing the ischemia/reperfusion experiment using neutrophil-depleted animals and examining whether the degree of tissue damage is reduced

The scientists were testing the hypothesis that toxins released by neutrophils are the cause of reperfusion injury. By repeating the experiment in neutrophil-depleted mice, they can assess whether the degree of tissue damage is reduced in the absence of neutrophils. A reduction in the amount of damage would implicate a role for neutrophils in the injury process. Thus, B is the best answer.

C ) Repeating the experiment with another antibody directed against the entire alpha/beta heterodimer, and examining whether the degree of tissue damage is reduced

D ) Repeating the experiment with another antibody directed against the beta subunit, and examining whether the degree of tissue damage is reduced


Solution

8. Information in the passage suggests most strongly that the function of the beta subunit involves: A ) adhering neutrophils to the endothelium.

The passage states that the beta subunit is part of a heterodimer receptor

Guess

present on the membrane of neutrophil cells that plays a role in the adhesion of neutrophils to endothelial cells. When antibodies were directed against the alpha subunit, no subsequent reduction in tissue damage occurred, implying that the neutrophils were still able to adhere to the endothelial cells. However, when antibodies were directed against the beta subunit of the receptor, a reduction in tissue damage was observed. This implies that when the beta subunit is blocked by antibodies, the receptor is no longer able to perform its function of helping the neutrophil adhere to endothelial cells. Thus, A is the best answer.

B ) transferring proteinases from endothelium to neutrophils.C ) hydrogen bonding with the alpha subunit.D ) the generation of antibody against the subunit.


Solution

Guess

9. The scientist applied antibody B to lab animals at different stages during ischemia and reperfusion. The animals were then studied for 10 days, and the results are shown in Figure 1.

Figure 1 Treatment of animals with antibody B at different time points during an ischemia/reperfusion experiment

The figure shows that the animals can be protected from tissue injury if the antibody treatment is received: A ) at any time during ischemia or reperfusion.B ) after reperfusion only.C ) during reperfusion only.D ) before reperfusion only.

The graph shows that when the antibody was administered during reperfusion the amount of tissue damage that occurred was nearly the same as that of the control, which had not received antibody treatment. However, both times that antibody treatment was administered prior to reperfusion, there was a substantial reduction in tissue damage as compared to that of the control. Thus, D is the best answer.


Solution

Guess

10. The scientist claimed that antibody B offers a better means for preventing organ injury than agents such as free radical or protease inhibitors. Which of the following reasons offers the best support for this claim? A ) Antibody B is a high-affinity antibody; therefore, it will not be rejected by

the patient. B ) Antibody B can block the initiation of events that result in the release of

harmful, biologically active molecules.

Administration of antibody B reduced tissue damage following ischemia, presumably by preventing neutrophils from being able to adhere to the vascular endothelium. If this truly is the case, antibody B treatment would prevent neutrophils from invading the ischemic tissue, avoiding the release of the toxic chemicals such as free radicals and proteases. This should be more effective than trying to minimize damage from the toxins after they have already been released. Thus, B is the best answer.

C ) Antibody B is a very specific antibody; therefore, it will not recognize anything else other than the beta subunit.

D ) Antibody B exhibits a high half-life and can be used at any dosage at any time.


Solution

Guess

11. The scientist wanted to use antibody B clinically (i.e., to treat humans), but this proposal was rejected. Which of the following is the most logical reason for the rejection? A ) Because the antibody was generated in the mouse, it can never be used in

humans. B ) Because the antibody was generated in the mouse, repeated usage in the

same patient would elicit the production of human anti-mouse antibodies.

Antibody B was produced in mice; therefore, the human body would recognize it as a foreign antigen and mount an immune response, producing anti-mouse antibodies. Because an initial immune response to a specific antigen usually takes longer and is weaker than subsequent responses to the same antigen, a particular patient may not have a severe reaction to antibody B until after multiple exposures. Thus, B is the best answer.

C ) Because the antibody was generated in the mouse, it will not recognize human antigens.

D ) Because the antibody was generated in the mouse, it can only be used in vitro.


Solution

12. Which of the following situations would most likely occur in patients with defects in the neutrophil adhesion receptors during bacterial infections? A ) A presence of pus at sites of bacterial infectionsB ) An absence of pus at sites of bacterial infections

Pus is usually generated when neutrophils destroy cells by phagocytosis.

Guess To do so, neutrophils must first adhere to the cells that they are attacking. If the neutrophil has a defect in its adhesion receptors, it will not be able to bind to and engulf bacterial cells, and no pus will be generated. Thus, B is the best answer.

C ) An increase in the amount of circulating red blood cellsD ) A decrease in the amount of circulating white blood cells


Solution

Guess

13. A neutrophil has point mutations in the genes coding for the alpha and the beta subunits of the adhesion receptor. However, this cell can still migrate through endothelium. Which of the following conclusions about the effect of this mutation can be drawn? A ) The cell cannot release toxic products such as prostaglandins. B ) The cell has only functional beta subunits.C ) The cell can bind to endothelium.

Cell adhesion is required for migration to occur. The neutrophils containing these particular point mutations can still migrate through the endothelium. This implies that the neutrophils can still bind to the endothelial cells. Thus, C is the best answer.

D ) The cell has a defective cell membrane. 14. Which of the following rationales

explains which compound is the product of kinetic control? A ) Its semicarbazone has the lower

melting point. B ) It forms faster at all temperatures

because its rate of formation is independent of its path.

C ) Its reaction profile has the lower energy of activation.

The kinetically controlled product is formed faster than the thermodynamically controlled product because the energy of activation for the formation of the kinetically controlled product is lower. Thus, C is the best answer.

D ) Its reaction profile has the higher energy of activation.


Solution

Guess

15. Which semicarbazone is the product of thermodynamic control?A ) Cyclohexanone’s, because the more stable product forms fasterB ) Cyclohexanone’s, because it contains more alkyl substituents to a double

bond than does the other product C ) 2-Furaldehyde’s, because it is produced under equilibrium conditions and

is more stable than the other product

The melting point data in Table 1 indicate that the thermodynamically controlled product is the semicarbazone of 2-furaldehyde. The thermodynamically controlled product is the one that is formed under equilibrium conditions and is more stable. Thus, C is the best answer.

D ) 2-Furaldehyde’s, because its potential energy is higher I missed this question because I...

16. Based on its structure (shown below), the aldehydic proton in 2-furaldehyde should appear in its 1H NMR spectrum as a:

Solution

Guess

A ) singlet.

Because there are no hydrogens on the carbon next to the carbonyl carbon, the signal for the aldehydic proton is not split. The aldehydic proton will appear as a singlet. Thus, A is the best answer.

B ) doublet.C ) triplet. D ) quartet.


Solution

Guess

17. In each experiment, the crystals are collected on a Hirsch funnel and washed with two portions of cold water in order to: A ) recrystallize the product.B ) remove soluble impurities.

Rinsing the crystals with cold water will wash away the impurities soluble in cold water. Thus, B is the best answer.

C ) remove insoluble impurities.D ) spread the crystals evenly over the surface of the funnel.


Solution

18. What is the hybridization of the carbonyl carbon in cyclohexanone?A ) s B ) sp C ) sp2

Guess The carbonyl carbon in cyclohexanone is bonded to three other atoms (two carbons in the ring and the carbonyl oxygen). There are three hybrid orbitals; the carbon is sp2 hybridized. Thus, C is the best answer.

D ) sp3

19. Assuming Hypothesis B to be

correct, which of the following endocrine disorders would cause hypertension that could NOT be rectified by physiologically normal kidneys? A ) An excess of aldosterone

Aldosterone is a hormone released by the adrenal glands. Physiologically normal kidneys respond to aldosterone by increasing the reabsorption of both sodium and water. This leads to an increase in blood volume and therefore blood pressure. Thus, A is the best answer.

B ) An excess of glucagon C ) A shortage of thyroxineD ) A shortage of insulin


Solution

20. What mechanism probably would be responsible for the increased urine output induced by hypertension according to Hypothesis B? A ) Increased blood flow to the bladder

Guess

B ) Increased renal tubular reabsorption of solutes and water C ) Increased collecting duct permeability to waterD ) Increased glomerular filtration rate

The glomerular filtration rate is proportional to the glomerular capillary blood pressure minus the sum of the plasma osmotic pressure and the Bowman’s capsule hydrostatic pressure. An increase to the systemic blood pressure would initially increase the glomerular capillary blood pressure, which would increase the glomerular filtration rate. Without a corresponding increase in the rate of tubular reabsorption of water, this would lead to an increase in urine output. Thus, D is the best answer.


Solution

Guess

21. If restriction of blood flow to the kidneys (by placing clamps on the renal arteries) resulted in an immediate but small increase in blood pressure, followed by the gradual development of severe hypertension, which hypothesis would these results best support? A ) Hypothesis A, because the clamps increased the vascular resistance to

blood flow B ) Hypothesis A, because the clamps caused the kidneys to receive less bloodC ) Hypothesis B, because the kidneys were responding to decreased

glomerular blood pressure

The reduced flow of blood through the renal arteries due to the clamps would cause a decrease in glomerular blood pressure. The kidneys respond to this drop in pressure by activating the renin–angiotensin system of hormones. This increases the amount of sodium and water that is reabsorbed by the kidneys, therefore increasing blood volume and pressure. Thus, C is the best answer.

D ) Hypothesis B, because the volume of body fluids was probably decreasing I missed this question because I...

Solution

Guess

22. If blood pressure doubled and the resistance to blood flow increased by 50%, the amount of blood pumped by the heart would have: A ) increased by 1/3.

The passage states that P = CO × VR. Solving the equation for cardiac

output (CO) and letting the original CO equal , the new CO would

then equal . Reducing this equation shows that the new CO is

the old CO, or an increase of . Thus, A is the best answer.

B ) increased by 1/2.C ) decreased by 1/3. D ) decreased by 1/2.


Solution

Guess

23. According to Hypothesis A, enhanced activity of which of the following basic muscle types would be most likely to cause hypertension? A ) StriatedB ) Smooth

Enhanced activity of smooth muscles in blood vessels would cause vasoconstriction, and according to Hypothesis A, increased vasoconstriction is a major cause of hypertension. Thus, B is the best answer.

C ) CardiacD ) Multinucleate

24. Synthesis of antibody proteins in

eukaryotic cells is associated with what organelle?

A ) NucleusB ) MitochondrionC ) Endoplasmic reticulum

As secreted proteins, antibodies are translated by ribosomes attached to the rough endoplasmic reticulum. Thus, C is the best answer.

D ) Golgi apparatus


Solution

Guess

25. In the human "knee-jerk" reflex, the knee is struck and the lower leg jerks forward. Which of the following represents the complete pathway that the nerve impulse travels in effecting this response? A ) Sensory neuron, motor neuron

The knee-jerk reflex is a simple monosynaptic stretch reflex. A tap to the tendon that connects the quadriceps to the patella activates a sensory neuron that directly synapses with a motor neuron in the spinal cord, causing the quadriceps to contract. Thus, A is the best answer.

B ) Sensory neuron, brain, motor neuronC ) Sensory neuron, associative neuron, brain, associative neuron, motor

neuron D ) Sensory neuron, associative neuron, motor neuron, associative neuron,

motor neuron I missed this question because I...

26. The most effective way to remove triethylamine during the workup of an organic reaction would be to extract the reaction mixture with aqueous:

Solution

Guess

A ) sodium bicarbonate.B ) sodium bisulfite.C ) sodium sulfate.D ) hydrochloric acid.

Triethylamine would most likely be soluble in an organic solvent. Washing with aqueous hydrochloric acid will result in the formation of triethylamine hydrochloride. This salt is water-soluble and will be removed with the water wash. Thus, D is the best answer.


Solution

Guess

27.

Compound 1

What is the orientation of the tert-butyl and chloro substituents, respectively, in the predominant conformation of Compound 1? A ) Axial, axial B ) Axial, equatorial C ) Equatorial, axial

Compound 1 is cis-1-tert-butyl-4-chlorocyclohexane. The predominant conformation of this compound is also the most stable one and has the cyclohexane ring in the chair conformation with the larger tert-butyl substituent in the equatorial orientation and the smaller chloro substituent in the axial orientation. Thus, C is the best answer.

D ) Equatorial, equatorial 28. Uric acid enters the urine both

through filtration and secretion in the kidney. The process of filtration of uric acid in the kidney takes place in the: A ) glomerulus.

Filtrate is formed as fluid passes from the glomerular capillaries through the glomerular membrane into the Bowman’s capsule. This region of the nephron is known as the glomerulus. Thus, A is the best answer.

B ) loop of Henle.C ) distal convoluted tubule. D ) proximal convoluted tubule.


Solution

Guess

29. Colchicine most likely relieves gout symptoms through what mechanism?A ) Prevention of uric acid diffusion through cell membranes B ) Inhibition of leukocyte phagocytosis of uric acid crystals

Phagocytosis requires that the cell change shape dramatically as it surrounds and engulfs large extracellular particles. Microtubules are one of the cytoskeletal elements that help determine cell shape. This function relies on the ability of the microtubules to disassemble and reorganize. The drug colchicine inhibits microtubule reorganization and would therefore inhibit phagocytosis of uric acid crystals by leukocytes. Thus, B is the best answer.

C ) Inhibition of uric acid crystal formationD ) Maintenance of the pH optimum for PRPP synthetase


Solution

Guess

30. What nitrogenous base would promote the formation of uric acid crystals in gout? A ) CytosineB ) UracilC ) Guanine

The passage states that uric acid is formed by the breakdown of purines to xanthine, a uric acid precursor. Guanine is one type of purine that is found in cells. Thus, C is the best answer.

D ) Thymine I missed this question because I...

Solution

Guess

31. In the patient described in the passage, the likely genetic basis of the increased levels of uric acid is a mutation: A ) affecting an allosteric site of PRPP synthetase.

The passage indicates that the patient produces the normal amount of PRPP synthetase, but its activity in vivo is 3 times the normal level. However, PRPP synthetase purified from the patient does not show this increased enzymatic activity in vitro. The fact that the enzyme binds the substrate and converts it to product at normal levels in vitro suggests that the active site of the enzyme has not been altered, but rather that an allosteric site on the enzyme has been affected. The mutant PRPP can most likely bind an intracellular molecule at an allosteric site, which changes the shape of the enzyme, enhancing its activity. This activity-enhancing molecule most likely is not present in the in vitro reaction mix. Thus, A is the best answer.

B ) affecting the active site of PRPP synthetase.C ) in a promotor gene regulating the rate of transcription of the PRPP

synthetase gene. D ) in a gene coding for a transcription factor for the PRPP synthetase gene.


Solution

Guess

32. Some animals have developed the ability to excrete nitrogenous waste largely in the form of uric acid, which is nontoxic and does not require large amounts of water for its excretion. Considering its lifestyle, what animal would excrete nitrogen primarily in the form of uric acid? A ) Wild pig B ) Flying bird

Species of flying birds have evolved many characteristics that reduce their overall body mass, including the ability to excrete nitrogenous waste in the form of uric acid. Because the excretion of uric acid does not require large amounts of water, the amount of water that birds must ingest is decreased, making the birds lighter. Thus, B is the best answer.

C ) Carnivorous sharkD ) Herbivorous bony fish

33. The first step of the Hofmann

rearrangement involves the abstraction of one of the protons on the nitrogen. The amide N–H proton is slightly acidic because the: A ) resulting anion is resonance-

stabilized.

The removal of the amide proton results in the formation of an anion with the negative charge on the nitrogen. The adjacent carbonyl group stabilizes the anion through resonance. A

resonance structure of the anion can be drawn in which the negative charge resides on the oxygen. Thus, A is the best answer.

B ) N–H bond is polar. C ) aromatic ring is electron-

donating. D ) amide is not basic.


Solution

Guess

34. What is the major product when 2-phenylacetamide, below, is treated with bromine and aqueous base under the conditions in the passage?

A )

B )

C )

According to the passage, a Hoffman rearrangement of a primary amide results in the loss of the carbonyl carbon and the formation of an amine with one fewer carbon. Thus, C is the best answer.

D )


Solution

Guess

35. It has been reported that under some conditions hydrolysis of the amide can compete with the Hofmann rearrangement. What product would be expected if this side reaction were important for Compound 1? A ) Benzamide B ) 3-Nitroaniline C ) 3-NitrobenzamideD ) 3-Nitrobenzoic acid

Hydrolysis of Compound 1 would result in the formation of the corresponding carboxylic acid, 3-nitrobenzoic acid. Thus, D is the best answer.


Solution

Guess

36. The students monitored the conversion of Compound 1 to m-nitroaniline by infrared spectroscopy. The disappearance of which band would indicate that the starting material had been consumed? A ) 1550 cm–1 B ) 1650 cm–1

As Compound 1 is consumed, the IR spectrum will show the loss of absorptions attributed to the carbonyl group. The C=O stretch of an amide appears at approximately 1650 cm–1. Thus, B is the best answer.

C ) 2200 cm–1 D ) 3300 cm–1


Solution

Guess

37. The conversion of Compound 1 to m-nitroaniline can also be monitored by 13C NMR spectroscopy. The disappearance of the signal at which frequency accompanies the consumption of the starting material? A ) 65 ppm B ) 107 ppm C ) 120 ppm D ) 165 ppm

As Compound 1 is consumed, the 13C NMR spectrum will show the loss of the signal attributed to the carbonyl group. The carbon of the carbonyl group appears at approximately 160–170 ppm. Thus, D is the best answer.

38. When glycerol reacts with three

different fatty acids, how many stereogenic centers does the product triacylglycerol contain? A ) 0B ) 1

When glycerol reacts with three different fatty acids, only carbon 2 in the resulting triacylglycerol is attached to four different groups. Thus, B is the best answer.

C ) 2D ) 3


39. Which of the following formulas represents a general structure of a fatty acid salt produced in Reaction 1? (Note: Rn = R1, R2, or R3.)

Solution

Guess

A ) Rn—CH2– Na+

B ) Rn—CH2O– Na+ C ) Rn—C(O)– Na+ D ) Rn—CO2

– Na+

A fatty acid would have the general formula Rn–CO2H. The corresponding fatty acid salt would have the general formula Rn–CO2–Na+. Thus, D is the best answer.


Solution

Guess

40. Which of the following is the most plausible explanation for the fact that the saponification of the triacylglycerol in the passage resulted in four different fatty acid salts? A ) The triacylglycerol molecule consisted of four different fatty acid units.B ) Glycerol was transformed into a fatty acid salt under the reaction

conditions. C ) One of the fatty acid salts was unsaturated, and it completely isomerized

under the reaction conditions. D ) One of the fatty acid salts was unsaturated, and a small percentage

isomerized under the reaction conditions.

If one of the R groups in the triacylglycerol contained a carbon–carbon double bond and if isomerization of the double bond occurred during the saponification reaction, four fatty acids would be obtained instead of three. Thus, D is the best answer.


41. A triacylglycerol can also be accurately described as a:A ) triacid of glycerol.B ) triether of glycerol.

Solution

Guess

C ) triester of glycerol.

The three acyl groups are joined to the glycerol backbone through ester linkages. Thus, C is the best answer.

D ) trihydroxy glycerol. I missed this question because I...

Solution

Guess

42. How much sodium hydroxide is needed to completely saponify a triacylglycerol? A ) A catalytic amount, because OH– is continuously being regenerated during

saponification B ) One-third of an equivalent, because each OH– ion reacts to form three fatty

acid salts C ) One equivalent, because each OH– ion reacts to produce one molecule of

glycerol D ) Three equivalents, because one OH– ion is required to saponify each of the

three fatty acid groups

One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide will be needed to completely saponify the triacylglycerol. Thus, D is the best answer.


Solution

Guess

43. Which of the following statements most accurately describes the solubility properties of fatty acid salts? A ) They are soluble in polar media only.B ) They are soluble in nonpolar media only.C ) They can partially dissolve in both polar and nonpolar media.

A fatty acid salt contains a long hydrocarbon chain, which is soluble in

nonpolar solvents. The salt also contains the charged group –CO2–Na+,

which is soluble in polar solvents. Thus, C is the best answer.

D ) They are completely insoluble in both polar and nonpolar media. 44. A male taking excess testosterone

may become infertile because of reduced spermatogenesis. According to Figure 2, this could result directly from: A ) an increase in inhibin

concentration. B ) a reduction in inhibin

concentration. C ) a reduction in FSH concentration.

Figure 2 indicates that testosterone is part of a negative feedback loop that acts on the hypothalamus to prevent the release of GnRF. In the presence of testosterone, less GnRF would be present to stimulate the release of FSH from the pituitary gland, causing a decrease in FSH available to act on the Sertoli cells. Therefore, FSH is less able to promote and maintain spermatogenesis. Thus, C is the best answer.

D ) a reduction in LH concentration.


45. The cell type in the male reproductive system that is most analogous to the female ovum is the:

Solution

Guess

A ) spermatogonium. B ) primary spermatocyte.C ) spermatid. D ) spermatozoon.

The mature ovum is the female gamete that has completed meiosis and contains the haploid number of maternally derived chromosomes. This makes it most analogous to spermatozoa, the mature male gametes that contain the haploid number of paternally derived chromosomes. Thus, D is the best answer.


Solution

Guess

46. Some drugs used in cancer chemotherapy kill proliferating cancer cells by selectively inhibiting various stages of the life cycle. Which of the following normal reproductive processes is likely to be most affected by the use of chemotherapy? A ) Sertoli cell functionB ) Testosterone productionC ) Spermatogenesis

Spermatogenesis is the production of spermatozoa by the meiotic division of spermatocytes. Therefore, it would be the most affected by the use of chemotherapy. Thus, C is the best answer.

D ) Inhibin production I missed this question because I...

47. Which of the following hormones is(are) directly required for spermatogenesis?

I. Luteinizing hormone (LH)

Solution

Guess

II. Follicle-stimulating hormone (FSH) III. Inhibin IV. Testosterone

A ) IV onlyB ) I and IV only C ) II and IV only

Sertoli cells support and nourish the spermatocytes and promote the process of spermatogenesis. Spermatogenesis would not occur without Sertoli cells. The two hormones that directly stimulate Sertoli cells are FSH and testosterone. Thus, C is the best answer.

D ) I, II, and III only I missed this question because I...

Solution

Guess

48. On the basis of their function as “nurse” cells, which of the following organelles are most likely to be prominent in Sertoli cells? A ) Golgi apparatus

As nurse cells, Sertoli cells secrete many proteins and other nutrients for the developing spermatocytes to use. The organelle responsible for sorting and packaging proteins into vesicles for exocytosis is the Golgi apparatus. Thus, A is the best answer.

B ) Lysosomes C ) Mitochondria D ) Cilia


49. Which of the following statements correctly describes the distinction between the exocrine and endocrine portions of the testis?

Solution

Guess

A ) The exocrine portion secretes only peptides; the endocrine portion secretes only steroids.

B ) The exocrine portion releases its products into ducts; the endocrine portion releases its products into the blood.

Exocrine glands secrete their products through ducts; endocrine glands release their products into the bloodstream. Thus, B is the best answer.

C ) The exocrine portion secretes only cellular elements; the endocrine portion secretes only chemical substances.

D ) The exocrine portion is the target tissue for the products of the endocrine portion.

50. The pancreas produces which of the following substances for the digestive system? A ) Bile saltsB ) EmulsifierC ) Gastric juices D ) Proteolytic enzymes

The pancreas produces several proteolytic enzymes, which are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase. Thus, D is the best answer.


51. Which of the following characteristics clearly marks fungi as eukaryotes?A ) They have cell walls.B ) They contain ribosomes.

Solution

Guess

C ) They contain mitochondria.

One characteristic that distinguishes eukaryotic cells from prokaryotic cells is that eukaryotic cells contain membrane-bound organelles such as mitochondria. Thus, C is the best answer.

D ) They exhibit sexual reproduction. I missed this question because I...

Solution

Guess

52. If the ester shown below were hydrolyzed in acidic H218O, which product

would be expected to contain 18O?

A ) CH3CO2H

The first step of the hydrolysis would be the protonation of the carbonyl oxygen. The H2

18O would then act as a nucleophile, attacking the protonated carbonyl carbon. Cyclopentanol is the leaving group. The products are cyclopentanol and 18O-labeled acetic acid. Thus, A is the best answer.

B ) CH3OH C )

D )


53. From which germ layer(s) do the tissues of the heart and blood vessels differentiate?

I. Ectoderm

Solution

Guess

II. Mesoderm III. Endoderm

A ) II only

The heart and blood vessels both differentiate from the mesoderm. Thus, Ais the best answer.

B ) III onlyC ) I and II only D ) I and III only


Solution

Guess

54. The most effective method for producing an increase in the total amount of water lost through the skin during a certain period would be: A ) inhibiting kidney function.B ) decreasing salt consumption.C ) increasing water consumption.D ) raising the environmental temperature.

Water is lost through the skin primarily as a means to keep the body at normal temperatures. Therefore, raising the environmental temperature would cause a person to perspire, releasing water to the environment where it can evaporate and cool down the body. Thus, D is the best answer.

55. To determine the significance of UV

reflectance by the dewlap, it would be most useful to compare the behavior of: A ) sighted and sightless lizards, in

response to flashing of the dewlap.

B ) lizards responding to flashing of normal dewlaps versus treated dewlaps that absorb UV.

To determine the significance of UV reflectance by the dewlap, it would be most useful to compare the response of lizards to dewlaps that reflect UV light to the response of lizards to dewlaps that do not reflect UV light. If the reflectivity of the dewlap is kept as the only variable, it can be determined whether a correlation exists between the lizards’ behavior and the reflectivity of the dewlap with which the lizards were flashed. Thus, B is the best answer.

C ) the five lizard species, when they are placed together in the same habitat.

D ) the five lizard species under illumination by red light only.


Solution

Guess

56. If these lizards use UV light in communication, a mutation that eliminated UV photoreceptors would probably cause the LEAST disadvantage to: A ) species A.B ) species B.C ) species D.D ) species E.

Assuming that the lizards use the UV-reflectivity of the dewlap primarily as a means of intraspecies communication, species E would most likely be least affected by a mutation that eliminated UV photoreceptors. Its dewlaps are the least UV-reflective of the five lizard species, which indicates that species E is least likely to rely heavily upon this form of communication in the first place. Thus, D is the best answer.


Solution

Guess

57. If Anolis lizards have X-Y chromosomal sex determination, the locus of a gene for the UV reflectance pigment: A ) must be on the X chromosome.B ) must be on the Y chromosome.C ) must be on an autosome.D ) could be on a sex chromosome or on an autosome.

Based on the information presented, the gene encoding UV-reflectance pigment could be on a sex chromosome or an autosome. The fact that the pigment is expressed in the dewlap, a structure found only in males, is not sufficient to eliminate any chromosome as the location of this gene. Thus, D is the best answer.


Solution

Guess

58. Two neighboring lizard populations would be considered separate species if:A ) one population inhabited the forest and the other lived in a field. B ) one population had a UV-reflective dewlap and the other did not.C ) they did not communicate with each other.D ) they did not interbreed and produce fertile offspring.

One of the key factors that determines a species is the ability to successfully breed and produce fertile offspring. Two organisms that do not meet this criteria are considered separate species. Thus, D is the best answer.


Solution

Guess

59. Which of the following conclusions about dewlap reflectance is supported by information in the passage? A ) Lizard habitat is determined by dewlap reflectance for each species.B ) High dewlap reflectance is most important in brightly lit habitats.

Figure 1 shows that the three species of lizards that live in unshaded fields possess dewlaps that are significantly more capable of reflecting UV light than do the two species of lizards that live in the shaded understory. This supports the conclusion that high dewlap reflectance is most important in brightly lit habitats. Thus, B is the best answer.

C ) High dewlap reflectance is most important in dimly lit habitats.D ) Dewlap reflectance is highest at the blue end of the visible spectrum.


Solution

Guess

60. Dewlaps that reflect UV light would evolve by natural selection only if:A ) individuals with UV-reflective dewlaps produced more offspring than did

individuals without them.

Although many different types of adaptations may help an individual organism survive, they will not be passed on to the next generation unless the organism produces offspring, passing on the genes that cause the advantageous phenotype. To evolve by natural selection and become a general characteristic of the species, the genes that cause dewlaps to reflect UV light must become a significant portion of the gene pool, which will most likely occur if individuals with UV-reflective dewlaps produce more offspring than do individuals without them. Thus, A is the best answer.

B ) individuals with UV-reflective dewlaps were better able to communicate than individuals without them.

C ) individuals with UV-reflective dewlaps were less subject to predation than

individuals without them. D ) individuals with UV-reflective dewlaps mated more frequently than did

individuals without them. I missed this question because I... 61. At what concentration of free actin

will the + end of the microfilament grow faster than the – end? A ) Exactly at 1 µM B ) Only between 1 µM and 4 µMC ) At any concentration greater than

1 µM

Figure 1 shows that at free actin concentrations greater than 1µM, actin is added to the + end of a microfilament. At concentrations lower than 1µM, the + end loses actin subunits. The – end of the microfilament does not begin to add actin until the free actin concentration is greater than 5.5µM. The rate at which actin is added to the + end is greater than the rate it is added to the – end, implying that at any concentration greater than 1µM, the + end of the microfilament grows faster than the – end. Thus, C is the best answer.

D ) At any concentration


62. Below is a diagram of a muscle sarcomere. Based on the passage, which statement best explains why the microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction?

Solution

Guess

A ) The – ends of the microfilaments are capped by Z lines, and the actin

subunit concentration is kept above 1 µM in muscle cells. B ) The – ends of the microfilaments are capped by Z lines, and the + ends are

capped by another protein.

Within a sarcomere, the microfilament length remains stable. Because one end of the microfilament is anchored in the Z line, actin monomers are prevented from being added to or subtracted from that end. This rules out the possibility of treadmilling. Therefore, to retain a stable length, both ends of the microfilament must be capped. Thus, B is the best answer.

C ) The actin subunit concentration is kept above 4 µM in muscle cells. D ) The – ends polymerize and the + ends depolymerize at the same rate.


Solution

Guess

63. The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion? A ) Amoeboid movement stops upon exposure to cytochalasins.

The passage proposes that force is generated as a microfilament elongates and pushes against a structure such as the plasma membrane. This is representative of how an amoeba moves. Cytochalasins are drugs that inhibit the growth of microfilaments. Therefore, if amoeboid movement stops upon exposure to cytochalasins, microfilaments and their ability to elongate are both implicated as being necessary to generate the force for movement in an amoeba. Thus, A is the best answer.

B ) Amoeboid movement cannot occur if mitosis is blocked. C ) Moving Amoeba cells produce more troponin than do stationary ones.D ) The rate of movement is inversely proportional to the viscosity of the

medium in which the Amoeba moves. I missed this question because I...

Solution

Guess

64. Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will the microfilament treadmill? A ) 0.25 µM B ) 1.0 µM C ) 1.5 µM

Figure 1 shows that at a free actin concentration of 1.5µM, the rate at which actin subunits are added to the + end of the microfilament is equal to the rate at which actin filaments are removed from the – end. This fits the definition of treadmilling. Thus, C is the best answer.

D ) Any concentration between 1.0 µM and 4.0 µM I missed this question because I...

Solution

Guess

65. Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will both the + and – ends of the microfilament experience a net loss of subunits? A ) At any concentration below 1 µM

Figure 1 shows that at free actin concentrations less than 1µM, both the + and the – ends of the microfilament experience a net loss of actin subunits. Thus, A is the best answer.

B ) Exactly at 1 µM C ) At any concentration above 1 µM

D ) Only between 1 µM and 4 µM I missed this question because I...

Solution

Guess

66. Which of the following observations supports the hypothesis that microfilaments are involved in the release of viral particles? A ) Exocytosis of viral particles from an infected cell is proportional to the rate

of microfilament polymerization.

During exocytosis of viral particles the plasma membrane goes through many distortions of its shape. This requires polymerization of microfilaments, making it likely that the rate at which the viral particles can be expelled from the cell is dependent upon the rate of microtubule polymerization. Thus, A is the best answer.

B ) Treatment with phalloidin does not prevent the exocytosis of virus particles from the infected cell.

C ) No known virus carries genes coding for actin subunits.D ) Some viruses have capsules composed of myosin.


Solution

Guess

67. Assuming that Amoeba uses microfilament-generated forces for locomotion, which of the Amoebas pictured below will move from left to right? A )

B )

C )

D )

Diagram D shows a cell with a free actin concentration of 2µM with the + end of the microfilament oriented toward the right side of the cell. At this concentration, the microfilament should show a net increase in length toward the right. When it reaches the plasma membrane it should continue to lengthen and generate a force that pushes the membrane outward, causing the cell to move in a left-to-right direction. Thus, D is the best answer.

68. In which of the experiments is a

rearrangement of the carbon skeleton observed? A ) 1 only

In Equation 1, one of the methyl groups moves to the adjacent carbon. This rearrangement is not seen in equations 2 and 3. Thus, A is the best answer.

B ) 2 onlyC ) 3 onlyD ) 2 and 3 only


Solution

Guess

69. Which set of reagents or condition could be used to prepare the alcohol in the following reaction?

A ) H2SO4/H2O B ) Hg(OAc)2/THF-H2O; NaBH4/OH–

C ) THF:BH3; H2O2/OH–

Among the methods represented by the three equations in the passage, the one represented by Equation 3 is the only method of alcohol formation that results in the hydroxyl group adding to the less substituted carbon. Thus, Cis the best answer.

D ) Heat I missed this question because I...

Solution

Guess

70. Alcohols have higher boiling points than hydrocarbons of comparable molecular weight. This is a result of: A ) hydrogen bonding.

Alcohols contain hydroxyl groups, which enable them to undergo hydrogen bonding. Alcohols have higher boiling points because hydrogen bonds are strong intermolecular forces and more energy is needed to overcome them. Thus, A is the best answer.

B ) van der Waals forces.C ) covalent bonding. D ) resonance.


Solution

Guess

71. Another way to prepare an alcohol is via a Grignard synthesis. Which of the following reactants can be used in a Grignard reaction to produce the same alcohol that was produced in Experiment 1? A )

2,3-Dimethyl-2-butanol was prepared in Experiment 1. As shown in answer choice A, the reaction of isopropylmagnesium bromide (a Grignard reagent) with acetone will result in the addition of the isopropyl group to the carbonyl carbon and the formation of 2,3-dimethyl-2-butanol. Thus, A is the best answer.

B )

C )

D )

I missed this question because I... 72. The discovery that the amount of

thymine equals that of adenine and the amount of guanine equals that of cytosine in a given cell provides supporting evidence that: A ) the Watson and Crick model of

DNA is correct.

In the Watson and Crick model of DNA structure, the nitrogenous bases form hydrogen bonds with each other in a 1:1 ratio: guanine pairs with cytosine, and adenine pairs with

thymine. This implies that the amount of guanine and cytosine would be the same, and the amount of adenine and thymine would be the same. Thus, A is the best answer.

B ) DNA is the genetic material.C ) the genetic code is universal.D ) the code for one amino acid must

be a triplet of bases.


Solution

Guess

73. Consider an organism that has three pairs of chromosomes, AaBbCc, in its diploid cells. How many genotypically different kinds of haploid cells can it produce? A ) 4B ) 8

The number of different possible gametes that can be formed by diploid organisms as a result of independent assortment of chromosomes during meiosis can be calculated using the formula 2n where n is the haploid number of chromosomes. In this case, the haploid number is 3, making the number of different haploid cells 23, or 8. Thus, B is the best answer.

C ) 16D ) 32


74. What is the net volume of fresh air that enters the alveoli each minute,

Solution

Guess

assuming that the breathing rate is 10 breaths/min, the tidal volume is 800 mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL? A ) 65 mL B ) 95 mL C ) 6500 mL

The amount of air entering the lungs in a single breath, or tidal volume, is given as 800 mL/breath. Of that 800 mL only 650 mL reaches the alveoli per breath (800 mL of air inhaled minus 150 mL of nonalveolar respiratory volume). Therefore the net volume of air that reaches the alveoli each minute is equal to 650 mL/breath multiplied by 10 breaths/min, or 6500 mL. Thus, C is the best answer.

D ) 7850 mL I missed this question because I...

Solution

Guess

75. In mammals, which of the following events occurs during mitosis but does NOT occur during meiosis I? A ) SynapsisB ) The splitting of centromeres

One of the key differences between mitosis and meiosis occurs during their respective anaphases. During anaphase of mitosis, sister chromatids are pulled apart at the centromeres, each becoming an independent chromosome in the two diploid daughter cells. During anaphase I of meiosis I, homologous pairs of chromosomes are separated into the two daughter cells. However, each chromosome still consists of two sister chromatids joined to each other at the centromere. It is not until anaphase II of meiosis II that the centromere is split and the sister chromatids separate. Thus, B is the best answer.

C ) The pairing of homologous chromosomesD ) The breaking down of the nuclear membrane


Solution

Guess

76. In a particular species of plant, tall vine depends on a dominant gene (T), and a pink flower is the result of the heterozygous condition of the genes for red and white flowers (Rr). What fraction of the offspring from the cross of a tall, pink plant (heterozygous for height) with a short, pink plant would be expected to be pink AND tall? A ) 3/4B ) 1/2C ) 3/8D ) 1/4

Based on the given information, the tall pink plant would have the genotype TtRr and could form the following gametes: TR, Tr, tR, and tr. The short pink plant would have the genotype ttRr and could form the following gametes: tR and tr. The possible genotypes of the offspring are displayed in the following Punnett square.

TR Tr tR tr tR TtRR TtRr ttRR ttRr tr TrRr Ttrr ttRr ttrr

Of these, 1/4 would be both pink (Rr) and tall (TT or Tt). Thus, D is the best answer.


Solution

Guess

77. During prokaryotic protein synthesis, translation begins as soon as the newly synthesized mRNA strand begins to extend from the DNA strand. This situation differs from that in eukaryotes, because eukaryotes: A ) carry out translation without using ribosomes.B ) transcribe mRNA molecules without using DNA.C ) destroy most mRNA as soon as it is synthesized.

D ) localize the processes of transcription and translation in the nucleus and cytoplasm, respectively.

Eukaryotes have membrane-bound organelles including the nucleus, which contains the DNA. Transcription of DNA into RNA occurs in the nucleus. The RNA is then transported to the cytoplasm where ribosomes translate it into proteins. Thus, D is the best answer.

biological sciences 8

Documents

brightly lit habitats

require large amounts

eliminated uv photoreceptors

reflect uv light

endocrine portion secretes

tr ttrr ttrr

exocrine portion secretes

fatty acid salts