MCAT Biology Practice Test 2022: (59 Questions Answers)

AAMC MCAT Biology Practice Test 2022: (59 Questions Answers) Biological and Biochemical Foundations of Living Systems: Association of American Medical Colleges (AAMC) Medical College Admission Test (MCAT) Biological and Biochemical Foundations of Living Systems practice test questions answers. In addition, you can download a printable copy of the PDF for better MCAT Test Prep & Review.

The MCAT Biology Practice Test questions are designed to match the real exam as closely as possible in format and degree of difficulty. If you approach these tests as you would the real exam, your results will give you a good idea of how well you would score if you took the real MCAT Biology Test today. Following each set of Answers is an Evaluation Chart. Use it to measure your MCAT Biology Practice Test-taking readiness and to pinpoint content areas that you may need to review again before taking the real exam.

MCAT Biology Practice Test 2022

Test Name MCAT Practice Test 2022
Subjects Biological and Biochemical Foundations of Living Systems
Topics Cardiovascular Systems, Cells Digestive Systems, Embryogenesis and Development Endocrine Systems, Excretory Systems, Genetics and Evolution Immune Systems, Nervous Systems , Musculoskeletal Systems, Reproduction Respiratory Systems
Questions Format Multiple Choice Question
Total passage-based sets of questions  10 sets 4-6 questions per set
Total independent questions 15
Total Questions 59 Questions
Time Limit: 95 Minutes

Passage 1:

Leukemia is a term that describes a wide variety of blood cell cancers. Most leukemia cases involve elevated leukocyte counts. In acute cases of leukemia, immature leukocytes proliferate quickly, which leads to many abnormal immature cells in the bone marrow. This inhibits the ability of the bone marrow to produce healthy new leukocytes. This form of the cancer progresses rapidly, and the cells metastasize easily as they move into the bloodstream from the bone marrow. It can be fatal within weeks or months if not treated promptly. In contrast, chronic cases of leukemia typically involve the proliferation of abnormal mature leukocytes. This form of the disease progresses more slowly, and treatment need not always be immediate.

Once classified as acute or chronic, leukemia can be further subdivided into lymphocytic or myelogenous. This distinction depends on which type of leukocyte is abnormal in the patient. In lymphocytic leukemia, the lymphoid cells, such as the lymphocytes, are abnormal. In the myelogenous form, the myeloid cells, such as basophils, neutrophils, and eosinophils, are affected.

Chemotherapy drugs are often used in the management of leukemia cases. There are numerous categories of these drugs, and they have varied mechanisms of action. In treatment, several drugs are used in combination because some of the drugs are cell-cycle specific while others are not. Alkylating agents are capable of disrupting the function of DNA. Tubulin-binding agents are designed to interfere with microtubule formation. Other drugs are designed to interfere with DNA polymerase, DNA repair mechanisms, or protein synthesis, or to act as purine analogs. Many of these drugs cannot enter into the brain when injected or ingested, so if the cancer has spread to
this area, special methods must be used to introduce the drugs to the brain.

Q1. What would be the BEST explanation as to why people with chronic forms of leukemia can live longer untreated than those with acute cases of leukemia?

  • A. Chronic cases involve types of leukocytes that have less important functions in the body.
  • B. Chronic cases of leukemia involve abnormal cell division of mature cells, so there are still some immature cells that can proliferate into normal mature cells.
  • C. The immune system is less likely to recognize and destroy abnormal immature leukocytes than it would be to recognize and destroy abnormal mature leukocytes.
  • D. all of the above
View Correct Answer
 Answer Key: B  

Q2. What is the most logical explanation for why certain drugs may NOT be able to reach the brain if they have been ingested or injected?

  • A. These drugs have a chemical nature that does not allow them to cross the blood-brain barrier.
  • B. These drugs are so toxic that they are broken down before they reach the brain.
  • C. Most of these drugs target only cancerous cells that normally would not be in the brain.
  • D. The drugs that have been ingested have likely been denatured by the acids in the stomach.
View Correct Answer
 Answer Key: A  

Q3. A variety of risk factors exist for leukemia. Some of these factors include exposure to radiation, certain viral infections, and inherited tendencies.What do all of these factors have in common?

  • A. All involve problems with protein translation in the cells.
  • B. All can cause an increase in the production of cellular growth factors that enhance cell proliferation.
  • C. All cause changes to the DNA that can lead to the production of altered proteins that can influence the development of cancer.
  • D. All cause problems with transcription in the cells.
View Correct Answer
 Answer Key:  C 

Q4. Many leukemia patients require antibiotic treatments frequently. Why would this be the case?

  • A. Some antibiotics are known to be toxic to certain types of cancer cells and may help to damage the cancerous cells.
  • B. The antibiotics might help to increase the function of the patient’s immune system.
  • C. The antibiotics can help prevent bacterial infections in the patient whose leukocyte function is compromised by the cancer.
  • D. The antibiotics can help prevent viral infections in the patient.
View Correct Answer
 Answer Key:  C 

Q5. Of the types of drugs mentioned in the passage, which category is most likely to be cell-cycle specific?

  • A. tubulin-binding agents
  • B. protein synthesis inhibitors
  • C. purine analogs
  • D. alkylating agents
View Correct Answer
 Answer Key:  A 

Q6. It is not uncommon for leukemia patients to receive stem cell transplants. Why might this be a productive treatment option?

  • A. The stem cells are capable of differentiating into new blood cells, which include the leukocytes, erythrocytes, and platelets.
  • B. The stem cells can damage and replace the cancerous leukocytes.
  • C. The stem cells can act as immune system modulators, enhancing the function of the entire system.
  • D. The stem cells can temporarily replace the function of the cancerous cells and slow down the progression of the disease.
View Correct Answer
 Answer Key:  A 

Questions 7–12 are based on the following passage.

Passage 2

The kidneys are responsible for filtering certain components of blood and producing urine from them. In an adult, the kidneys are typically capable of processing about 1 to 1.5 liters of water per hour depending on the individual and conditions within the body. In a condition known as water intoxication, a person ingests more water (usually 3 or more liters in a single sitting) in a shorter period of time than the kidneys can process. In a child, the amount of water required to induce water intoxication is much less.

As the excess water moves from the digestive system, it accumulates in the plasma and extracellular fluids. Electrolytes such as sodium are drawn out of the cells in an attempt to equilibrate the concentrations inside and outside of the cells. As more and more water enters the extracellular fluids, the plasma sodium concentration drops, leading to a condition called hyponatremia, defined as a plasma sodium concentration less than 130 mmol/liter. Due to the effects of osmosis, water begins entering the cells. The cells will begin to swell and may eventually burst if the condition is not reversed in a timely manner.

The situation becomes increasingly problematic as the cells and tissues of the central nervous system take in more water. Pressure builds in the brain that causes symptoms that resemble alcoholintoxication. Lightheadedness, nausea, vomiting, and headaches are typical as well. Damage can also occur in the lungs and heart. If not treated promptly, seizures, coma, and death can occur within several hours.

Q7. During water intoxication, the concentrations of the intracellular and extracellular fluids are not at equilibrium. In this situation, the extracellular fluid would be to the cells.

  • A. hypertonic
  • B. isotonic
  • C. hypotonic
  • D. hyperosmotic
View Correct Answer
 Answer Key:  C 

Q8. Imbalancesin sodiumion concentrations wouldMOST likely cause problems with:

  • A. neuron function
  • B. aerobic cellular respiration
  • C. cell division
  • D. muscle contraction
View Correct Answer
 Answer Key:  A 

Q9. Of the options listed, what would be the best treatment for someone suffering from water intoxication?

  • A. administering a saline solution that is more concentrated than the cells
  • B. administering an isotonic solution
  • C. administering a solution low in solute concentration
  • D. performing dialysis to remove excess water from the plasma
View Correct Answer
 Answer Key:  A 

Q10. Due to the excess water consumed that causes water intoxication, which of the following responses to this excess water would NOT be expected in the body?

  • A. an increase in blood volume and pressure
  • B. an increase in the secretion of antidiuretic hormone
  • C. an increase in urine volume
  • D. an increase in glomerular filtration rate
View Correct Answer
 Answer Key:  B 

Q11. The regulation of water and sodium levels in the nephrons is adjusted in two regions of the kidneys where water and sodium can be reabsorbed from the nephron and returned to circulation. Those regions are the:

  • A. proximal convoluted tubule and distal convoluted tubule
  • B. the loop of Henle and the distal convoluted tubule
  • C. the distal convoluted tubule and the collecting duct
  • D. the proximal convoluted tubule and the loop of Henle
View Correct Answer
 Answer Key: B  

Q12. Death that occurs due to water intoxication involves swelling and damage to certain parts of the central nervous system. Damage to which of these structures
in the central nervous system would be MOST likely to have fatal consequences?

  • A. the cerebrum
  • B. the spinal cord
  • C. the cerebellum
  • D. the medulla oblongata
View Correct Answer
 Answer Key:  D 

Passage 3

Sperm count in human males is influenced by a variety of factors. One of the most important variables affecting sperm count is the number of Sertoli cells located in seminiferous tubules of the testes. The Sertoli cells’ primary function is to nurture cells through the process of spermatogenesis by providing them with nutrients and an environment conducive to spermatogenesis. Each Sertoli cell has a set number of spermatozoa that it can support. Sertoli cells have an additional role in forming the blood-testis barrier that provides immune privilege for the seminiferous tubules. This forms a physical barricade that keeps the developing sperm from coming into contact with cells of the immune system.

While it is not clear as to all the factors involved in determining Sertoli cell number, it is apparent that their development is most critical during the first nine months of life and prior to puberty. Events that affect the development of Sertoli cells in young males can influence fertility later in life. There may also be genetic factors involved in determining the number of Sertoli cells formed in a particular male.

The following data were collected from two groups of men. The men in group 1 had a normal sperm count while the men in group 2 were seeking infertility treatment due to a low sperm count. Levels of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the pituitary were measured as were the testicular hormones inhibin-B and testosterone in an attempt to determine which hormones seem to be linked to reduced sperm count.

Q13. In which group would you expect to see a higher number of Sertoli cells?

  • A. group 1
  • B. group 2
  • C. both groups 1 and 2
  • D. impossible to tell from these data
View Correct Answer
 Answer Key: A  

Q14. Based on the data provided from both groups in the table, it is suggested that there is:

  • A. a positive feedback relationship between LH and FSH
  • B. a negative feedback relationship between inhibin-B and FSH
  • C. a positive feedback relationship between testosterone and FSH
  • D. no relationship between these hormones
View Correct Answer
 Answer Key: B  

Q15. An elevated level of which hormone seems to be the best indicator of sperm count?

  • A. LH
  • B. FSH
  • C. inhibin-B
  • D. testosterone
View Correct Answer
 Answer Key:  C 

Q16. Why would it be important for the Sertoli cells to establish a blood-testis barrier and immune privilege?

  • A. to prevent the immune cells from attacking the sperm cells, which display different antigens from the other cells in the body
  • B. to prevent the developing sperm cells from leaving the reproductive system
  • C. to prevent blood from flowing into the testes
  • D. to keep the developing sperm cells from entering into the epididymis before they are mature
View Correct Answer
 Answer Key:  A 

Q17. The hormones FSH and LH are also involved in oogenesis in women. When LH surges in the ovarian cycle, what event will result?

  • A. endometrium proliferation
  • B. ovulation
  • C. menstruation
  • D. corpus luteum degradation
View Correct Answer
 Answer Key:  B 

Q18. Fully differentiated Sertoli cells lose the ability to proliferate to produce more Sertoli cells. This would mean that they cannot perform:

  • A. mitosis
  • B. meiosis
  • C. apoptosis
  • D. binary fission
View Correct Answer
 Answer Key:  A 

Passage 4

Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.

Osteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.

A major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor–related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.

Q19. Which of the following potential drugs might be helpful in preventing the death of osteoblasts caused by S. aureus infection?

  • A. one that increases capsase-8 activity
  • B. one that increases TRAIL activity
  • C. one that is an antagonist to TRAIL
  • D. all of the above
View Correct Answer
 Answer Key:  C 

Q20. Osteoblasts are unusual in that they actively engulf S. aureus, which survives in the osteoblast and eventually triggers apoptosis. In engulfing the bacteria, the osteoblasts resemble the normal activities of which type of leukocyte?

  • A. basophils
  • B. lymphocytes
  • C. eosinophils
  • D. macrophages
View Correct Answer
 Answer Key: D 

Q21. Osteomyelitis begins when S. aureus enters the bloodstream and moves to the bones. Under normal conditions, the nonspecific immune defense that would be activated as bacteria enter the blood would be:

  • A. antibody production by plasma cells
  • B. cytotoxic T-cell activation
  • C. the complement system
  • D. interleukin secretion
View Correct Answer
 Answer Key: C 

Q22. Many strains of S. aureus that cause osteomyelitis are termed methicillin-resistant Staphylococcus aureus (MRSA). These strains have developed resistance to the
antibiotic methicillin as well as to other antibiotics. The mechanism by which this antibiotic resistance occurs is by:

  • A. the overuse of antibiotics, which causes patients to become resistant to the antibiotics
  • B. the overuse of antibiotics, which selects for bacteria that have mutations that allow them to survive in the presence of the antibiotics
  • C. mutations, which occur as the bacteria are exposed to antibiotics that can serve as mutagens
  • D. all of the above
View Correct Answer
 Answer Key: B 

Q23. The strains of S. aureus associated with osteomyelitis have a variety of cell adhesion molecules called adhesins that help them to attach to and be internalized by
osteoblasts. These molecules are likely found in which of the following bacterial structures?

  • A. the nucleoid
  • B. the ribosomes
  • C. the capsule
  • D. spores
View Correct Answer
 Answer Key:  C 

Q24. Osteoblasts are normally involved in building new bone matrix. Their activity is stimulated by the hormone made by the .

  • A. calcitonin; thyroid
  • B. parathyroid hormone; parathyroid glands
  • C. osteotonin; bone marrow
  • D. none of the above
View Correct Answer
 Answer Key:  A 

Passage 5

Endothelin is a vasoactive peptide involved in a variety of processes, including the control of blood flow. Endothelin is capable of binding to two different receptors. There are receptors for endothelin located on smooth muscle (ETA) and receptors for endothelin located on vascular endothelium (ETB). When endothelin binds to ETA, it initiates vasoconstriction by narrowing blood vessels. However, when endothelin binds to ETB, vasodilation occurs due to the widening of blood vessels. The activity of the ETB receptor seems to counterregulate the ETA effects and prevents excessive vasoconstriction.

Drug X is a selective receptor antagonist for endothelin. It is designed to bind preferentially to certain endothelin receptors. Drug Y is another selective receptor antagonist for endothelin. It is also designed to preferentially bind to one of the endothelin receptors. Researchers are hopeful that drugs X and Y will be of value in the treatment of conditions such as pulmonary arterial hypertension (PAH). In PAH, excess endothelin is produced, which causes an increase in blood pressure within the pulmonary arteries. Normal pressure in the pulmonary arteries is expected to be about 14 mm Hg, and pressure at or above 25 mm Hg constitutes PAH. Both drugs have been tested in patients with PAH. There were five patients in each group. At the end of the study, the percent decrease in mean pulmonary artery pressure was measured for each patient and reported in the table.

Q25. Based solely on the data provided, which conclusion should be drawn concerning the effectiveness of drugs X and Y?

  • A. Drug X is a more effective treatment for PAH because most patients had a slight or no increase in pulmonary blood pressure while taking the drug.
  • B. The placebo is the best treatment for PAH because some patients taking the placebo had a decrease in pulmonary blood pressure.
  • C. Drugs X and Y should be used as combined therapy to maximize their effects.
  • D. Drug Y is a more effective treatment for PAH because the majority of patients taking the drug had a decrease in pulmonary blood pressure.
View Correct Answer
 Answer Key:  B 

Q26. Which endothelin receptor is drug X selectively binding to?

  • A. ETA only
  • B. ETB only
  • C. both ETA and ETB
  • D. neither ETA nor ETB

Q27. If a patient has PAH, the heart can become overly stressed which can ultimately lead to heart failure. Which area of the heart is MOST likely to initially be affected
by increased blood pressure in the pulmonary arteries?

  • A. the right and left atria
  • B. the right and left ventricles
  • C. the right ventricle
  • D. the left ventricle
View Correct Answer
 Answer Key: B 

Q28. A patient overexpresses the ETB receptor. Explain what should happen to the patient’s pressure.

  • A. It should remain the same as long as the ETA receptor is also expressed.
  • B. It should decrease due to elevated levels of vasodilation.
  • C. It should increase since overexpression of ETB will block available ETA receptors.
  • D. It should stay stable since other systems in the body will compensate.
View Correct Answer
 Answer Key:  B 

Q29. Suppose a patient suffers from chronic systolic and diastolic hypertension due to an inherited defect in an endothelin receptor. How would the kidneys attempt to
alleviate the hypertension?

  • A. by decreasing water absorption in the nephrons
  • B. by decreasing urine output thus increasing blood volume
  • C. by increasing the amount of aldosterone secreted from the adrenal glands
  • D. by increasing the amount of antidiuretic hormone secreted from the pituitary gland
View Correct Answer
 Answer Key: C 

Q30. In addition to regulation of vessel diameter, ETA is known to have other roles in the body. When activated by a specific type of endothelin, ETA can synergize with growth factors to cause rapid cell proliferation. Based on this information, it is MOST likely that ETA can be involved with:

  • A. an increase in metabolism
  • B. the development of cancer
  • C. immune system hypersensitivities
  • D. increased stress to the kidneys and liver
View Correct Answer
 Answer Key: B 

Passage 6

An endocrinologist has been studying an enzyme that she suspects is the rate-limiting step in the conversion of cholesterol to estrogen. It is reported in the literature that female mice with a mutation in the gene coding for this enzyme reach sexual maturity earlier than mice without the mutation. Controlled experiments are completed using breeding pairs of wild-type mice and the mutant strain of mice. The plasma estrogen concentration is measured in female offspring at 3 weeks of age, prior to sexual maturity, which typically occurs by 6 weeks of age. The results, in estrogen pg/mL of plasma, are seen in the following table.

Q31. What conclusion can be made based on the data?

  • A. The presence of the wild-type enzyme increases estrogen concentration.
  • B. The presence of the wild-type enzyme increases cholesterol concentration.
  • C. The presence of the mutant enzyme increases estrogen concentration.
  • D. The presence of the mutant enzyme increases cholesterol concentration.
View Correct Answer
 Answer Key: C

Q32. The rate-limiting step in a metabolic pathway:

  • A. requires the lowest activation energy
  • B. is the fastest step in a metabolic pathway
  • C. is always the last reaction in the pathway
  • D. is the slowest step in a metabolic pathway
View Correct Answer
 Answer Key: D

Q33. Recent studies have shown that cholesterol levels in women can vary significantly depending on where a woman is in her reproductive cycle. Most notable is the fact that total and LDL cholesterol levels typically decline at the points in the cycle where estrogen levels are at their highest. Based on this, we would expect that total and LDL cholesterol levels would be the lowest:

  • A. following menopause
  • B. during menstruation
  • C. during ovulation
  • D. prior to puberty
View Correct Answer
 Answer Key: C

Q34. In addition to serving as a precursor for steroid biosynthesis, cholesterol has a variety of other roles in the body including:

  • A. maintenance of structural integrity and fluidity in plasma membranes
  • B. nucleotide biosynthesis
  • C. transport of molecules through the plasma membrane
  • D. maintaining osmotic balance between cells and their surrounding environment
View Correct Answer
 Answer Key: A

Q35. Based on the data presented, it is likely that:

  • A. Wild-type mice reach sexual maturity faster than the mutant strain.
  • B. The mutant strain mice will not reach sexual maturity.
  • C. The mutant strain mice have a lower cholesterol level than wild-type mice.
  • D. Wild-type mice will not reach sexual maturity.
View Correct Answer
 Answer Key: C

Q36. If plasma cholesterol levels were to be measured, what would you expect to see in female mice that have reached sexual maturity as compared to female mice that
are far from reaching sexual maturity?

  • A. Cholesterol levels would be increased in wild-type mice that have reached sexual maturity as compared to wild-type mice that have not reached sexual
    maturity.
  • B. Cholesterol levels would be the same in wild-type mice that have reached sexual maturity as compared to wild-type mice that have not reached sexual
    maturity.
  • C. Cholesterol levels would be increased in mutant strain mice that have reached sexual maturity as compared to mutant strain mice that have not reached sexual maturity.
  • D. Cholesterol levels would be decreased in mutant strain mice that have reached sexual maturity as compared to mutant strain mice that have not reached sexual maturity.
View Correct Answer
 Answer Key: D

Passage 7

Human immunodeficiency virus (HIV) is known to infect T cells and macrophages possessing the CD4 receptor. In addition to the CD4 receptor, there are a variety of coreceptors that are also needed for certain strains of HIV to infect cells. One of these coreceptors is called CCR5 and is needed for the most common strains of HIV (called R5) to infect their host cells. The more CCR5 receptors a cell has, the greater the rate of infection for the cell. Some individuals have the delta-32 allele of CCR5 that leads to a decreased risk of HIV infection.

Women are at a much higher risk for contracting HIV from a male partner than a male is for contracting HIV from a female partner. Some of the increased risk for women is dependent on sex hormones. It has been hypothesized that women have different risks of contracting HIV at different points in their reproductive cycles. For example, women may be at a greater risk of contracting HIV following ovulation during the last 2 weeks of their cycle as opposed to the first 2 weeks of their cycle prior to ovulation. The difference between risk of infection before and after ovulation relates to sex hormones produced at various points in the reproductive cycle. It seems that the estrogen that is produced prior to ovulation provides a somewhat protective, although certainly not absolute, role against HIV infection. For women who do become infected with HIV, studies have shown that women in the first 3 to 5 years after HIV infection carry lower HIV viral loads than men, perhaps due to the influence of estrogen. This suggests a major role for sex hormones in the infection and progression of HIV.

Q37. According to the hypothesis presented concerning estrogen’s role in HIV infection, which of these individuals would be at LEAST risk for HIV infection?

  • A. a female producing elevated levels of testosterone
  • B. a female with an ovarian tumor that causes increased secretion of estrogen
  • C. a female who has had her ovaries removed
  • D. a male with a reduced testosterone level
View Correct Answer
 Answer Key: B

Q38. Women are at a higher risk of contracting HIV following ovulation. Based on the normal events of the female reproductive cycle, which hormone might account for the increased risk of HIV infection in women following ovulation?

  • A. increased levels of progesterone
  • B. increased levels of follicle-stimulating hormone
  • C. increased levels of estrogen
  • D. increased levels of luteinizing hormone
View Correct Answer
 Answer Key: A

Q39. As a result of chronic infection with HIV, the CCR5 gene is generally up-regulated over time. This would MOST likely account for:

  • A. an increased number of host cells becoming infected as the result of prolonged infection
  • B. the elimination of HIV from the body in some individuals
  • C. increased protection in women as the result of estrogen
  • D. resistance of cells carrying the CCR5 receptor to infection
View Correct Answer
 Answer Key: A 

Q40. Some populations with the delta-32 CCR5 mutation seem to have close to 100% protection against HIV, while other populations with the delta-32 CCR5 mutation seem to have only partial protection. What could account for this difference?

  • A. There is an environmental influence on expression of the delta-32 CCR5 phenotype.
  • B. The delta-32 CCR5 allele exhibits incomplete dominance. Homozygotes have complete protection while heterozygotes have only partial protection.
  • C. The expression of the delta-32 CCR5 allele can be masked by epistasis.
  • D. Those that have only partial protection against HIV infection have been exposed to larger doses of HIV.
View Correct Answer
 Answer Key: B

Q41. HIV belongs to a unique category of viruses termed retroviruses. Which of the following enzymes would be unique to retroviruses?

  • A. DNA polymerase
  • B. RNA polymerase
  • C. reverse transcriptase
  • D. helicase
View Correct Answer
 Answer Key: C

Q42. Drugs such as AZT and ddI are nucleoside analogs used in HIV-infected patients. Both are chemically modified versions of the nucleotides that make up DNA and
RNA. These modified nucleosides interfere with normal replication and transcription. The MOST important goal of these drugs would be:

  • A. to prevent the replication of host cell DNA so that the host cell cannot function
  • B. to prevent the transcription of host cell RNA so that no proteins can be expressed
  • C. to prevent the viral nucleic acid from entering the host cells
  • D. to prevent replication of the viral genome
View Correct Answer
 Answer Key: D

Passage 8

In the past, there has been a great deal of controversy over the existence of viable but nonculturable (VBNC) states in bacteria. While normal culturing methods cannot detect these VBNC cells, there are a variety of molecular assays that can be used to indicate metabolic activity in the cells. Additionally, these cells can be resuscitated from VBNC states to a culturable state under appropriate conditions. The VBNC state has been observed in numerous species of bacteria and most recently in yeasts.

Escherichia coli is one of many species of bacteria that are able to enter and resuscitate from VBNC states. Certain strains of E. coli cause a variety of infections in humans. Urinary tract infections (UTIs) in women are frequently caused by E. coli that exist as harmless residents in the intestinal tract. These infections can be treated using antibiotics to kill the E. coli. However, about 30 percent of women treated for UTIs end up with a recurrence. For those who support the VBNC theory, it has been hypothesized that recurrent UTIs are caused by VBNC E. coli that resuscitate to initiate a new infection. These VBNC cells may not be killed using antibiotic therapies normally used to treat UTIs. Only culturable cells cause the symptoms of a UTI.

The following graph shows the number of E. coli bacteria present in a patient following antibiotic therapy for a UTI. Day 1 indicates the number of bacteria present at diagnosis. A urine sample was collected each day and cultured for the presence of E. coli as measured in colony-forming units per milliliter of urine (cfu/ml). An assay was also used to determine the number of metabolically active cells that were potentially alive but not culturable (VBNC). The data are as follows:

Q43. What can you conclude from the graph?

  • A. Antibiotic therapy is highly effective at eliminating all E. coli.
  • B. While the current infection has been effectively treated, this patient may be at risk for a recurrent UTI.
  • C. The antibiotic therapy used for this patient was not sufficient to eliminate the infection.
  • D. Antibiotic therapy has increased the total number of E. coli cells in this patient.
View Correct Answer
 Answer Key: B

Q44. Which piece of information would be the BEST support for the existence of the VBNC state?

  • A. the presence of DNA in the alleged VBNC cells
  • B. the presence of an intact plasma membrane in the alleged VBNC cells
  • C. the presence of specific proteins in the alleged VBNC cells
  • D. the detection of transcription in the alleged VBNC cells
View Correct Answer
 Answer Key: D

Q45. Why would women be more likely to contract a UTI caused by E. coli as compared to men?

  • A. It is likely that the intestinal tracts of women contain a higher number of the E. coli bacteria as compared to men.
  • B. The urethra of a female is closer to the intestinal tract than in a male, making it easier for the E. coli bacteria from the intestinal tract to gain access to the
    urinary tract.
  • C. The E. coli bacteria in women are more likely to be resistant to antibiotic treatments.
  • D. The immune system is less efficient in women, making infection more likely to occur.
View Correct Answer
 Answer Key: B

Q46. If the antibiotic treatment used in the graph was continued for a period of 30 days, the MOST likely prediction would be that the:

  • A. Colony-forming units per milliliter of E. coli cell counts would begin to increase.
  • B. Metabolically active cell counts would eventually decrease.
  • C. Colony-forming units per milliliter of E. coli cell counts would remain low and the count of metabolically active cells would gradually decrease.
  • D. Colony-forming units per milliliter of E. coli cells would remain low and the count of metabolically active cells would remain about the same.
View Correct Answer
 Answer Key: D

Q47. The primary concern with the extended use of antibiotics to treat infections is that:

  • A. People become immune to the antibiotics so that when they really need them, the drugs will be useless.
  • B. The selective pressures of continued antibiotic use will induce resistance in the bacteria.
  • C. The antibiotics routinely cause toxicity in the body that can damage the liver and kidneys.
  • D. The antibiotics encourage the overgrowth of resident bacteria in the body.
View Correct Answer
 Answer Key: B

Q48. One theory concerning VBNC cells is that they are simply on their way to dying. A piece of evidence that would refute that theory would be:

  • A. if some of the VBNC cells did die
  • B. if the VBNC cells resuscitated
  • C. if apoptosis was observed in the cells
  • D. if no metabolic activity was observed
View Correct Answer
 Answer Key: B

Questions 49–59 are not associated with a passage.

Q49. Which of the following explanations could account for the fact that dominant alleles that cause fatal disorders are less common than recessive alleles that cause
fatal disorders?

  • A. Every person carrying a single fatal dominant allele dies, whereas most individuals who carry a single recessive lethal allele live and reproduce.
  • B. Recessive fatal alleles must cause sterility.
  • C. Dominant alleles that cause fatal disorders are more serious than recessive lethal disorders.
  • D. Dominant alleles are generally less common than recessive alleles.
View Correct Answer
 Answer Key: A

Q50. A certain type of cell makes antibodies that are secreted from the cell. It is possible to track the path of these as they leave the cell by labeling them with radioactive
isotopes. Which of the following might be the path of antibodies from where they are made to the cell membrane?

  • A. Golgi complex to lysosomes to cell membrane
  • B. rough endoplasmic reticulum to Golgi complex to cell membrane
  • C. nucleus to Golgi complex to cell membrane
  • D. smooth endoplasmic reticulum to lysosomes to cell membrane
View Correct Answer
 Answer Key: B

Q51. A scientist suspects that the food in an ecosystem may have been contaminated
with radioactive phosphates over a period of months. Which of the following substances could be examined for radioactive phosphate to test the hypothesis?

  • A. the carbohydrates produced by plants in the area
  • B. the DNA of the organisms in the area
  • C. the lipids produced by organisms living in the area
  • D. all of the above
View Correct Answer
 Answer Key: B

Q52. Neurons contain a high concentration of potassium ions relative to the fluids surrounding them. How could a neuron acquire even more potassium?

  • A. active transport
  • B. osmosis
  • C. endocytosis
  • D. diffusion
View Correct Answer
 Answer Key: A

Q53. At the end of the electron transport chain in aerobic cellular respiration, the final acceptor of the electrons is , which will then produce molecule of .

  • A. CO2; O2
  • B. NAD+; NADH
  • C. O2; H2O
  • D. ADP; ATP
View Correct Answer
 Answer Key: C

Q54. A geneticist allows a cell to replicate in the presence of radioactive nucleotides. Which of the following would occur?

  • A. The DNA in one of the daughter cells would be radioactive but not in the other daughter cell.
  • B. The mRNA made by the daughter cells would be radioactive.
  • C. The DNA in each of the daughter cells would be radioactive.
  • D. The DNA would not be radioactive in either of the daughter cells.
View Correct Answer
 Answer Key: C

Q55. How does DNA differ from RNA?

  • A. DNA nucleotides use deoxyribose sugar; RNA uses ribose sugar.
  • B. DNA uses the bases A, U, C, and G; RNA uses the bases A, T, C, and G.
  • C. DNA is produced in transcription; RNA is produced in translation.
  • D. RNA is a double-stranded molecule; DNA is a single-stranded molecule.
View Correct Answer
 Answer Key: A

Q56. A human cell undergoes mitosis. Each of the resulting cells have chromosomes.

  • A. 8
  • B. 23
  • C. 46
  • D. 92
View Correct Answer
 Answer Key: C

Q57. When a nonsteroid (peptide) hormone binds to a receptor on the cell surface:

  • A. the hormone moves into the nucleus where it influences gene expression
  • B. the hormone-receptor complex moves into the cytoplasm
  • C. a second messenger will form within the cell
  • D. the cell becomes inactive
View Correct Answer
 Answer Key: C

Q58. The attachment site for RNA polymerase on the DNA template is called the:

  • A. promoter
  • B. regulator
  • C. operon
  • D. repressor
View Correct Answer
 Answer Key: A

Q59. Consider a bacterial cell that performs anaerobic respiration. If that bacterial cell had access to 6 molecules of glucose, how many molecules of ATP would it be able
to produce?

  • A. 2
  • B. 6
  • C. 12
  • D. 36
View Correct Answer
 Answer Key: C

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