imone/ARB huggingface dataset:
Question: 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.
\begin{tabular}{ccccc}
\hline Sample Group & LH & FSH & Inhibin-B & Testosterone \\
\hline 1 & $5.3 \mathrm{IU} / \mathrm{L}$ & $5.0 \mathrm{IU} / \mathrm{L}$ & $190 \mathrm{pg} / \mathrm{mL}$ & $18.0 \mathrm{nmol} / \mathrm{L}$ \\
2 & $3.1 \mathrm{IU} / \mathrm{L}$ & $17.6 \mathrm{IU} / \mathrm{L}$ & $51 \mathrm{pg} / \mathrm{mL}$ & $17.8 \mathrm{nmol} / \mathrm{L}$ \\
\hline
\end{tabular}
The hormones FSH and LH are also involved in oogenesis in women. When LH surges in the ovarian cycle, what event will result?

Options:
A. endometrium proliferation
B. ovulation
C. menstruation
D. corpus luteum degradation
Answer: This question relies on your knowledge of the female reproductive cycle. In women, as FSH levels climb, follicles are stimulated to grow. These follicles produce estrogen, which exerts positive feedback on LH, whose levels begin to climb. When LH levels peak on the 14 th day of the cycle, the follicle ruptures, releasing the egg from the ovary and leaving the corpus luteum behind in the ovary. After ovulation, the corpus luteum secretes both estrogen and progesterone to inhibit GnRH. When GnRH is inhibited, FSH and LH will also be inhibited. Choice B correctly indicates that the LH surge causes ovulation. The correct answer is B.
Question: Questions 18-20 are based on the following data.
The generalized events in DNA replication can be depicted by the following flowchart:
Double helix $\rightarrow$ Nicking of a strand $\rightarrow$ Unwinding $\rightarrow$ Destabilization and relief of pressure $\rightarrow$ Initiation $\rightarrow$ Elongation $\rightarrow$ Closing of nicks
Between which two steps would the enzyme DNA polymerase be used?

Options:
A. Double helix $\rightarrow$ Nicking of a strand
B. Nicking of a strand $\rightarrow$ Unwinding
C. Initiation $\rightarrow$ Elongation
D. Elongation $\rightarrow$ Closing of nicks
Answer: Chain growth in DNA replication is initiated by a primase and is extended by a DNA polymerase. Thus the enzyme acting between Initiation $\rightarrow$ Elongation is DNA polymerase. The correct answer is C.
Question: 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.
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:

Options:
A. antibody production by plasma cells
B. cytotoxic T-cell activation
C. the complement system
D. interleukin secretion

Answer: Complement proteins circulate in the blood in an inactive form. When bacteria enter the blood, the complement proteins are activated, which causes a cascade of reactions that ultimately cause lysis and death of the bacteria. Antibody production or cytotoxic T-cell activation could certainly be stimulated by bacteria in the blood, but the question stipulated that this must be a nonspecific defense and these defenses are specific. Interleukins are cytokines secreted by immune cells, and they would not directly be able to defend against bacteria in the bloodstream. The correct answer is C.
Question: 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?

Options:
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.
Answer: This question relies on a knowledge of dominant and recessive patterns of inheritance. Dominant alleles are expressed, even in the heterozygous individual. In this case, our example is a dominant allele that causes a fatal condition. Since every individual carrying a single allele of this type would likely not survive, we would expect the frequency of the allele to decrease over time. In contrast, disorders that follow recessive patterns of inheritance require two copies of the allele for expression. Those who are heterozygous do not express the disorder and thus would survive. The correct answer is A.
Question: The hormone responsible for maintaining control of water balance in the human body is:

Options:
A. antidiuretic hormone
B. luteinizing hormone
C. adrenocorticotropic hormone
D. growth hormone
Answer: This problem can be solved through knowledge of each of the various hormones given in the four options. Antidiuretic hormone maintains control of water balance in the human body by stimulating increased water reabsorption by the kidneys. Luteinizing hormone stimulates secretion of sex hormones by ovaries and testes. Adrenocorticotropic hormone stimulates the adrenal glands. Growth hormone regulates a wide variety of activities involved in growth and metabolism. The correct answer is A.
Question: 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.
\begin{tabular}{ccccc}
\hline Sample Group & LH & FSH & Inhibin-B & Testosterone \\
\hline 1 & $5.3 \mathrm{IU} / \mathrm{L}$ & $5.0 \mathrm{IU} / \mathrm{L}$ & $190 \mathrm{pg} / \mathrm{mL}$ & $18.0 \mathrm{nmol} / \mathrm{L}$ \\
2 & $3.1 \mathrm{IU} / \mathrm{L}$ & $17.6 \mathrm{IU} / \mathrm{L}$ & $51 \mathrm{pg} / \mathrm{mL}$ & $17.8 \mathrm{nmol} / \mathrm{L}$ \\
\hline
\end{tabular}
Fully differentiated Sertoli cells lose the ability to proliferate to produce more Sertoli cells. This would mean that they cannot perform:

Options:
A. mitosis
B. meiosis
C. apoptosis
D. binary fission
Answer: This question relies on your knowledge of cell division. Mitosis is the process by which differentiated cells produce identical daughter cells. The correct answer is A.
Question: The DNA doubles and chromosomes replicate during which phase of the cell cycle?

Options:
A. $\mathrm{G} 1$
B. metaphase
C. $\mathrm{S}$
D. G2
Answer: The majority of the cell cycle is spent in interphase, which consists of three stages: G1, S, and G2. In the G1 stage of interphase, the cell organelles are doubled, and materials required for DNA synthesis are accumulated for the onset of cell division. The $\mathrm{S}$ stage of interphase is the stage of the cell cycle where the amount of DNA doubles with the replication of chromosomes. In the G2 stage of interphase, which follows the synthesis of DNA, proteins required for the next cell division are synthesized. Metaphase is the stage in mitosis characterized by the precise lineup of the chromosomes along the equatorial plane. The correct answer is C.
Question: A geneticist allows a cell to replicate in the presence of radioactive nucleotides. Which of the following would occur?

Options:
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. MCAT
Answer: This question relies on your knowledge of semiconservative DNA replication where a single double helix is separated in order for both strands of the double helix to serve as templates for the production of new strands. The correct answer is C.
Question: 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.
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?

Options:
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.
Answer: Of the items listed in the question, the thing they all have in common is that they are mutagens or involve some genetic change. These mutations to the DNA can lead to the production of altered proteins that can be associated with cancer. While some cancers are associated with changes in cellular growth factors, this would not be the case for all cancers. The translation of proteins is directly from the mRNA, which is derived from the DNA in transcription. There would have to be a change in the DNA for the translation of proteins to be affected. The correct answer is C.
Question: \section{
In October 2012, Felix Baumgartner attempted a world-record sky-diving jump from just over 128,000 ft (39,014 m) above the Earth's surface. He was fitted with a customized, pressurized spacesuit that brought his total weight to $260 \mathrm{lbs}$ ( $1156 \mathrm{~N}$ ). The spacesuit was designed to collect and transmit important monitoring, communications, and tracking information throughout the jump. Baumgartner then ascended to the jumping altitude housed within a pressurized capsule of weight 2900 lbs ( $12900 \mathrm{~N})$ attached to a $30 \times 10^{6}$ cubic feet $\left(8.5 \times 10^{8} \mathrm{~L}\right)$ helium-filled weather balloon. The weather balloon with the attached capsule ascended at a rate of $1000 \mathrm{ft} / \mathrm{min}(5.1 \mathrm{~m} / \mathrm{s})$, reaching the jumping altitude in approximately two hours.
Once the capsule reached the jumping altitude, the door to the capsule was opened and Baumgartner stepped out and fell forward, beginning his free-fall descent. He continued falling until he reached a maximum speed of $834 \mathrm{mph}(373 \mathrm{~m} / \mathrm{s})$, which exceeded the speed of sound. This speed was Baumgartner's terminal velocity-the velocity attained in which he was no longer being accelerated downward due to gravity. Once he reached this speed, Baumgartner deployed his parachute and glided safely to Earth's surface.The distance traversed by Mr. Baumgartner before he reached maximum velocity was approximately:

Options:
A. $4800 \mathrm{~m}$
B. $5480 \mathrm{~m}$
C. $6350 \mathrm{~m}$
D. $7100 \mathrm{~m}$
Answer: The initial velocity $\left(v_{i}\right)$ of Felix Baumgartner in the vertical direction is 0 , his final velocity $\left(v_{f}\right)$ is $373 \mathrm{~m} / \mathrm{s}$, and the acceleration due to gravity is $9.8 \mathrm{~m} / \mathrm{s}^{2}$ (although in reality it is probably slightly lower because of the significantly large jumping altitude). The equation of motion that can be used to determine the distance traversed by Baumgartner is: The correct answer is D.
Question: Genetically modified (GM) foods contain DNA from more than one source. The goal of creating GM foods is to modify the organisms to express new and useful traits. A few examples involve engineering crops to be resistant to bacterial or viral infections, to be resistant to herbicides, to ripen more slowly, to survive extreme temperature changes, to contain vaccines, or to create crops that are more nutritious.
One GM food that has received media attention over the years is golden rice. This GM product was first introduced in 2000 and was engineered to provide carotene that is a precursor to vitamin $\mathrm{A}$. The carotene in the rice provides the golden color. Vitamin A deficiency is a major problem in underdeveloped parts of the world, killing more than one million people a year, most of whom are children. This vitamin is critical for the development and maintenance of epithelial tissues, for vision, and for growth.
Rice plants normally produce carotene in the inedible parts of the plant. To be able to access and use the carotene, it was necessary to engineer the rice to produce it in the edible grains. In order to achieve this goal, two genes from daffodil plants, psy and $l y c$, and a bacterial gene called $c r t 1$ were added to the rice genome. These genes engineered a pathway that produces lycopene, which is eventually converted to carotene. Each of these genes was placed under the control of a grain-specific promoter so that the products were produced in the rice grains.
While golden rice was produced to provide carotene, which the human body can convert into the essential vitamin A, the rice provided much less carotene than was anticipated. Golden rice 2, a modified version released in 2005, contains a much larger amount of carotene and, as such, is a better source of the vitamin. Field tests were conducted in 2009, and human consumption trials were completed in 2012. However, neither version has been grown in large-scale quantities at this time.
The engineering of golden rice to have several exogenous genes that can be induced under the control of a single promoter would MOST resemble:

Options:
A. an operon system
B. hybridized DNA
C. a vector
D. Okazaki fragments
Answer: A series of several genes under the control of a single promoter is most similar to the inducible operon system in bacteria. Okazaki fragments are produced during semiconservative replication on the lagging strand of DNA. Strands of DNA that are hybridized are complementary to each other. A vector is used to carry DNA from one source into another. The correct answer is A.
Question: Venous blood coming from the head area in humans returns to the heart through which major vessel?

Options:
A. superior vena cava
B. aorta
C. hepatic portal vein
D. carotid artery
Answer: The superior vena cava is the vein that carries deoxygenated blood from the upper half of the body to the heart into its right atrium. The aorta is the largest and longest of the arteries of the human body that transports oxygenated blood directly from the left ventricle to all other organs and tissues in the body. The hepatic portal vein is the major vein that transports blood from the digestive organs into the liver. The carotid artery is one of two types of arteries (the other being vertebral) that feed blood from the heart and into the head/ brain area. Answers and Explanations No passage MCAT The correct answer is A.
Question: 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. Of the options listed, what would be the best treatment for someone suffering from water intoxication?

Options:
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
Answer: In this case, administering a solution that is hypertonic to the cells would help. A hypertonic extracellular solution would draw water out of the swollen cells to counteract the effects of water intoxication. An isotonic solution would not help. A solution low in solute concentration would be high in water relative to the cells making the solution hypotonic, which would cause more water to enter the cells. A saline solution would be high in solutes and hypertonic to the cells. The correct answer is A.
Question: \section{
In October 2012, Felix Baumgartner attempted a world-record sky-diving jump from just over 128,000 ft (39,014 m) above the Earth's surface. He was fitted with a customized, pressurized spacesuit that brought his total weight to $260 \mathrm{lbs}$ ( $1156 \mathrm{~N}$ ). The spacesuit was designed to collect and transmit important monitoring, communications, and tracking information throughout the jump. Baumgartner then ascended to the jumping altitude housed within a pressurized capsule of weight 2900 lbs ( $12900 \mathrm{~N})$ attached to a $30 \times 10^{6}$ cubic feet $\left(8.5 \times 10^{8} \mathrm{~L}\right)$ helium-filled weather balloon. The weather balloon with the attached capsule ascended at a rate of $1000 \mathrm{ft} / \mathrm{min}(5.1 \mathrm{~m} / \mathrm{s})$, reaching the jumping altitude in approximately two hours.
Once the capsule reached the jumping altitude, the door to the capsule was opened and Baumgartner stepped out and fell forward, beginning his free-fall descent. He continued falling until he reached a maximum speed of $834 \mathrm{mph}(373 \mathrm{~m} / \mathrm{s})$, which exceeded the speed of sound. This speed was Baumgartner's terminal velocity-the velocity attained in which he was no longer being accelerated downward due to gravity. Once he reached this speed, Baumgartner deployed his parachute and glided safely to Earth's surface.The drag force, $F$ drag, can be expressed in terms of the terminal velocity, $v t$, as $F$ drag $=C v_{t}^{2}$, where $C$ is a drag constant. The value of the drag constant for Mr. Baumgartner is:

Options:
A. $0.008 \mathrm{~kg} / \mathrm{m}$
B. $0.680 \mathrm{~kg} / \mathrm{m}$
C. $1.320 \mathrm{~kg} / \mathrm{m}$
D. $5.640 \mathrm{~kg} / \mathrm{m}$
Answer: When an object in free fall reaches terminal velocity, it is in equilibrium. In other words, the drag force acting upward becomes equal in magnitude yet opposite in direction to the object's weight, or: The correct answer is A.
Question: Iodine-131, a radioisotope typically employed in the diagnosis of thyroid disorders, has a mass number, $A$, of 131 and an atomic number, $Z$, of 53 . The number of neutrons, $N$, of iodine-131 is:

Options:
A. 131
B. 78
C. 53
D. 25
Answer: The number of neutrons is the mass number $A$ minus the atomic number $Z$. Thus, the number of neutrons of iodine-131 is $N=A-Z=$ $131-53=78$ The correct answer is B.