MCAT BIO

Which organisms are dependent on the electron transport chain to create ATP?

A: All obligate aerobes Obligate aerobes, whether prokaryotic or eukaryotic, can't survive without creating ATP through aerobic means. Aerobic respiration involves the electron transport chain. - ETC is inhibited by certain antibiotics (i.e cyanide, azide, and carbon monoxide)

Viruses

- Enveloped viruses bud off the host's membrane whereas non-enveloped viruses cause the host to burst to release viral particles. Viruses don't have any organelles or a nucleus; the genetic material is simply packed inside a protein coat. Viruses are roughly 100 times smaller than bacteria, and 1000 times smaller than eukaryote cells. The host's ribosomes make the necessary protein coats and polymerases that replicate the viral genetic material and host's ATP provides necessary energy (the host also provides the raw materials like nucleotides and amino acids). The coat proteins and viral genetic material will assemble into viral particles all by themselves.

A student is designing a DNA probe to detect the location of various sequences within the mouse genome. To make visualization as easy as possible, he adds a 5' poly(A) tail constructed with radiolabeled adenine to each primer. One of the genes of interest contains the sequence 5' GTTCGCGTTAAGG 3'. A possible primer used to identify this gene is:

5' AAAACCTTAACGCGAAC 3'

Which proper is most suitable for PCR?

A: "5'-GCATAGAAGCATTCCGC-3' (I picked 5'-ATTACGTTAACATGAAG-3' because of the start codon AUG) The answer to this question must have a suitable primer with a high GC content (high temperature resistivity) and G or C base pairs at the 5' and 3' ends.

Which of the following amino acid residues is most likely to be found in the transmembrane domain of EGFR?

A: Val The membrane-spanning domain of EGFR is in contact with the nonpolar tails of the phospholipid molecules that comprise the bilayer membrane. This region would be expected to be abundant in nonpolar amino acid residues, such as Val (valine)

Which technique CANNOT be used to analyze gene expression? A. Western blotting B. Northern blotting C. Southern blotting D. Reverse transcription PCR

Answer: C Souther blotting is a technique used to detect a particulaR sequence in a sample DNA

Vmax

The rate of the reaction cannot exceed Vmax; Vmax represents the rate that is reached when the enzyme is saturated with substrates. When the enzyme is not yet saturated, the rate can reasonably fall along a range of values that are all less than Vmax, even 0.

Single-strand binding protein

responsible for keeping the DNA unwound after the helicase. SSBs stabilize single-stranded DNA by binding to it.

Chain termination of transcription

there are 2 ways that transcription can terminate; 1) intrinsic termination: specific sequences called a termination site creates a stem-loop structure on the RNA causing the RNA to slip off the template & 2) Rho (ρ) dependent termination: a protein called the ρ factor travels along the synthesized RNA and bumps off the polymerase

Transcription in eukaryote

- Regulation of transcription in eukaryotes: Transcription factors bind to enhances or silences to affect transcription. The main difference in eukaryotes that sets them apart from prokaryotes is that enhancers/silencers can be very far away from the actual promoter and can be upstream/downstream. The DNA must loop back on itself so that the transcription factor bound to enhancer/silencer can actually make contact with the promoter. Eukaryotes lack the bacterial transcription regulation mechanisms such as the operon (exists but very rare) and attenuation. - Common DNA-binding domains include helix-turn-helix (HTH), zinc finger, basic-region leucine zipper (bZIP).

Different types of capillary

1) Continuous capillary: no pores on endothelial cells where exchange occurs through the clefts or by vesicle trafficking through endothelial cells (found in skins and muscles like BBB), 2) Fenestrated capillary: small pores that are large enough for molecules but not blood cells to leak through. Found in small intestines to facilitate nutrient absorption and also endocrine and kidney. 3) Sinusoidal capillary: large pores (large enough for blood cells to lea

All of the following statements regarding reverse transcriptase are false EXCEPT: A. it is produced by viral ribosomes. B. it is carried within the retroviral capsid and released into the cytosol following viral penetration. C. it is unable to use DNA as a template. D. it is unable to use RNA as a template.

A- B: it's carried within the retroviral capsid and released into the cytosol following viral penetration Reverse transcriptase must be carried intact within the capsid to process the RNA genome. Without reverse transcriptase, the RNA genome cannot be converted into DNA. - A is false because Viruses don't have ribosomes. All viral proteins are produced at some point during the replication cycle by host ribosomes - C is false because reverse transcriptase first synthesizes a single stand of DNA from the viral RNA template. It then synthesizes a second complementary strand of DNA using the first DNA strand - Reverse transcriptase uses the viral ssRNA genome as the initial template for DNA

A researcher concludes that the sample represents single-stranded DNA and NOT another form of nucleic acid. Which of the following compositions most likely represents this sample?

A. 17% A, 17% T, 33% G, 33% C B. 29% A, 14% U, 11% G, 46% C C. 4% A, 4% U, 46% G, 46% C D. 12% A, 12% T, 30% G, 46% C B and C can be eliminated because they contain Uracil that is only present in RNA. Because choice A has equal proportions of A&T and G&C, they could easily represent dsDNA through base-pairing. Therefore, the answer is D.

Which event is directly mediated by a ligand-gated ion channel?

A. Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane (I picked "release of Ca2+ from the sarcoplasmic reticulum of a muscle fiber to initiate muscle contraction) The influx of Na+ across the motor end plate occurs when Na+ ion channels bind the ligand acetylcholine. - Ligands don't directly influence the product of cAMP, hydrolyze GTP->GDP, and aren't involved in any intracellular signaling. It just allows for rapid response to a stimulus (often allosterically bound); thus, cells that need quick responses like neurons possess ligand-gated ion channels.

Which of the following acts to decrease blood pressure?

A: Atrial natriuretic peptide Atrial natriuretic peptide is released from the atria of the heart in response to high blood pressure (exactly opposite of aldosterone) It antagonizes aldosterone and cause kidney to excrete both more sodium and water

Which cellular features increase in the direction of the arrow (multipotent -> pluripotent -> totipotent)

A: "Open" chromatin and lineage potential As cells progress totipotency -> pluripotency -> multipotency, their lineage potential decreases (they lost the ability to give rise to a broad range of tissue types) & their DNA becomes increasingly methylated, decreasing the prevalence of euchromatin. Moving in the opposite direction, it increases lineage potential and chromatin's tendency to fall into a loose arrangement

In Drosophila, the white gene for eye color is a recessive mutation on chromosome X. Affected Drosophila exhibit white eyes, while wild-type Drosophila have red eyes. If a white-eyed female Drosophila is crossed with a wild-type male Drosophila, what percent of their female offspring will have white eyes?

A: 0 If a wild-type allele for white eye is designated as "X" and the mutated allele as "x", the white-eye female is xx. The wild type male is XY. A female offspring will receive a healthy X chromosome from the father and a mutated one from the mother. However, because the trait is recessive and a female has two X chromosomes, none of the female offspring will be white-eyed. Notice that for male offspring, they will all be white-eyed, because they receive the mutated X from mother and a Y from father. Since the males have only one copy of the X chromosomes, the trait will be expressed and all males will be white-eyed

In humans, traits A and B are autosomal and located on separate chromosomes. Trait A is dominant over a, and trait B is codominant with b. If a man with a genotype of aabb has children with a woman with a genotype of AABB, what proportion of their children would be expected to be homozygous for either trait?

A: 0 The question tells us that the traits are AUTOSOMAL; so we know there will be NO difference between boys and girls. If you cross AA and aa, all the children will be Aa. If you cross BB and bb, then all of the children will be Bb. Thus, none of the children will be homozygous

What is the probability that the daughter of an unaffected father and a mother who is a carrier for the DMD (DUchenne muscular dystrophy) gene is affected by the disease?

A: 0% The passage states that DMD is X-linked recessive, which means that a woman MUST have a mutated copy of the dystrophin gene on BOTH of her X chromosomes to express the trait. Because the father is unaffected and will pass his X chromosome onto his daughter, there is 0% change that she will be affected. - X-linked recessive diseases are extremely rare in women

In the U.S. population, the frequency of the allele for colorblindness, Xc, is 8%. Which of the following is the frequency of colorblind women and colorblind men in the population, respectively?

A: 0.64%, 8% For a woman to be colorblind, she must get two copies of the Xc gene. Since the frequency of this gene is 0.08, the odds of being homozygous for the gene are 0.0064, or 0.64%. Males only need a single copy of the Xc gene to be colorblind (since their other gene is Y), making the odds of a male being colorblind 8%.

In oxidative phosphorylation, cytochrome c acts as

A: 1-electron carrier Because cytochrome c is a heme protein that only cycles between a ferrous and ferric state during oxidative phosphorylation, only single electron transfers are possible

According to the passage, what total length of DNA is found in the average human cell before it is condensed?

A: 1.8m This question asks us to determine the total length of the nuclear material in a cell. Prior to being condensed, the DNA will be duplicated, giving us ~6 billion base pairs (bp) of DNA. If each one has a length of 0.3 nm (paragraph 3), this gives: (6 x 10^9 bp)(0.3 x 10^-9 m/bp) = 1.8 m.

Suppose an organism contains an enzyme that can insert one random string of nucleotides into the coding portion of a gene. If this enzyme performs its function once on a particular gene, what is the probability that the resulting protein will be translated in the same frame as it would be if unmodified?

A: 1/3 If an enzyme inserts a random number of bases, it will only add a multiple of three bases 1/3 the time. Whenever it inserts a different number (2,4,5,7..etc), the reading frame changes, leaving 1/3 change that the frame will remain unmodified while 2/3 change of a shifted frame.

Hemophilia, a disease in which the time required for blood to clot is greatly prolonged, is determined by a sex-linked gene. Suppose a man with normal blood clotting marries a woman with normal blood clotting whose father was a hemophiliac. If this could has three sons, what is the probability that hemophilia will be transmitted to all three of them

A: 1/8 The woman's inherited X chromosome from her father contained the hemophilia gene. The mother will pass on one X chromosome to her sones, either the X chromosome containing the normal (wild) gene or chromosome containing the hemophilia gene. The probability of a son receiving the hemophilia gene and being a hemophiliac is 1/2. Thus, it's 1/2*1/2*1/2 = 1/8

An inactive tetramer of IN (integrase) is expected to have approximately what molecular weight?

A: 128 kDa The passage indicates that an integrase monomer is composed of 288 amino acids, which will have an approximate molecular of 32kDa (*the average weight of an amino acid is 110 Da*). Thus, a tetramer will have an approximate molecular weight of 128 kDa.

Assuming no alternative splicing occurs, approximately how many base pairs will be in a mature mRNA produced from the FAM83H gene?

A: 1441 A mature mRNA consists of a 5'-cap (one base pair), the 5' UTR (untranslated region), coding regions, the 3' UTR, and a poly(A) tail. Therefore, the length of a mature mRNA using the numbers given in the table is : 1+130+300+250+150+180+200+90+140=1441 - Untranslated regions (UTR) have a significant influence on translation; therefore, they must be transcribed and end up in the mature mRNA product. These regions are different from the coding regions since the UTRs typically do not end up in the final protein product

Meiosis I results in

A: 2 haploid cells with 23 chromosomes, each chromosome consisting of 2 sister chromatids Meiosis I results in 2 haploid cells, each with 23 chromosomes consisting of 2 sister chromatids per chromosome. In the male, the sister chromatids are split into 4 gametes during meiosis II. For females, meiosis I results in a secondary oocyte (a gamete) and a polar body. Penetration of the secondary oocyte by a sperm brings on anaphase II. Telophase II produces a zygote and a second polar body. *Remember for MCAT* : Mitosis results in diploid (46 chromosomes) daughter cells each chromosome with a single chromatid while meiosis results in haploid cells (23 chromosomes) to produce gametes.

758 of 1000 individuals have a dominant trait in which they cannot differentiate between the colors periwinkle and lavender. What fraction of the population consists of people homozygous for this allele?

A: 25% The 758 individuals described are those who express the dominant allele, meaning that they can be either homozygous or heterozygous for the trait. If we round 758 to 750, we can estimate that 75% of the population has the inability to differentiate these colors; in other words, p2 + 2pq = 0.75. While this may be difficult to solve for p, we can easily use it to find q2 using the equation p2 + 2pq + q2 = 1. If 0.75 + q2 = 1, then q2 = 0.25 and q = 0.5. Thus, p + 0.5 = 1, and p = 0.5. Finally, we can use this value to find that p2 (the proportion of homozygous dominant individuals) is 0.25, or 25% of the population.

A farmer crosses two distinct variants of corn. One is sweet with yellow kernels, while the other is not sweet and displays red kernels. The first crop yields only sweet corn with yellow kernels. What percentage of the next season's crop can the farmer expect to have red kernels?

A: 25% The first crop (or cross) yields sweet, yellow corn only. We can thus infer that these traits are dominant and that every plant in the first generation is a heterozygote. Conducting a dihybrid cross with two heterozygotes yields a Mendelian ratio of 9:3:3:1. In other words, 3/16 of the population should display red kernels and a sweet taste, while 1/16 will exhibit both recessive traits (red kernels and a non-sweet flavor). Our answer, then, is 4/16 or 25%.

Which of the following nucleotide sequences describes an antisense molecule that can hybridize with the mRNA sequence 5'-CGAUAC-3'?

A: 3'-GCUAUG-5' When the RNA molecules hybridize, the antisense molecule lines up in an ANTIPARALLEL fashion with the sense molecule (3' end lines up with the 5' end of its complement).

DNA polymerase catalyzes the replication of chromosomal DNA in bacteria as shown below. A double stranded DNA molecule contains bases with a ration of (A+T)/(G+C) = 3:1. This is replicated with DNA polymerase in the presence of the four deoxynucleoside triphosphates with a molar ratio of (A+T)/(G+C)=1:1. What is the expected ratio of (A+T)/(G+C) in the double-stranded daughter DNA?

A: 3:1 The double stranded daughter DNA molecule would be an exact duplicate of the parent molecule; thus, it has the same (A+T)/(G+C) ration, giving the correct ratio of 3:1

In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Assume mendalian inheritance pattern)

A: 4/9 (I picked 1/2) Given mendelian inheritance patterns, a 3:1 ratio of F1 offspring means that the original crosses beetles are both heterozygotes, and the F1 offspring are 25% red (homozygous dominant), 50% red (heterozygous), and 25% brown (homozygous recessive). If two red F1 beetles are crossed and both red and brown beetles appear in the F2 generation, the F1 red beetles that were crossed must both be heterozygotes. The probability that, of the red F1 beetles, both were heterozygous is 2/3*2/3 = 4/9 (only red beetles were selected from and 2/3 of the red F1 beetles were heterozygous)

The sequence below is found in the sense strand of a human gene, upstream from the region that codes for the bulk of the associated protein's active site. 5' CCCGTATAC 3' A researcher desires to render this protein entirely nonfunctional. A mutation into which of the following sequences is most likely to accomplish this goal?

A: 5' CCCGTATAA 3' To knock out the protein's functionality, the associated mRNA should be truncated with a premature stop codon. If TAC is mutated into TAA, it corresponds to UAA, which Is a stop codon in mRNA. Note that the question specified that this DNA is the sense strand which has the same sequence as the product mRNA translated (if we were given the antisense strand, we must find the complementary strand).

For the siRNA (5'-CUAGGUCGAUCAAU-3') to work, the EOLA1 gene must contain which antisense strand?

A: 5'-CTAGGTCGATCAAT-3' If the siRNA sequence is 5'-CUAGGUCGAUCAAU-3', then it must bind the following complementary mRNA sequence: 3'-GAUCCAGCUAGUUA-5'. The mRNA strand represents the same sequence as the sense strand on the DNA with the exception of the replacement of uracil (U) for thymine (T). DNA is complementary and anti-parallel; therefore, the antisense strand will be complementary to the mRNA sequence given above, yielding 5'-CTAGGTCGATCAAT-3' (Working backward: siRNA -> complementary RNA -> sense DNA -> anti-sense DNA)

A partial DNA sequence of the coding strand of a gene is shown: 5'-GACATGGACTCGCTA-3' Which sequence corresponds to the mRNA for this DNA?

A: 5'-GACAUGGACUCGCUA-3' (Picked: "5-UAGCGAGUCCAUGUC-3') For any given gene, the nucleotide sequence of both coding strand and the mRNA are complementary to the sequence of the template strand

If vDNA sequence encoding a protein is inserted into a host genome by IN. The protein is translated from the hypothetical mRNA sequence shown: 5'-GGCAACUGACUA-3' Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence?

A: 5'-GGCAACUGACUA-3' (I picked 5'-TAGTCAGTTGCC-3') According to the passage, viral DNA integrated into a host cell genome by integrase (IN) would originate from a retrovirus. mRNA transcribed from retroviral DNA is either used to synthesize viral proteins or used as the RNA genome for progeny viruses. Thus, the sequence of the nucleotide in the original viral genome will be the same as that of the transcribed mRNA.

A male Myshkin mouse is crossed with a wile-type female mouse. What proportion of the male offspring would be expected to show the Myshkin behavior pattern?

A: 50% The passage tells us that the Myshkin mice are heterozygotes for an autosomal mutation (M/+). A cross between a Myshkin mouse (M/+) and a wile-type mouse (+/+) would result in half the offspring being Myshkin and half being wild-type; the sex of the mice is irrelevant as this is an autosomal mutation

Blood type in humans is a classic example of codominant inheritance. An AB+ male has a child with an O+ female. What is the likelihood that their first child will be able to receive a blood transfusion from a B- donor?

A: 50% The child has a 50% chance of inheriting his father's A allele, and a 50% chance of inheriting his father's B copy. On the mother's side, the child is guaranteed to inherit an O allele (as type O individuals are homozygous for that recessive allele). In other words, the child has a 50% chance of being type A and 50% change of being type B. Only a type B (or O) individual can receive blood from a type B donor without complications so the answer is 50%.

Phenylketonuria is a rare autosomal recessive disorder with a carrier rate of 1/50 within Caucasian populations. A family enters a clinic with a sick child. Although neither parent is experiencing symptoms, you suspect PKU after a family history revels that the child's paternal great-grandmother had this condition. What are the chances, respectively, that the father is a carrier and that the child expresses this trait?

A: 50%, <1% Since the child's great-grandmother had PKU, she must have possessed two copies of the recessive allele. She passed one of these alleles to her offspring (the father's mother or father), making him or her an obligate carrier. This individual then had a 50% chance of passing it to the father of the sick child. (Note that we do not need to consider the father's other parent in this calculation, as the carrier rate is extremely small.) Now, we know that the father had a 50% chance of inheriting one allele and thus has a 50% chance of giving it to his child. As we lack any specific information about the child's mother, we can assume that she, like the population in general, runs a 1/50 risk of being a carrier. If so, she, too, has a 50% chance of giving her disease allele to the child. Multiplying these proportions yields (1/2)(1/2)(1/50)(1/2), or a 1/400 chance that the child is affected by PKU. A one-in-four-hundred risk is less than 1%.

CMIIs are typically inherited in an autosomal recessive fashion in which 75% of the alleles in the population are normal. What proportion of the population would exhibit CMIIs?

A: 6% The question requires one to know the Hardy-Weinberg formulas: p+q = 1 (alleles in a population) and p^2 + 2pq + q^2 = 1 (phenotypes in a population. Where p is the frequency of the dominant allele and q is the frequency of the recessive allele. The problem states that p = 0.75. Thus, q = 0.25. The proportion of the population exhibiting an autosomal recessive phenotype would be equal to q^2 = 0.25^2 ~ 6%

What is the net volume of fresh air that enters the alveoli each minute, assuming that the breathing rate is 10 breathes/min, the tidal volume is 800mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL?

A: 6500 mL The amount of air entering the lungs is a single break (tidal volume), given as 800mL. Of 800mL, only 650mL reaches the alveoli per breath. Thus, the net volume of air reaching the alveoli each minute is 650mL/breath multiplied by 10 breathes/min

By approximately what percentage of its original length is a free DNA strand shortened by the coiling around a single histone? (Note: A histone has a diameter of 11 nm; assume inter-histone length is neglible.)

A: 75% This question asks us to determine how much each histone shortens the DNA strand. We can imagine this linking as winding a string around a tennis ball. From paragraph 3, we know that 200 nucleotides wrap around 1 histone. Initially, a 200-nucleotide sequence will be 200 (0.3 nm) = 60 nm in length. Once coiled around the protein, we can assume (without getting into unnecessary surface area calculations) that a single layer of DNA strand will be wrapped around the protein (strand thickness = 2 nm), meaning we now have the strand condensed to a ~ 4-nm layer (2 nm on each side) around the 11-nm-diameter protein. Thus, about 60 nm of DNA is replaced by 15 nm, meaning there is approximately a 45-nm decrease in length. 45/60 = 0.75 = a 75% decrease in length.

The instance of nondisjunction for the X chromosome in females over the age of 30 is about one out of every 130 live births. If a woman over 30 gives birth to a viable baby, assuming the risk of nondisjunction from the father is negligible, what's the likelihood that it will have a normal phenotype?

A: 99.2% This asks for the probability of a mother giving birth to have a phenotypically normal child. The chances of nondisjunction are 1/130 (0.008) ~ 0.8%. If nondisjunction occurs, there is a 100% chance that she will pass on either no copies of her X chromosome and have a child with XO or pass two copies of X and have a baby with XXX. The only viable monosomy is the XO. Since the question asks us to assume a viable birth, 1-(probability of nondisjunction) = 1 - 0.008 = 99.2%

A lower-than-normal blood pressure will cause which of the following effects on the rate of plasma clearance of substance A?

A: A decrease, because the decreased rate of urine output will allow more reabsorption by the kidney Low blood pressure has an effect on ADH levels, affecting the amount of substance A in the urine (but the passage doesn't provide enough information). Low blood pressure decreases the glomerular filtration rate, allowing more time for reabsorption and decreasing the amount of substance A in the urine. Blood pressure is the source of the energy that forces fluid into the capsular space. If the heart stopped and the blood in the glomerular capillaries had no hydrostatic pressure, fluid in the space around the glomerulus would flow back into the capillary bloodstream because the protein-rich blood would be hypertonic. Thus, the fluid would flow down the osmotic gradient into the blood.

Which cell is LEAST likely to be lysed by a T cell upon presentation of an antigen in the context of MHC?

A: A dendritic cell presenting a peptide from E.coli Dendritic cells present antigen to helper T cells to activate the adaptive immune system; they are not lysed by the T cells to which they present - Dendritic cells are antigen presenting cells in tissues (common points for initial infectioN) that identify threats and act as messengers via antigen presentation; most effective at eliciting adaptive immune system

A nucleotide consists of

A: A five-carbon sugar, a nitrogenous base, and 1-3 phosphate groups - Nucleoside: a five-carbon sugar and a nitrogenous base only

Which of the following most likely pushes a cardiac myocyte above the threshold membrane potential level?

A: A neighboring cell's depolarization allowing some positive ions to enter through gap junctions We know that electrical signals are propagated through the heart and that a cardiac myocyte will not be able to achieve an action potential on its own. Therefore, it must rely on one or more neighboring cells.

A biologist views the myosin head of a single thick filament, which is protruding in the form of a cross-bridge that is near, but not attached to, a thin filament. The cross-bridge is currently positioned at a 90° angle to the thin filament. This moment in time occurs immediately after:

A: ATP is hydrolyzed to form ADP and inorganic phosphate. The question stem mentions that actin and myosin are not attached, but that they are positioned perpendicularly, implying that they are preparing for future binding. Hydrolysis of ATP into ADP and inorganic phosphate restores the myosin head to a 90° angle, "cocking" it for new attachment and the ensuing power stroke. - ATP binding of myosin does detach it from actin, but the myosin head remains bent until ATP is hydrolyzed. - Immediately after myosin binds actin and before the power stroke, ADP is released from myosin.

Physicians usually diagnose CMIIs using postnatal magnetic resonance imaging. If this technology were not available, which symptom might the study scientists look for to determine if a hydrocephalus event is occurring in a patient?

A: Abnormal heart rhythm According to the second paragraph, hydrocephalus leads to fluid accumulation near the brainstem. This would exert pressure on, and impair the function of, the brainstem. The brainstem is comprised of the medulla oblongata, the pons, and the midbrain. The medulla oblongata is involved in regulation of breathing, heart rate, and blood pressure. Therefore, disrupting the function of the medulla oblongata may cause abnormal heart rhythm.

Which solvent will best promote an SN2 mechanism?

A: Acetone SN2 is favoted by polar aprotic solvents such as acetone or DMSO. - Water and methanol are protic solvents and toluene is a nonpolar solvent

A patient with persistent difficulties fighting off bacterial infections is found to have a genetic mutation that adversely affects the speed of phagocytosis in macrophages. This mutation most likely impairs which type of cytoskeletal polymer?

A: Actin filaments Macrophages must undergo rapid actin reorganization during phagocytosis; if a macrophage can't engulf bacteria in this manner, its overall function is considered impaired.

Pinocytosis involves the engulfment of extracellular fluid and solutes by one or more vesicles. It is a form of

A: Active transport and endocytosis Pinocytosis is a specific form of endocytosis that requires large amounts of ATP classified and is classified as active transport

Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD?

A: Allosteric The passage states that high levels of NADPH inhibit G6PD. Because the structures of G6P and NADPH are very different, it's unlikely that NADPH competes with G6P at the active site (thus it's not competitive inhibition). It is also not an irreversible inhibition; rather, it is dynamic based on the amount of NADPH available; thus, NADPH binds to a site that is not the active site (characteristic of allosteric inhibition).

In a negative inducible operon, a repressor protein binds to the operon and the genes are not actively transcribed. In such an operon, which of the following could be added to restore the operon to a transcriptionally active state?

A: An activator In a negative inducible operon, transcription is inhibited by a repressor, resulting in a basal transcription rate near 0. However, transcription can be "switched on" by the addition of an activator, which blocks the repressor from binding to the operon. - Inducers bind to activators. They upregulate transcription in POSITIVE operons; addition of an inducer couldn't remove the repressor protein from the operon - Inhibitors deactivate activator proteins and decrease the level of transcription

Women over 35 years of age have an increased risk of nondisjunction due to errors in what phase of meiosis?

A: Anaphase II Nondisjunction is the failure of chromosomes to separate properly during anaphase I of meiosis or the failure of sister chromatids to separate properly during anaphase II of meiosis.

In which of these stages do homologous chromosomes separate into distinct cells?

A: Anaphase of meiosis I Remember the difference between homologous chromosomes and sister chromatids! Homologous chromosomes are the pairs of non-identical chromosomes that are present in every somatic cell. (For example, you inherit one version of chromosome 2 from your father and one from your mother; these structures are homologous.) These pairings separate during meiosis I, causing the daughter cells generated by this process to be haploid before the start of meiosis II.

Some bacterial toxins are able to bind with high affinity to MHC, displacing whatever peptide was previously bound. As a result, the immune response is systemically activated without a clear target. The resulting pathology would most resemble:

A: Anaphylaxis The question describes a systemic reaction in which the immune system responds inappropriately to all cell types that express MHC (essentially, all somatic cells). This closely resembles anaphylaxis, an extreme systemic allergic reaction that can be fatal.

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 can block the initiation of events that result in the release of harmful, biologically active molecules The passage shows that the presence of antibody B (directed at subunit B) significantly reduces tissues injury. This suggests that the function of subunit B is to adhere the neutrophil to vascular endothelium, which occurs before the release of toxic molecules by the neutrophil. Thus, the use of antibody B would be a proactive treatment that prevents neutrophil adherence and the subsequence release of free radicals and proteases, rather than a reactive treatment, such as free radical and protease inhibitors. - Antibody binds to antigen to bring about: 1) neutralization: pathogen can't adhere to host cell, 2) opsonization: makes it easier for phagocytosis, 3) complement activation: kills infected cell by punching holes in cell membrane

Oncotic pressure is mainly generated by serum albumin. How can this molecule best be described?

A: As a soluble 65kd protein that serves to draw fluid into the venous ends of capillary beds Oncotic pressure is another term for the osmotic pressure that exists due to proteins. This is necessary to promote the influx of water into the distal regions of capillary beds. Albumin (like other proteins) is too large to diffuse in/out of a capillary. Instead, it stays in the capillary bed and draws water with its associated toxins and waste products from the tissues to the bloodstream. - Remember that osmotic pressure involves the attraction of WATER to regions that are relatively solute-rich. - Vessels (arteries, etc) have walls that are thick to facilitate gas and fluid exchange so diffusion normally doesn't occur - Albumin is a protein made by the liver that keeps fluid from leaking out of blood vessels, nourish tissues, transport substances throughout the body

Which statement describes the change in osmolarity throughout the loop of Henle?

A: As the filtrate progresses through the descending portion, its osmolarity increases The descending limb of the loop of Henle is permeable to water, but not salt. As the filtrate progresses further down the limb, water will continue to flow outward as the filtrate enters an increasingly hypertonic environment. Thus, filtrate osmolarity increases because water exits the nephron while the total amount of solute remains constant. -As the filtrate progresses through the ascending portion, its osmolarity will decrease as salts are removed.

The scientist wanted to use antibody B clinically (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, repeated usage in the same patient would elicit the production of human anti-mouse antibodies. (I picked: "Because the antibody was generated in the mouse, it will not recognize human antigens") Human immune system will recognize the mouse antibody as a foreign substance (antigen) and generate an immune response towards it, which may include a mild to severe allergic response. The immune response will generate antibodies against the most antibody, limiting its usefulness as a treatment.

Embryonic mouse cells divide every 10 hours at 37 degree. How many cells would be produced from an egg after 3 days

A: Between 50-500 Given 72 hours, 1 -> 2 -> 4 -> 8 -> 16 -> 32 -> 64 -> 128 -> 256; there would be 128 cells

During rat embryogenesis, researchers noticed the development of a fluid-filled cavity in cells that had previously undergone morulation. That cavity was most likely

A: Blastocoele It's asking to identify the developmental stage after the morula (a mass of 16 undifferentiated cells). To remember the development of the zygote.. *"More blasting, I am nervous"* Morula -> blastula -> gastrula -> neurulation The blastocoele is a fluid filled central region in the blastocyst during mammalian embryogenesis. The blastocyst consists of an inner cell mass (ICM) along with an outer cell layer called the trophoblast, which surrounds both the ICM and blastocoele - Gastrulation is when the single-layered blastula becomes the three-layers gastrula (ectoderm, mesoderm, and endoderm) - The inner cell mass is the mass of cells within the blastocyst that will eventually give rise to the fetus

If some but not all of the offspring from repeated matings of the same pair of fruit flies show the recessive traits of vestigial wings (vv) and ebony color (ee), which of the following could have been the genotypes of the individuals mated?

A: Both VvEe To display both recessive phenotypes, the offspring must have INHERITED a recessive allele for each gene from both parents (each parent must have at least one copy of a recessive allele for both genes). Some of the offspring did not display the recessive phenotypes for these traits; this implies that at least one parent has a dominant allele for the dominant gene and the ebony gene.

Herpes simplex virus alters the DNA of a host cell, which can then be passed on to daughter cells via mitosis. In some cases, several cell generations may go by with the instructions being passed on and no virus being produced. The viral insertion to the host DNA may then be activated by an environmental trigger, resulting in a number of viruses being produced within the cell, which then escape to spread through the host body or infect new hosts. Herpes simplex is:

A: Both lytic and lysogenic Viral DNA being passed on to daughter cells through ordinary means (mitosis) occurs in a lysogenic cycle. The production of large numbers of virus particles within a cell that then escape it (with or without destroying the cell) is an example of a lytic life cycle

Control of heart rate, muscle coordination, and appetite is maintained by the

A: Brain stem, cerebellum, and hypothalamus

A resident of a famine area who appears undernourished and extremely emaciated has eaten only starches for the past 3 months. A urine analysis shows that a large amount of nitrogen is being excreted. This is most likely evidence of

A: Breakdown of the body's own structural proteins to provide energy (I picked: "an abnormally high rate of glycogen breakdown in the liver") In starvation, the body uses up its stores of carbohydrate and lipids, and then begins to break down body proteins for metabolic energy. A byproduct of the metabolism of the amino acids from protein is nitrogen. Therefore, the nitrogen in the urine comes from breakdown of the body's proteins. Neither lipids nor carbohydrates (such as glycogen) contain nitrogen so the excess nitrogen couldn't come from these sources.

Which statement below most accurately describes the roles of the proteins actin and myosin during muscular contraction?

A: Bridges between actin and myosin form, break, and re-form, leading to a shortening of muscle sarcomeres Neuronal impulses cause the release of calcium from the sarcoplasmic reticulum within muscle cells. The calcium then binds to troponin, which causes a shift in the troponin/tropomyosin complex, revealing the binding site for myosin. Myosin binds to actin, causing a conformational change in myosin that "cocks" the head of the myosin molecule and slides the actin filament relative to myosin. ATP binds to myosin, causing it to detach from actin and "recharge". If another binding site is available on actin, myosin will bind again and slide the actin filament even further.

Which of the following best explains the differences in adhesion between CA and AA macrophages (CA's is higher)?

A: CA macrophages have higher integrin levels than AA macrophages Integrins are transmembrane receptors that modulate cell-to-extracellular matrix interactions. These proteins often attach the cell to collagen and fibronectin fibers. Passage indicates that CA have higher adhesion than AA on collagen and fibronectin, but not on plastic, suggesting that the difference in adhesion is caused by higher integrin levels in the CA macrophages - Cadherins are transmembrane proteins which play a primary role in cell-to-cell adhesion, forming adherens junctions to bind cells within tissues together. They don't play a role in the cell-to-extracellular matrix interactions required to bind the macrophages to collagen/fibronectin CA macrophages (in comparison to AA), will be 1) less motile, 2) found in greater number in the blood vessels while 3) lower levels of dynein. Macrophages found in the blood are derived from monocytes that originate in the bone marrow while macrophages in the tissues are derived from the embryonic yolk-sac. CA macrophages have a higher affinity for collagen. Collagen is a primary component in connective tissues (found in the tendons, cartilage, and blood vessels). Thus, CA are more likely to have greater number in the blood vessels. - Microglia acts like macrophages and scavenge the CNS for plaques, damaged neurons, and infectious agents. They do this by 1) Secreting cytotoxins, 2) phagocytosis, 3) Antigen presentation - 3 types of cartilage: 1) articular (absorb shock), 2) elastic (shape/support), 3) fibrous (rigidity)

Which statement explains how two forms of CREB (cAMP response-element binding protein) can be generated in cells?

A: CREB mRNA transcripts with different combinations of exons are generated Different protein isoforms are synthesized from the same gene through alternative splicing, during which sections of the full transcript (both introns and exons) are spliced. Different combinations of exons can produce different protein isoforms

Patients in respiratory arrest typically require assistance via positive-pressure breathing. In contrast to negative-pressure breathing, this technique

A: Can exert excess pressure on the great vessels, reducing venous return to the heart When air is forced into the lungs, pressure in the mediastinum increases. Because fluid travels from high to low pressure, this occlude blood flow back to the heart. Positive breathing forces air into the lungs to provoke expansion.

Given passage information, it would be logical to expect VEGF dysregulation to play a role in which other disease?

A: Cancer Passage states that VEGF plays a role in pathological angiogenesis (inappropriate development of new blood vessels). This is an important aspect of the development of cancer.

An alternative method for examining the effects of fatty acids on blood flow would be to measure changes in blood pressure. If blood pressure were measured, one would predict that it would be lowest in which of the following?

A: Capillaries Compared to the heart, arteries, or arterioles, blood pressure is lower in the capillaries. Pressure will be even lower in veins. *As blow flows, it flows from high -> low pressure*

Gap junctions are most likely to be found in large numbers in which of the following tissues?

A: Cardiac muscle Gap junctions allow the flow of ions and solutes that would normally be impermeable to the membrane. It's important that cardiac muscle cells have gap junctions so that act potentials can travel through them to coordinate muscle contraction

At the end of the initial hospital stay, a few E. coli cells remained in the patient's colon, even though he was taking antibiotics. These cells were most likely present because

A: Chance mutations in a few E.coli before the treatment made these cells and their descendants antibiotic-resistant Mutations making cells drug resistant are very rare, and the few drug-resistant bacteria that do develop such mutations flourish when the nonresistant cells are killed by antibiotics. Antibiotics are unlikely to have been the source of the mutations; nor do bacteria develop "immune" reactions to antibiotics. Any resistance of the patient's own colon cells to antibiotics is irrelevant to the resistance of bacteria to antibiotic

DDT would most likely initiate cancer of cause a mutation if which of the following structures is damaged?

A: Chromosome Mutations are heritable changes in the sequences of the nucleic acid component CHROMOSOMES.

Methanol poisoning occurs when the body converts a large amount of methanol to harmful chemicals that attack the optic nerves, leading to acute blindness. Ethanol is the antidote for methanol poisoning because ethanol is preferentially processed by the body. Ethanol is most likely

A: Competitive inhibitor Competitive inhibitors bind to the active site of the enzyme, blocking the substrate from attacking. The question states that ethanol is "preferentially processed"; therefore, it's highly likely that ethanol binds to the active site where methanol otherwise would, making it a competitive inhibitor - Allosteric and noncompetitive inhibitions do not involve binding to the active site; thus, they don't explain why ethanol is "preferentially processed" in the place of methanol. - Irreversible inhibition occurs when the inhibitor interacts with the enzyme in a way that CHEMICALLY ALTERS it.

Bats, birds, and butterflies all developed wings in order to utilize the sky and all of its associated advantages. What type of evolution is involved here?

A: Convergent evolution Convergent evolution occurs when entirely separate lineages gradually appear more similar over time. Here, bats, birds, and butterflies are very distantly related, but all evolve to possess wings through different mechanisms. - Parallel evolution happens when closely related species evolve in a similar way over time - Coevolution requires that two species evolve in response to each other. - Divergent evolution: two closely related lineages gradually become more dissimilar.

If the boy is producing insufficient insulin, down regulation of which of the following hormones would help lower plasma glucose concentration?

A: Cortisol One of the functions of cortisol is to increase blood sugar. Down regulating cortisol would thus reduce blood sugar

Latrunculin, a toxin produced by marine sponges of the genus Latrunculia, acts by binding to actin monomers and preventing their polymerization. Which cell function would be most directly sabotaged by this toxin?

A: Cytokinesis Cytokinesis involves contracting ring of actin microfilaments that pinches the two daughter cells apart. Preventing actin monomers from polymerizing will inhibit the filament reorganization necessary for cytokinesis. - Centrosome polarization involves microtubules not actin. - The formation of the mitotic spindle also involves microtubules

If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which would be expected to be most similar?

A: DNA synthesis DNA sequences that are common among different species, phyla, or even kingdoms are called "conserved sequences" which tend to remain that way due to the fact that they code for a vital function common among disparate species - Protein modification is something that's unique to each phylum or even species.

Administration of S.aureus-specific antibiotics to patients in shock is meant to

A: Decrease the translation of SEB An antibiotic will act directly to kill the S. aureus present in the tissue. This will prevent bacteria from compromising the mucosal barrier and stop them from synthesizing new SEB proteins; translation is the process of protein synthesis - The antibiotic is meant to kill bacteria, not stop the binding of SEB to T cell receptors

The chemical valinomycin inserts into membranes and causes the movement of K+ into the mitochondria. If mitochondria are treated with valinomycin, the rate of ATP synthesis in the mitochondria will likely

A: Decrease, because movement of K+ into the mitochondrial compartments will disrupt proton movement into the intermembrane space Any disruption of mitochondria decreases ATP production. An influx of another positively charged ion into the compartment disrupts the electrochemical gradient responsible for the necessary flow of protons

Fever in septic shock leads to which of the following compensation mechanisms?

A: Dilation of capillary beds in the skin To compensate for the increased body temperature in a fever, skin capillaries dilate to dissipate heat through skin. Skeletal muscle activity in the form of shivering may occur, which generates heat, instead of dissipating heat.

A type of plant uses different insect pollinators depending on its height. Unfortunately, due to an increase in the use of pesticides, the medium-height pollinator has recently become extinct. What type of selection will occur in the plant species over the next several generations?

A: Disruptive selection; future generations will include short and tall plants only. Due to the extinction of the associated insect species, medium-height plants now cannot be pollinated. Over time, these plants will die out before they can reproduce, leaving only the short and tall alleles to pass to the next generation. This represents disruptive selection, as only the extreme phenotypes are being evolutionarily favored. - Directional selection favors either of the extremes, not both.

What might be a consequence of a dysfunctional synaptonemal complex?

A: During meiosis, problems might occur with recombination and synapsis The synaptonemal complex is a protein-based linkage that appears during meiosis and connects homologous chromosomes. As these chromosomes are closely situated to facilitate crossing over, issues could arise involving genetic recombination or synapsis.

A couple trying to conceive without using IVF methods would most likely have their highest chance of pregnancy

A: During the luteal phase of the menstrual cycle This is the only option where the egg would be even be accessible to the sperm. During the luteal phase, the egg would be traveling down the Fallopian tube until it reached the uterus. If it were fertilized, it would implant in the uterine wall at that point.

After a molecule has been engulfed via endocytosis, in what order does it progress through membrane-bound compartments before it's degraded

A: Early endosome, late endosome, lysosome During endocytosis, external molecules/pathogens are first engulfed in an invagination of the cell membrane known as a vesicle. These vesicles initially deliver their contents to endosomes and fuse with lysosomes for degradaton - Lysosomes function to digest toxins, proteins, and unusable cellular macromolecules. They are membrane-bound and function at low pH

Compared to WT, what is the most likely effect of the W140L substitution on the stability of the PRR-prorenin complex?

A: Elimination of a pi-stacking interaction decreases stability of the complex In W140L variant, the replacement of tryptophan (a hydrophobic, aromatic amino acid) by leucine (a hydrophobic, non-aromatic amino acid) results in a decrease in stability of the PRR-prorenin complex as indicated by the increase in Kd (dissociation). This is most likely due to the elimination of a pi-stacking interaction with the side chain of tryptophan

Which type of signaling exerts its effect on the cells that are most distant from the signal's origin?

A: Endocrine The endocrine system uses hormones which are signaling molecules traveling via the circulatory system. Therefore, they can move throughout virtually the entire body and generally act on distant targets. - Autocrine agents bind to autocrine receptors on the same cell from which they are released; they act on the nearest possible targets (themselves) - Paracrine factors are secreted into the surrounding extracellular environment. From here, they diffuse to act on nearby cells - Exocrine products (generally enzymes) are secreted through ducts rather than the bloodstream; signaling affects targets that are somewhat close

Proteins that are encapsulated in Arfl-COP derived vesicles are bound for the

A: Endoplasmic reticulum The passage states that Arf1-COP undergoes retrograde transport. Protein transcription principally occurs in the endoplasmic reticulum (ER). The proteins are then transported to the Golgi apparatus for post translational modification, processing, and packing for proper localization. Retrograde transport would reverse the motion and return the protein to the ER.

Untreated type I diabetes and unmanaged type II diabetes lead to increased risk of stroke or heart attack. Why?

A: Endothelial cells in the blood vessel lining take in an excess of glucose, which causes them to overexposes surface glycoproteins, weakening the basement membrane (I picked "increased blood glucose levels create an osmotic gradient, drawing in water; this expands blood vessels, increasing the change of a rupture") Blood vessels don't rupture from a momentary increase in glucose. The heart pumps blood through the arteries at high pressure so the vessels can't rupture under the pressure increase from a small osmotic volume increase. Nor do the ordinary increases and drops in blood glucose regulated by glucagon and insulin in a healthy person lead to constant cell rupture as they would have to just as surely if this occurred in diabetes

The G2-M checkpoint

A: Ensures that the DNA has been replicated accurately - A cell has dramatically shortened length of S phase. As a result, this cell would be prevented from passing the G2-M checkpoint, which ensures correctness in DNA replication. Instead of quickly entering mitosis, it will spend additional time in the G2 phase to correct the errors generated by the traumatic event. - G1-S checkpoint assesses whether it has enough organelles and is large enough to proceed to DNA replication

In what way does heterochromatin differ from euchromatin?

A: Euchromatin is associated with increased levels of transcription while heterochromatin is related to the downregulation of transcription. - Euchromatin is less tightly packed around nucleosomes and thus associated with higher levels of transcription. - Heterochromatin is densely-organized form of DNA that appears dark when viewed under a light microscope. Its tight structure impairs access by RNA polymerase and other enzymes.

Which of the following best explains the failure of doxycycline to inhibit human protein synthesis?

A: Eukaryotic ribosomes lack a binding site for the drug (I picked: "Eukaryotes have lysosomes, which degrade the drug") The passage tells that doxycycline works via targeting the 30S ribosome subunit whereas eukaryotes have 80S ribosomes with 60S and 40S subunits. Thus, the aminoacyl-tRNA binding site in eukaryote is different enough that doxycycline can't interfere with it. - Although eukaryotic cells have lysosomes and prokaryotic cells don't, the lysosome does not participate in drug metabolism. If the drug was degraded by lysosomes, it would not be a good antibiotic against endosymbiotic organisms

An enzyme is more effectively inhibited by uncompetitive inhibitors when: I. The substrate concentration is decreased II. The substrate concentration is increased III. The inhibitor concentration is increased

A: II and III (Picked "III only") Uncompetitive inhibitors bind their target enzymes only when the substrate is first bound to the enzyme. Since at higher substrate concentrations, the substrate-enzyme complex are more abundant, the uncompetitive inhibitor will work most effectively when the substrate concentration is the highest. Also, an increase in the inhibitor concentration results in increased enzyme binding and inhibition

According to the passage information, what is most likely to be observed in an individual whose parents both carried an allele that prevents the function of ubiquitin?

A: Examination of a random cell sample from the individual's body might reveal either maternal/paternal mitochondria, or both (I picked: "The individual would be more likely to have a chromosomal abnormality") Ubiquitin is used to mark the smaller number of paternal mitochondria for destruction in the fertilized zygote. Lacking this, reproduction of both lines of mitochondria might continue, independent of the cell and each other. Therefore, any cell in the adult body might have either type of mitochondria, resulting in a short of mitochondria mosaicism

Why is increased urination an early sign of type II diabetes onset?

A: Excess glucose cannot be entirely reabsorbed by the kidneys; as a result, the urine is hypertonic and tends to increase in volume The urine has a higher concentration of glucose, which makes it hypertonic (more concentrated with solute) to other fluids. Instead of some of the fluid being reabsorbed in the nephron, the volume is maintained and increased by the osmotic gradient with water tending to flow into the tubule lumen to balance the filtrate's high solute concentration. The greater urination volume leads to more frequent urinaton - Reabsorption of filtrate would increase blood volume and decrease urine output

The unfolded proteins that accumulate in response to ER homeostatic disruption most likely:

A: Exhibit disrupted tertiary structure due to a lack of proper covalent bond formation Completely unfolded proteins lack secondary, tertiary, and quaternary structure. Each of these levels of protein structure is dictated by: secondary structure forms due to hydrogen bonding between the amide hydrogen oaf one residue and the carbonyl oxygen of another while tertiary structure is largely the result of hydrophobic interactions and disulfide bondings.

The amount of NE released by sympathetic nerve terminals will be most strongly influenced by a change in which of the following?

A: Extracellular [Ca2+] When an AP reaches a nerve terminal, it triggers the opening of Ca+ channels in the neuronal membrane. Because the extracellular concentration of Ca+ is greater than the intracellular Ca+, Ca+ flows into the nerve terminal, triggering a series of events causing the vesicles containing NE to fuse with the plasma membrane and release NE into the synapse

Which statement describes a characteristic of FSH?

A: FSH does not require transport proteins to remain soluble in the bloodstream Peptide hormones are hydrophilic and soluble in blood. Hormones that must bind transport proteins are steroid proteins, which are lipophilic.

In human females, mitotic divisions of oogonia that leads to formation of presumptive egg cells (primary oocytes) occur between

A: Fertilization and birth only Mitotic divisions that lead to primary oocytes occur are: fertilization, birth, puberty, menopause. Primary oocytes occur prior to birth

What's the difference between filtration and secretion

A: Filtration refers to the original movement of fluid and solute into the bowman's capsule while secretion relates to the transport of specific toxins and protons into later regions of the nephron. - Filtration is extremely general with virtually any molecule smaller than protein allowed passage into the filtrate. - Filtration is accomplished by physically forcing fluid to enter the nephron (it's a passive process that relies on diffusion). In contrast, secretion uses antiporters, a type of channel participating in secondary active transport. - The presence of microvilli reminds of the small intestine where nutrient absorption is a main function; the only region of the nephron heavily involved in absorption is the PCT.

At very low CTP concentrations, kinetic data fitted to the Michaelis-Menten equation predicts that the initial rate of the CCT-catalyzed reaction is most nearly what order with respect to CTP?

A: First oder For an enzyme catalyzed reaction with a very low substrate concentration, menten equation is simplified to V=Vmax[S]/Km. where Vmax/Km is constant. Also, graph in the passage indicates that it grows in a linear fashion with CTP concentration. - Zero-order kinetics is when an increase in CTP concentrations did not lead to an increase in the reaction velocity (usually true under saturating conditions). - 2nd and 3rd order relationship at low substrate concentration would show a nonlinear relationship, which isn't the case.

Folate is a common supplement given to pregnant women to prevent neural tube defects. The most likely reason for this is: A.folate is involved in action potential transmission from mother to fetus. B. folate is directly involved in erythropoiesis. C.folate is critical for proper development of the mesoderm. D. folate plays a role in ectodermal induction

A: Folate plays a role in ectodermal induction Neural tube defects are defects in the central nervous system that is derived from the ectoderm. Folate is important for neuralation (or the induction of the ectoderm to differentiate into the nervous system). - A is false because the maternal nervous system doesn't communicate directly with the fetal nervous system. Also, if the fetus already had a nervous system, folate wouldn't help prevent neural tube defects. - C is false because the mesoderm gives rise to structures such as bone, cardiac muscle, skeletal muscle, smooth muscle, and tissues involved in the excretory and reproductive systems where folate is not crucial for this development.

Which of the following correctly describes the conversion of AMP to IMP by AMPdeaminease?

A: G < 0 and ATP could be produced A high activation energy barrier usually causes hydrolysis of a "high-energy bond" to be very slow in the absence of an enzyme catalyst. This kinetic stability is essential to the role of ATP and other compounds with these bonds. If ATP would rapidly hydrolyze in the absence of a catalyst, it couldn't serve its important roles in energy metabolism and phosphate transfer. Phosphate is removed from these molecules ONLY when the reaction is coupled via enzyme catalysis to some other reaction useful to the cell (i.e transport of an ion, phosphorylation of glucose, regulation of an enzyme by serine phosphorylation). In the passage, it's given that Keq > 100. When Keq > 1, it indicates that the free energy change for the reaction is negative; thus, the reaction is spontaneous, and energy (in the form of ATP) is not required to drive it forward. Therefore, the reaction could be used to generate ATP, a molecule used biologically to store chemical potential energy.

A stable, differentiated cell that will NOT divide again during its lifetime would most likely be found in which of the following stages of the cell cycle

A: G1 The cell cycle stage where a stable, differentiated, NONDIVIDING (diploid) cell will be most likely found in G0 or G1 where the cell remains metabolically active but is not replicating its DNA (S) or segragating its duplicated chromosomes and dividing (M).

A follow-up experiment assaying for cell cycle arrest with radiolabeled [3H]-thymine indicated that CRC157 cells transfected with pC27-53 did not incorporate [3H]-thymine during DNA synthesis. Based on this information, at what regulation point in the cell cycle does WT p53 arrest cell growth?

A: G1/S (Picked S/G2) Given that uracil, not thymine, is incorporated during transcription, the radiolabeled [3H]-thymine would only be incorporated during DNA (not RNA) synthesis. DNA synthesis only occurs during the S phase of the cell cycle. For p53 to prevent the uptake of [3H]-thymine, the protein must arrest the cell cycle prior to the S at the G1/S regulation check point.

Under normal conditions in the absence of VP infection, what is the structure of the molecule that binds and activates Rac?

A: GTP (GuanosineTriphosphate) This question is about transduction (secondary).

People who are born without sweat glands are likely to die of heat stroke in the tropics. This indicates that, under tropical conditions, the human body may

A: Gain, rather than lose, heat by radiation If people lack sweat glands, they are unable to make sweat nor to capitalize upon the evaporative cooling of sweat. They are forced to rely solely on vasodilation (radiation) for responding to elevated external temperatures.

A physician is analyzing a particular digestive molecule, but (due to an unfortunate memory lapse) forgets which molecule he is studying. If he knows only that this molecule is not initially secreted as a zymogen, he may be dealing with:

A: Gastrin Gastrin is secreted by G cells and promotes the release of hydrochloric acid by parietal cells in the stomach. Unlike many digestive substances (pepsin, trypsin, chymotrypsin), it's a hormone not an enzyme. Therefore, it's not initially present in zymogen form. - Enzymes are originally released as zymogens or proenzymes that must be cleaved or modified before their activation

H. pylori infection may cause increased proliferation of mucosal cells in the stomach. This may lead to gastric cancer if

A: Genetic mutations occur in proliferating somatic cells If H.pylori infection causes increased proliferation of mucosal cells in the stomach, this leads to gastric cancer if genetic mutations occur in proliferating somatic cells that line the stomach.

Lung capillaries are so narrow that RBCs must pass through them in single file. This feature aids respiration by

A: Giving maximum exposure of each RBC to diffusing gas Because gases enter and exit the RBC through diffusion across the RBC plasma membrane, if RBCs pass through capillaries single file, more RBC surface area is exposed, increasing the efficiency of gas exchange.

Blood sugar levels are influenced by both insulin and glucagon. All of these choices incorrectly describe a function of one of these hormones EXCEPT: A. glucagon increases gluconeogenesis in the liver. B. glucagon, a peptide hormone, decreases fatty acid breakdown in adipose tissue. C. insulin inhibits glycolysis in the liver. D. insulin increases ketone body formation in the liver.

A: Glucagon increases gluconeogenesis in the liver This question is a double negative; thus, we are looking for a correct statement. Glucagon acts to raise blood glucose levels; as such, it stimulates gluconeogenesis in liver cells so that glucose can be made from non-sugar substrates. - Glucagon stimulates, not inhibits, fatty acid breakdown in adipocytes. This increases the total amount of fuel sources available in the blood. - Ketone bodies are a last resort when energy sources are insufficient. In these, insulin is largely lacking in the blood because plasma glucose is extremely low

A synthetic "cell" contains 2.5M of glucose and 3.5M of sucrose, along with plenty of glucose and sucrose channels in its membrane. If placed in a solution of 5M glucose and 1M sucrose, in which direction will these initially diffuse?

A: Glucose will diffuse into the cell while sucrose will diffuse outward Although the total concentration of solutes is identical inside and outside the cell, glucose and sucrose each possess their own concentration gradients. Glucose will move down its gradient from the extracellular solution into the cell, and sucrose will travel in the opposite direction.

Which enzyme is used both in gluconeogenesis and glycogenolysis

A: Glucose-6-phosphatase (picked: "hexokinase") Glucose-6-phosphatase catalyzes the final step of both gluconeogenesis and glycogenolysis.

Activation of which enzyme would support the metabolism of newborn infants during the first 12 hours?

A: Glycogen phosphorylase (Picked: "glycogen phosphatase") The passage states that hepatic glycogen supports the basic supports the basic metabolism of newborn during the first 12 hours. Glycogen phosphorylase is the enzyme that catalyzes the rate-limiting step in glycogen breakdown (glycogenolysis)

Which of these is NOT a histone protein associated with human nucleosomes?

A: H5 H1, H2A, H2B, H3, and H4 histone proteins exist

BDNF is a plasma-soluble protein secreted after being manufactured in the endoplasmic reticulum. BDNF most likely

A: Has its effect through binding receptors on the external leaf of the plasma membrane A protein which is plasma soluble is one with many hydrophilic domains. Thus, it can't easily pass through the lipophilic core of the plasma membrane and must bind to external receptors and exert its effect through second-messenger systems.

A construction worker is taken to the hospital after an accident involving a nail gun. Upon examination, it is determined that a nail entered his torso at an angle that missed most vital organs and blood vessels. However, the nail lodged in his spine between two thoracic vertebrae and severed many of the frontmost nerves, though rear and middle nerves are unaffected. What symptoms will this injury likely cause for the worker?

A: He will lose mobility in parts of the body below the injury Sensory (or afferent) tract lie toward the rear (dorsal) side of the spinal cord while motor (or efferent) tracts are at near the front (ventral) and lateral sides.

A researcher is attempting to create an artificial cell membrane that retains its fluidity at extremely low temperatures. Which features should he incorporate in this membrane?

A: High sterols and unsaturated fatty acids When looking for characteristics of a membrane that increases fluidity especially in cold, unsaturated FA increase fluidity in general due to their "kinks". Cholesterol and similar molecules, which act as a type of buffer system for membrane fluidity, are slightly more complex. Large amounts of cholesterol decrease fluidity at high temperatures but increase it at low temperatures. Therefore, this researcher would need high levels of cholesterol to best accomplish his goal. - Saturated fatty acids, which stack easily due to their unkinked structures, tend to promote rigid plasma membranes.

Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge; which of the following conclusions will the researchers most likely reach?

A: Histone deacetylation generally decreases gene expression Lysine is a basic amino acid with a positively-charged side chain at physiological pH. Since DNA is negatively charged due to its phosphate backbone, the charge on lysine allows for tight histone-DNA interactions due to electrostatic attraction between the charged atoms on each molecule. However, acetylation of lysine makes the residue neutral, lessening the interactions and promoting a looser structure (loose chromatin structure is typically seen with euchromatin, the LESS DENSE, transcriptionally ACTIVE chromatin structure). In contrast, histone deacetylation restores the positive charge to the residue which allows the electrostatic attractions to return, leading to a denser chromatin structure with lowered transcription/gene expression.

To prolong submersion time, some divers will hyperventilate immediately prior to a dive. This behavior is particularly dangerous and can result in drowning because

A: Hyperventilation allows for the elimination of too much carbon dioxide, which can eliminate the urge to breath (it's not the retention of too much oxygen that eliminates the urge). Hyperventilation causes the loss of excess carbon dioxide. Extreme hyperventilation results in an arterial carbon dioxide level that is completely immeasurable. Because CO2 directly related to pH (primary mechanism behind the respiratory drive), a complete absence of carbon dioxide can eliminate the urge to breath. However, oxygen levels will continue to fall where he simply won't feel the compulsion to breath and loss consciousness due to hypoxia (oxygen deficiency). - Hyperventilation itself results in increased oxygenation because the individual is breathing more rapidly; however, in healthy individuals, oxygen levels DO NOT provide the primary drive to breathe.

An inhibition of the release of vasopressin may result in

A: Hypotension Vasopression (ADH) increases water reabsorption by inducing translocation of aquaporins into the cells of the collecting duck which increases the permeability of duct to water, which exits the nephron due to the surrounding hypertonic environment. If this is prevented, results in hypotension.

The sequence of events in the human menstrual cycle involves close interaction among which organs?

A: Hypothalamus-pituitary-ovary The hypothalamus exerts control over the pituitary hormones involved in menstruation by secreting hormone-releasing factors into the pituitary portal circulation. The gonadotropic hormones FSH and LH produced by the pituitary and the ovarian hormones estrogen and progesterone all have a role in regulating the human menstrual cycle.

Which of the following is least likely to be observed in a patient experiencing hyperventilation? A. Hypoxia B. Net exhalation of CO2 C. Increased blood pH D. Increased hemoglobin O2 affinity

A: Hypoxia During hyperventilation, there is a loss of CO2 and an increase in O2 in the blood, thus we experience hypoxia (oxygen deficiency) - During hyperventilation, CO2 is lost due to excess exhalation - Loss of CO2 corresponds with INCREASED blood pH

Fertility researchers noticed that some metaphase II-arrested oocytes contain an abnormal chromosome number. Nondisjunction involving which of the following would be likely to create this abnormality? I. Homologous chromosomes II. Ova III. Sister chromatids

A: I Nondisjunction occurs when homologous chromosomes or sister chromatids do not separate properly during meiosis or mitosis. This lack of proper separation can happen during anaphase I when a pair of homologous chromosomes fail to separate. Also, it can occur during anaphase II where sister chromatids do not separate. During oogenesis, a metaphase II-arrested oocyte already passed through meiosis I. An abnormal chromosome number, known as aneuploid, may have been caused by failure of homologous chromosomes to separate during meiosis I. - Mammalian gametes are mature egg and sperm cells. If the cell is a metaphase II-arrested oocyte, then it has yet to form the mature ovum (the final stage reached at the conclusion of meiosis). It has also failed to reach anaphase II, the stage in which sister chromatids separate.

Which of the following is NOT a type of proofreading found in humans? I. Photoreactivation II. Base pair excision III. Homologous recombination

A: I Photoreactivation is a prokaryotic process (still used in some eukaryotes) to reverse damage done by UV light; humans use a separate process. - Base excision repair is seen in both prokaryotes and eukaryotes - Homologous recombination is only seen in eukaryotes, useful in repairing double-strand breaks

Which of the following does the complementary nature of the DNA sense strand and antisense strand encompass? I. Every nucleotide base on the sense strand has a complementary base on the antisense strand. II. Every nucleotide base on the sense strand that is transcribed into RNA has a complementary base on the antisense strand which is also transcribed into RNA. III. Every codon on each strand that can be transcribed onto a parallel RNA strand can also be transcribed onto an antiparallel RNA strand.

A: I The sense and antisense strands of DNA wind around each other, with each nucleotide base having a complementary base on the opposite strand. - The sense strand is NEVER transcribed into RNA; only the antisense strand is - Transcription of DNA to RNA always takes place by transcribing DNA onto an antiparallel RNA strand, not a parallel one

Diabetes mellitus is characterized by dysregulation of the hormone insulin. Which of the following statements about insulin is/are correct? I. Insulin is produced by the beta cells of the pancreas. II. Insulin directly acts on glucose in the blood. III. Insulin is a peptide hormone.

A: I and III (I picked: "I, II, and III") Insulin is produced by the beta cells of the pancreas. Insulin is also a peptide hormone that has a quick-acting effect. - Insulin does NOT directly interact with glucose in the bloodstream; instead, it acts to reduce blood glucose levels by *promoting* the uptake of glucose into cells

While DNA and RNA are strikingly similar macromolecules, they also differ in a variety of ways. Which of the following statements regarding DNA and RNA is accurate? I. DNA contains a more stable sugar than RNA. II. RNA can catalyze biochemical reactions, but DNA cannot. III. DNA and RNA cannot base pair with each other. IV. DNA and RNA are found throughout the cell.

A: I and ii DNA contains a deoxyribose sugar while RNA is with ribose, a sugar with a free 2' hydroxyl group. This small difference makes RNA more reactive and unstable. It rapidly degrades once exiting the nucleus. Ribozymes and RNA-based enzymes are abel to catalyze biochemical reactions & virtually all other enzymes are composed of proteins (But DNA CAN'T CATALYZE). - III: DNA can hybridize with an RNA (i.e it must happen for transcription to occur) - IV: DNA is localized/found in nucleus or mitochondria <-> RNA is found throughout the cell

Oxytocin (Pitocin) is routinely administered to pregnant women during difficult labor. However, it can only be given when contractions have progressed to a certain point, or it will be ineffective. Why is this true? I. Oxytocin is normally not the initial stimulus for uterine contractions. II. Exogenous oxytocin reduces production of oxytocin by the pituitary due to negative feedback. III. Oxytocin and progesterone compete for the same receptor binding site.

A: I only Oxytocin perpetuates labor by increasing uterine elasticity leading to stimulation of uterine stretch receptors. These positively feed back to the pituitary to increase oxytocin production. However, increasing uterine elasticity has no effect if the uterus is not already in the process of contracting. - It's a positive feedback loop (exogenous oxytocin leads to further release)

The activity of a Class I transposon, also known as a retrotransposon, involves: I. a conversion of DNA to RNA. II. a conversion of RNA to DNA. III. the production of an additional copy of the transposon. IV. an end result of the same total number of transposons.

A: I, II, III Class I transposons are known as "copy-and-paste" transposons, as they involve the creation of a new copy of the transposable element. These elements first undergo transcription into RNA using RNA polymerase; as their name implies, they are then reverse transcribed back into DNA and placed in a distinct location elsewhere in the genome.

Physician use of doxycycline is in decline due to the rise of antibiotic resistance. Which of the following adaptations would render doxycycline ineffective? I. Modification of the 30S ribosomal subunit to allow aminoacyl-tRNA binding in the presence of the drug II. Expression of a pump that removes doxycycline from the cell III. Thickening of the bacterial capsule

A: I, II, and III - I is correct because doxycycline functions by competing for aminoacyl-tRNA for binding to the A site of the 30s ribosome subunit. If the ribosome can bind aminoacyl-tRNA even when the drug is bound, doxycycline is not able to stop protein synthesis. - II is correct (it is a common method of antibacterial resistance). Since doxycycline is a competitive inhibitor of the ribosome, keeping its concentration low will minimize its effect on translation. - III is true; passage indicates that Wolbachia is an endosymbiotic bacterium that doxycycline needs to able to get through the cell membrane to get in and kill its target via ribosome binding. Gram (+) and (-) bacteria have a capsule (polysaccharide layer) outside their cell envelope that protects them from insults. If the bacterium adapts to thicken this protective layer, it's unlikely that the doxycycline would continue to be effective against them

Studies of the enzyme structures involved in glycolysis reveal that the presence or absence of particular non-enzyme molecules will have strong effects on the reaction rate. Some of the molecules which would be capable of causing such effects are: I. vitamins. II. coenzymes. III. dissolved metallic ions.

A: I, II, and III Cofactors are molecules that assist enzymes with their function as catalysts of chemical reactions. Examples of cofactors include vitamins, coenzymes, and some inorganic molecules such as metallic ions

Which of the following tissues have cells that are in direct contact with the external environment or elements of the external environment? I. The lining of the reproductive tract II. The lining of the respiratory tract III. The lining of gastrointestinal tract

A: I, II, and III Tissues that are exposed to the external environments have MUCOSAL membranes. All of the tissues listed have mucosa (nasal cavity, vagina, and anus)

Which of these statements accurately identify a function of transmembrane proteins? I. They act as receptors for hormones and initiate signal transduction pathways. II. They allow for transport of charged molecules across the cell membrane. III. They are responsible for the production of the majority of the ATP synthesized in eukaryotic cells.

A: I, II, and III Transmembrane proteins, such G protein-coupled receptors, bind hormones and other ligands to activate signal transduction pathways. This activation ultimately results in a change in gene expression. Additionally, statement II is accurate, as such proteins may form channels that allow charged molecules to pass. Finally, while not embedded in the overall lipid bilayer of the cell, ATP synthase spans the inner mitochondrial membrane and is therefore a transmembrane protein.

Meiosis I and II are two parts of a process that results in the creation of diverse cellular progeny. Which of the following are differences between meiosis I and meiosis II? I. Meiosis I is a heterotypic division, while meiosis II is a homotypic division. II. Crossing over occurs only during meiosis I. III. DNA in meiosis I is highly conserved, but that in meiosis II often contains many mutations. IV. A long S phase precedes meiosis I, but no S phase comes before meiosis II.

A: I, II, and IV Meiosis I is a heterotypic (=reductional division). Parent cells begin as diploid while daughter cells formed during meiosis I are haploid. Meiosis II is a homotypic (=equational division). Crossing over also occurs during meiosis I.

Latrotoxin (LTX), a large globular protein produced by Latrodectus spiders, acts to perforate the presynaptic membrane at the axon terminal. The resulting channel is large enough to permit the free influx of calcium, as well as the passage of water and other small molecules. A Latrodectus bite would most likely result in: I. swelling of the axon terminal. II. fusion of docked vesicles with the plasma membrane. III. widespread release of glutamate within the CNS. IV. widespread release of acetylcholine within the PNS.

A: I, II, and IV Swelling results from the movement of water down its concentration gradient and into the neuron, which contains many large proteins and other solutes contributing to hypertonicity. Fusion of presynaptic vesicles is mediated by an increase in intracellular calcium, which the question mentions as a consequence of LTX action. Finally, in the peripheral nervous system, LTX will cause the widespread release of the principal motor neurotransmitter, acetylcholine. - We are told that the toxin is a large globular protein. Thus, it is unlikely to cross the blood-brain barrier, which is only permeable to small, hydrophobic molecules or those with a specific transporter. The toxic effects of LTX are most likely the result of peripheral activity

Which methods separate proteins based on their charge? I. SDS-PAGE II. Isoelectric focusing III. Ion-exchanging chromatography IV. Affinity chromatography

A: II and III (Picked I, II, and III) Isoelectric focusing separates proteins based on their isoelectric point (the pH at which the net charge of the protein is 0) and ion exchange chromatography separates proteins based on their net charge. In contrast, SDS-PAGE separates proteins based on their mass and affinity chromatography separates proteins based on their interactions with specific ligands

In a remote hospital in Asia, a patient is suspected to have dengue fever, a life-threatening disease that causes increased vascular permeability. As a result, her physician can expect to find: I. generalized edema. II. decreased heart rate. III. increased renin levels. IV. decreased blood pressure

A: I, III, and IV If our patient has increased vascular permeability, her capillaries will leak fluid into extracellular compartments. This fluid excess will show up as edema (swelling). With all that fluid leaving the circulation, the blood pressure will also drop. A drop in blood pressure will be sensed by the kidneys, which will activate the renin angiotensin-aldosterone system. This will cause fluid retention in an attempt to elevate blood pressure. - II is wrong: Recall that (cardiac output = heart rate * spoke volume). Decreased blood volume will cause the heart to compensate by increasing heart rate. Thus, our patient has tachycardia (an elevated heart rate)

Which of the following symptoms would be expected in a patient suffering from superantigen activation of T cells? I. Increased tyrosine kinase activity II. Decreased blood flow to the tissues III. Tissue hypoxia IV. Hypotension

A: I, III, and IV We are told that Sags activate T-cell receptors, relying on tyrosine kinase pathways for activation. According to the passage, super antigen activation causes massive cytokine release, resulting in vasodilation, increased capillary permeability, and decreased systemic vascular resistance, leading to a drop in blood pressure. The drop in systemic resistance will cause an increase in blood flow to the tissues. Finally, this increased flow combined with hypotension leads to hypo perfusion, a state of compromised nutrient and oxygen delivery to the tissues although the blood is circulating

Based on the passage, p16 normally acts as a(n): I. oncogene. II. tumor suppressor gene. III. transcription factor.

A: II The passage states that p16 is a cyclin-dependent kinase inhibitor. Cyclin-dependent kinases (CDKs) are responsible for promoting proliferation by allowing progression through the cell cycle. Thus, by inhibiting CDKs, p16 puts the brakes on proliferation, so it would be considered a tumor suppressor gene. - III is false; the passage states that p16 binds to other PROTEINS. Transcription factors bind to *DNA* and modulate the amount of transcript that is produced by a given gene. Since the passage says p16 binds proteins, not DNA, it is not a transcription factor

A newly-developed VDA is found to target only capillaries, effectively destroying them entirely. Which of the following cell or tissue types will directly be affected by treatment with this drug? I. Smooth muscle cells II. Endothelial cells III. Collagen-based connective tissue IV. Elastin-based connective tissue

A: II Capillaries are the smallest blood vessels in the body with walls that are composed only of a single endothelial cells. Thus, the new VDA must affect endothelial cells. - Smooth muscle is found in arteries and (to a lesser extend) veins, as well as arterioles and large venules. Capillary walls do not contain smooth muscle. - Both collagen and elastin are found in the walls of larger vessels (especially arteries), but neither is present in capillaries

In severe diabetic hyperglycemia (high blood sugar), insulin can't effectively induce the uptake of glucose by cells. Chronic hyperglycemia directly leads to the presence of which of these in the urine? I. Proteins II. Glucose III. Ketone bodies

A: II and III If cells can't take up glucose, it will remain in the blood and eventually be excreted in the urine when it builds up to the point that it can't be reabsorbed by the nephron. In hyperglycemia, the body relies on fat metabolism to generate energy, which produces ketone bodies that are also excreted in the urine. - Proteins in the urine are not the result of hyperglycemia; rather, they are damage to the glomerulus.

A hypothesis states that all mutations leading to hemophilia occur within the factor VIII gene. Which of the following pairs of inheritance patterns would fail to support this hypothesis? (#11) I. Normal sons born to hemophiliac fathers II. Normal sons born to hemophiliac mothers III. Hemophiliac sons born to normal fathers IV. Hemophiliac daughters born to normal fathers

A: II and III Neither II nor IV is consistent with the X-linked inheritance pattern observed when hemophilia is due to mutations in the factor VIII gene. Thus, these inheritance patterns would not support the hypothesis that all mutations leading to hemophilia occur within the factor VIII gene. pattern II is not consistent with X-linked inheritance because in X-linked inheritance, a mother homozygous for a recessive X-linked gene will always pass that recessive gene to her son, and a normal son cannot be produced. Pattern IV is not consistent with X-linked inheritance because a daughter inherits one X chromosome from her father; it would not be possible for a daughter that inherits a normal X-linked allele to have an X-linked recessive disorder as it requires two recessive X-linked alleles.

Which two of the following procedures served as controls in this experiment? I. Sham operation II. Pinealectomy III. Normal exposure IV. Hot exposure

A: II and III The experiment was designed to test the variables of pineal gland secretions and high environmental temperatures. The sham operation was a control for the pinealectomy as the hamsters were operated on but the pineal gland was not removed. Exposure to normal temperature was a control for the exposure to high temperature. The normal conditions are controls.

Researchers compare a DNA sequence and an identical sequence of RNA where all thymines are replaced with uracil. Which of the following describes changes that they can expect? I. The uracil base will be heavier than the thymine base because of an extra methyl group. II. The thymine base will be heavier than the uracil base because of an extra methyl group. III. The carbohydrate ring in RNA is heavier than the carbohydrate ring in DNA. IV. The carbohydrate ring in DNA is heavier than the carbohydrate ring in RNA.

A: II and III The ribose ring of RNA is heavier than the deoxyribose ring of DNA because the DNA lacks the 2' hydroxyl group that is present on the ribose of RNA

Upon physiological conditions, increased activity of succinyl-CoA synthetase will most likely result in: I. Increased levels of succinyl-CoA II. Increased levels of succinate III. Increased levels of GTP

A: II and III Increased activity of succinyl-CoA synthetase will result in greater levels of the reaction products, succinate and GTP. Succinyl-CoA is the substrate of the reaction that will likely decrease with increased succinyl-CoA function

Which of the following statements describe an obstacle researchers would face in implementing a similar transduction procedure to treat cancerous growth within patients? I. Gene incompatibility between patients with different endogenous copies of p53 II. Patient immunological response to the transduction virus III. Effective virus delivery into the entirety of the tumor

A: II and III only There are many obstacles facing successful gene therapy treatments for cancer, but gene incompatibility is not one of them. Critical genes are often conserved across species, and a wild-type copy of human p53 would certainly be genetically compatible with another human patient. (In fact, many genes are so highly conserved that human homologues can be used interchangeably with their bacterial counterparts.) For this reason, statement I is incorrect. Since statement II is in both remaining choices, we know that it must be correct. Transduction is mediated by viruses, which produce an immunological response when encountered by human cells. Finally, we turn to statement III. In order to arrest the growth of a tumor, nearly all (if not all) of the tumorous cells have must be successfully altered with the gene in question. Simply introducing the new gene to a small number of cells will be ineffective, as the remaining cancerous cells will repopulate the tumor and continue to grow uncontrollably. Thus, statement III is accurate as well.

Which of the following statements are true regarding spermatogenesis? I. Meiosis I marks the transition of a spermatogonium into a primary spermatocyte. II. In a secondary spermatocyte, sister chromatids are still paired in the same cell. III. Immediately before the second meiotic division, cells are diploid but chromosomes lack replicated copies.

A: II only Spermatogonia give rise to primary spermatocytes before meiosis even begins. Meiosis I, then, marks the division of a primary spermatocyte into two secondary daughter cells. At the end of meiosis I, which is also known as reductional division, cells are already haploid; however, chromosomes retain their identical copies in the form of attached sister chromatids.

After the depletion of hepatic glycogen in newborns, which compounds can be used as precursors to sustain the blood glucose level? I. Acetyl-CoA II. Lactate III. Oxaloacetate IV. a-Ketoglutarate

A: II, III, and IV Only lactate, oxaloacetate, and a-ketoglutarate areused as starting materials in gluconeogenesis

To determine if a small molecule acts like a LEDGIN with respect to IN, a researcher plans to incubate purified IN both with/without the small molecule and perform a Western blot to detect IN in each sample. Under which condition(s) should the gel electrophoresis step be performed? I. Denaturing II. Reducing III. Native

A: III (Picked I and II) The answer is III because the intent is to confirm that a small molecule induces the formation of integrase tetramers from integrase dimers, it's necessary to visualize the proteins in their native state. Use of a denaturing agent will disrupt the interactions between monomers. Use of a reducing agent only will disrupt any disulfide bonds

Suppose the cells in either the A culture or the B culture were taken from a Drosophila individual who possessed a gene associated with hemophilia (a sex-linked recessive blood disorder) on one and only one X chromosome. If this individual were to reproduce, what patterns of hemophilia inheritance would be manifested in offspring produced with a non-hemophiliac mate?

A: If the individual were female, male offspring would have a 50% change of manifesting hemophilia, and female offspring would never manifest hemophilia Here, the female Drosophila parent is a "carrier" of a hemophilia gene. Because hemophilia is recessive, this individual will not manifest the disorder, but can pass the hemophilia allele to offspring. Also, keep in mind that if the male parent is non-hemophiliac, that must mean he does not have a hemophilia-causing allele on his Y chromosome (because the prompt establishes that this allele is both sex-linked and appears on the X chromosome, we know that it won't ever appear on a Y chromosome). The male parent being non-hemophiliac also means that he won't have a hemophilia allele on his X chromosome (otherwise he would be hemophiliac rather than normal, since even a recessive allele predominates if it is on the X chromosome but not on the Y chromosome). Hence, male offspring might inherit from the mother either a hemophilia or non-hemophilia allele, which would determine whether he became hemophiliac. But female offspring of a carrier mother and normal father could only inherit one hemophilia allele, from the mother's X chromosome, and would have a normal X chromosome from the father; this means that every female offspring would have a normal, dominant allele and a recessive hemophiliac allele, and would not have hemophilia. This makes (A) the correct answer choice. - For a case of male, every male offspring will inherit a Y chromosome from the referenced male individual. This inherited Y chromosome will never contain a hemophilia allele, which is X-linked. The determining factor of hemophilia in males here is whether or not an X chromosome with a hemophilia allele is inherited from the mother. We know from the prompt that the mate (the mother, in this case) is non-hemophiliac, but because hemophilia is recessive, this could either mean that the mother is a carrier of a hemophilia allele or that she has no hemophilia alleles. In neither case is an X chromosome with a hemophilia allele guaranteed to be passed on to male offspring.

In rare cases, a person can be more resistant to bacterial infection due to the presence of antibodies in the small intestine that neutralize LPS. If biomedical researchers hope to use this natural immunity to develop an Ig-based therapy, which antibodies from these people should the researchers screen for?

A: IgA antibodies The question is asking which antibodies are involved in the immune response of the small intestine. IgA antibodies are present in mucosal areas such as the gut, respiratory tract, saliva, and urogenital tract - IgG antibodies provide most of the human immune response throughout the body - IgE antibodies are involved in allergies and anti-parasitic responses - IgM antibodies comprise the early immune response to pathogens before an effective IgG antibody response can be initiated

How does mitotic prophase differ from prophase II of meiosis?

A: In prophase II, half of the number of chromosomes are present in comparison to prophase. At the end of meiosis I, cells are already haploid, though they are still paired with their identical sister chromatids. For this reason, the 46 chromosomes present in a typical somatic cell have already been reduced to 23 by the time that prophase II is reached.

Unlike humans, birds use a ZW sex-determination system in which males have two Z chromosomes and females have one Z and one W chromosome. What is the most likely regulatory mechanism for the potential imbalance in gene expression between male and female birds?

A: Inactivation of one Z chromosome in males In humans, females have two of the same sex chromosome (XX), while males have only one X and one Y. However, the X chromosome is much larger and carries significantly more genes than the Y version. To balance their genetic load with that of males, one of each female's X chromosomes is inactivated. As far as we know, birds likely use the same mechanism but for the opposite gender. The male, who carries two Z chromosomes, must then inactivate one to balance his female ZW counterpart.

A hiker becomes lost and has no drinking water for 2 days. At the end of this time, which of the following changes in hormone production would be expected to be significant in this individual?

A: Increased antidiuretic hormone secretion Antidiuretic hormone (ADH/Vasopressin) secretion increases in times of stress and when the osmotic pressure in the blood stream increases as it would if the blood became more concentrated due to loss of water. - Glucocorticoid and aldosterone secretion increases with stress; thus it is unlikely that it would decrease in someone who was under the emotional/physical stress.

A patient is administered a drug which mimics the effects of the antagonist hormone of calcitonin. Which of the following is a likely direct effect of this drug?

A: Increasing osteoclast activity Calcitonin decreases plasma concentrations of Ca2+, increasing bone formation by increasing osteoblast activity and decreasing osteoclast activity. The antagonist hormone to calcitonin is parathyroid hormone, which acts to increase plasma Ca2+ concentrations by decreasing bone formation by osteoblasts and increasing bone degradation by osteoclasts.

An intravenous infusion causes a sharp rise in the serum level of albumin (the major osmoregulatory protein in the blood), which will likely cause

A: Influx of tissue fluid to the bloodstream The plasma proteins can't cross the walls of blood vessels, but water molecules can. The wall of the artery acts as a semipermeable membrane setting up conditions for osmosis. An increase in plasma albumin will upset the osmotic balance because the blood becomes hypertonic with respect to the tissue. Water will have to flow into the bloodstream to reestablish equilibrium. One of the causes of edema (increased fluid in body tissues) is a decrease in the plasma protein level, which occurs in starvation when the body is forced to use its albumin as an energy source. An increase in the plasma protein level has the opposite effect: fluid enters the bloodstream - Osmoregulation: Osmoregulation: blood plasma is mainly Na+ and Cl- (inside cell is mainly K+ and hydrogen phosphate ions); thus, blood osmolarity is predominantly determined by Na+ and Cl-. If blood osmolality too low, aldosterone reabsorbs Na+. Aldosterone reabsorbs Na+ and pee out K+. 1) High blood osmolarity -> water goes into blood -> high blood volume & 2) low blood osmolarity -> water goes into tissue -> lower blood volume

Expression of gene X has been shown to stimulate adipocyte growth and differentiation. Given this, gene X expression is likely to be increased by the effect of what hormone?

A: Insulin Insulin's anabolic effects include the stimulation of adipocyte (fat cell) growth, consistent with the anabolic needs of the body in the well-fed state.

Defects in which of the following proteins would most likely lead to loss of structural integrity in skin epithelial cells?

A: Intermediate filaments Intermediate filaments such as keratin are directly responsible for structural integrity in skin. - Microtubules are not involved in structural integrity, but in mitosis, meiosis, cell trafficking of vessels, and ciliar/flagellar motion

Which cytoskeletal component forms the majority of the outer layer of skin?

A: Intermediate filaments The outer layer of skin is made up of keratin accumulated in dead cells. Keratin is an intermediate filament that has great strength - Microfilaments and microtubules are cytoskeletal components largely used to give the cell motility and transport substances inside the cell. They do not provide a large amount of strength/rigidity - Elastin gives connective tissues elasticity; it is not found in the outer layer of skin

A congenital defect in the synthesis of the proteins cloudin and occludin is most likely to impact the function of the

A: Intestinal lining Cloudin and occludin are the proteins that form tight junctions between epithelial cells. Therefore, any organ that relies on tight junctions will be negatively affected if their synthesis is impaired. If the intestinal lining is to properly absorb nutrients while leaving behind undesired materials, free diffusion cannot be permitted between the digestive ECM and the lumen of the digestive tract. Tight junctions seal the gaps between epithelial cells in the intestine, allowing cells to selectively control what passes through the epithelium through the use of transmembrane transport proteins. Without tight junctions, nutrient concentrations would equalize due to diffusion through the epithelium, vastly decreasing the efficiency of the GI tract.

At physiological pH, what will be the principal interactions between Arg681 and Cys683?

A: Ion-dipole interactions between two chain B residues In order for arginine to participate in a salt bridge, it must be ionized, indicating that a nitrogen of the side chain guanidinium must be protonated. Cysteinine is not ionized at physiological pH but is polar. Thus, its interactions with the other residue are most likely to involve ion-dipole interactions.

Eukaryotic DNA polymerase requires a free 3'-OH group which is provided by a short RNA strand synthesized by primase enzymes. This -OH group is necessary because:

A: It binds to DNA polymerase, which otherwise cannot initiate synthesis DNA polymerase is INCAPABLE of starting initiation; instead, it is only able to elongate existing strands. Without a primer, DNA polymerase would have NO initial position to bind to.

Which of the following is most likely true of the 5'-UTR region of the FAM83H gene?

A: It is transcribed, but is typically not translated or is only partially translated (I picked: "it is cleaved from the pre-mRNA transcript as part of a post-transcriptional modification") The passage indicates that 5'-UTR is the region of mRNA that is directly upstream from the initiation codon. This region is important for the regulation of translation of a transcript.

One advantage of the degeneracy of the genetic code is that

A: It reduces the effect of point mutations because multiple codons may correspond to the same amino acid "Degeneracy" is a term for the fact that multiple distinct codons can correspond to the same amino acid. Thus, a point mutation in the "wobble" base often doesn't change the overall peptide.

Normally the immune system avoids attacking the tissue of its own body because

A: It suppresses cells specific to the body's own antigens The immune system is designed to attack foreign material in the body. It avoids attacking tissues of its own body because it suppresses cells that are specific to its own body's antigens (surface molecules that would otherwise initiate an immune response)

Hemoglobin's oxygen binding affinity is affected by the pH of its environment and the concentration of carbon dioxide. This relationship is described as the Bohr effect. Which of these conditions would increase the affinity between hemoglobin and oxygen?

A: Low carbon dioxide and high pH The Bohr effect characterizes the inverse relationship between binding affinity and plasma carbon dioxide concentration. An increase in CO2 generally correlated to an oxygen deficiency. i.e) when an individual exercises, hemoglobin must be able to release oxygen when it reaches the tissues. As a result, when the concentration of carbon dioxide increases, hemoglobin decreases its affinity for oxygen. In contrast, low CO2 concentrations facilitate an increase in binding affinity. Since CO2 in the plasma directly leads to the formation of carbonic acid, low [CO2] correlates to alkaline, or basic, blood.

Researchers selected candidate genes by identifying homozygous tracts shared by all three siblings. Along with the parental phenotype, what does this suggest regarding the inheritance pattern of DYT2?

A: It's recessive because the father was neurologically normal and it's autosomal. (I picked: "it's recessive because the father was neurologically normal and it's X-linked") The researchers identified candidate genes in part by searching the sibling's genotypes for shared homozygous regions. A relatively rare homozygous allelic variant not associated with a splice site found in all three siblings may be a homozygous gene mutation responsible for DYT2. Since the researchers focused on homozygous regions of DNA, researchers believed the inheritance pattern is recessive. Because the female siblings' father was neurologically normal, it DYT2 is a recessive condition, it must be autosomal. If the disorder were X-linked recessive, an affected daughter would need to have inherited copies of the mutant X-linked allele from both her mother and father. In such a situation, her father would show signs of DYT2 because his single X-chromosome would contain the defective allele. - Dominant conditions require only a heterozygous mutation to be displayed phenotypically

Underproduction of pulmonary surfactant in IRDS leads to decreased compliance of alveolar tissue. Based on this, which of the following must be true regarding pulmonary surfactant?

A: Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway During inspiration, contraction of the diaphragm and internal intercostal muscles leads to expansion of the thoracic cavity and a DECREASE in intra-pleural pressure. This NEGATIVE PRESSURE (relative to atmospheric pressure) at the entry of the upper airway, generates airflow through the respiratory tree and to its terminal extension, alveoli. Thus, inflation of the lungs is accomplished by negative pressure pumping action The elastic recoil force of the airway and the surface tension of the water lining the airway oppose expansion of the alveoli due to the influx of atmospheric pressure. Pulmonary surfactant adsorbs to the air-water-alveoli interface, reducing surface tension and the total force resisting expansion. This increases pulmonary compliance- a measure of lung volume change at a given pressure of inspired air and decreases the work required to expand the lungs at a given atmospheric pressure. - Increased surface tension would increase the respiratory effort required to inflate the lungs. - The efficiency of gas exchange across the alveolar membrane is primarily a function of the differences in partial pressures of gases in inspired air vs. the partial pressure of gases in pulmonary capillaries. - When surface tension exceeds the average alveolar radii, collapse of the small airways occurs - Oxygen movement from alveoli to capillary: Oxygen leaves the alveolus lumen and enters epithelial cells of the alveolus -> basement membrane -> endothelial cells -> enters lumen where it can bind a red blood cell

Which of the following cytoskeletal proteins contributes the most to the skin's resistance to stretching and tearing?

A: Keratin Keratin fibers extend across epithelial cells to link adjacent cells via structures called desmosomes. They possess high tensile strength and provide mechanical stress among the cells of an epithelial cells. Keratin contributes most to the skin's stretching ability - Collagen contributes to the tensile strength of CONNECTIVE tissue in the skin; it's a fiber in the extracellular matrix not a component of the cytoskeleton

A patient actively infected with human papillomavirus (HPV) was found to have high serum antibody levels. HPV is a non-enveloped DNA virus. Antibodies isolated from this patient would most likely show high affinity for:

A: L9, a viral capsid protein The capsid is the only choice that could possibly have been exposed to the surrounding environment before the virus entered the cell. Antibodies are involved in extracellular defense, meaning that an effective immune response would need to consist of antibodies specific for an antigen exposed prior to cellular penetration

Newly formed chylomicrons involved in the transport of intestinally absorbed lipids immediately pass from cells of the duodenum into:

A: Lacteals contained within intestinal villi Chylomicrons transport lipids absorbed from the intestine to cardiac, adipose, and skeletal muscle tissue, where lipoprotein lipase removes their triglyceride components by hydrolysis. Nascent chylomicrons are formed in enterocytes (columnar epithelial absorptive cells present in the small intestine) and released into lacteals (lymphatic vessels), originating in the small intestinal villi. - Chylomicrons won't enter the bloodstream from lymphatic circulation until reaching the junction of the thoracic duct and left subclavian vein - Lacteals are the lymphatic vessels in the S.I that allow chylomicrons to diffuse in, travel through the lymphatic system, and be deposited back into the circulatory system

Which of the following locations is expected to have the highest number of glucagon receptors?

A: Liver Glucagon is released FROM pancreas as a response to low blood glucose levels. Its main purpose is to increase glycogenolysis to increase blood glucose. Glycogen is stored in the liver; thus, most of glucagon's action occurs in the liver. - The kidney has glucagon receptors but not as many as the liver

As described in the passage, the kidney's response to alcohol consumption is most similar to its response to:

A: Low blood osmolarity and high blood pressure The question is asking us what physiological conditions would lead to the same response as what is observed during alcohol consumption. Based on the passage, we know that alcohol decreases the amount of water reabsorbed by the kidney, so it must increase the amount of water excreted in the urine. This is similar to the kidney decreasing water reabsorption, which it does in cases of high blood pressure or low plasma osmolality (our answer here). As a result of this decreased reabsorption, blood volume decreases and blood pressure becomes lower. - If blood pressure is low, blood volume is low (i.e dehydration). Thus, the kidneys would not reduce their reabsorption of water under these circumstances. - High blood osmolality means that solute concentrations in the blood are high (blood is very concentrated), provoking the body to attempt to conserve water, not excrete it

Delayed ovulation, as a cause of tubal pregnancy, would most likely be associated with delayed secretion of which of the hormones?

A: Luteinizing hormone Luteinizing hormone is the hormone responsible for triggering ovulation. The sex hormones progesterone and estrogen are secreted in response to the LH or actually trigger the LH surge; they are not directly involved in triggering ovulation.

HIV infects and kills helper T cells. What of the following cell types will function less effectively in a patient in the late stages of AIDS?

A: Macrophages, B cells, Cytotoxic T cells Helper T cells are critical for activation of the humoral immune response (b cells) and the cellular immune response (T cells). Helper T cells also produce cytokines that fully activate macrophages, allowing them to establish a more acidic phagosomal pH and kill pathogens. Loss of helper T cells compromises the entire immune response.

When human skin suffers a cut, the process of healing rapidly begins allowing for wound closure and healing within a few days. Keloids occur when skin around wounds continues to grow after the skin has healed. A disruption in the regulation of which cellular process is probably responsible for this condition?

A: Mitosis Certain tissues in the body can REGENERATE due to the ability of their cells to undergo mitosis when necessary. Skin grows by mitosis when injured.

The liver is different from many other organs in that it can at least partially regenerate following illness/damage. This regeneration is accomplished primarily through

A: Mitosis Unlike other organisms, the liver partially regenerates after illness/damage, accomplished by mitosis; mitosis is the process where human body cells reproduce.

A drug that binds to tubulin molecules of plant cells and prevents the cells from assembling spindle microtubules would most likely cause the resulting plants/plant cells to have

A: More than two sets of chromosome (I picked: "independent movement because of excess tubulin") The fibers that attach to the chromosomes in mitosis and form the mitotic spindle (spindle fibers) are microtubules, which are assemblies of tubulin proteins. Disruption of the mitotic spindle by drugs prevents the proper segregation of chromosomes into the daughter cells and results in unequal numbers being distributed to the two daughter cells (often results plant cells to have more than two sets of chromosomes). - Tubulin has only a minor role in motility, making independent movement unlikely

Which of the following is true of a typical viral capsid?

A: Most viruses have only one of two genes coding for capsid proteins Most viral capsids are composed of repeated identical monomers, reducing the need for a large genome with multiple different genes for structurally distinct capsid proteins. - The capsid is made from noncovalently-associated protein subunits - The capsid assembles spontaneously around the genetic material and sometimes requires viral enzymes for complete assembly, but never ribosomes.

Carpenter ants grow fungal gardens in their nests that provide them with nutrients and vitamins. In exchange, the ants carry leaves into their nests to feed these gardens. This relationship is most closely related to:

A: Mutualism Mutualism is a symbiotic relationship in which both parties benefit. Here, the ants feed the fungus and the fungus feeds the ants. - Commensalism involves a benefit to only one participating species; the other species is unaffected

When an odorous molecule binds to olfactory receptors, the cell transduces the information into an electrical signal that travels to the brain for processing. Which of the following accurately describes the state of the voltage-gated channels on this cell during the relative refractory period?

A: Na+ channels are de-inactivated, and K+ channels are activated (I picked: "Na+ channels are inactivated, and K+ channels are activated") The absolute refractory period lasts nearly the entire duration of an action potential, during which time a second action potential can't be generated. During this time, voltage-gated sodium channels are "inactivated" If the "inactivation gate" is closed, the channels are inactivated. If the inactivation gate is open but the activation gate is closed, the channel is "de-inactivated"; it is neither inactivated nor open. This "De-inactivation" occurs once the action potential nears its end and the membrane voltage becomes sufficiently low (generally during hyperpolarization). At this time, the inactivation gate opens and the activation gate closes. Since the channel is not inactivated, a stimulus could produce an action potential, but since the cell is hyperpolarized, this would need to be larger than normal. This is termed "relative refractory period". At this time, sodium channels are de-inactivated; potassium channels are still activated, allowing potassium to continue flowing out of the cell to finish the action potential

Capillaries in the kidney and elsewhere in the body maintain fluid homeostasis by balancing hydrostatic and osmotic pressures. Which of the following is the initial effect of a blood clot forming on the venous side of a capillary bed?

A: Net fluid flow in the direction of interstitial spaces will increase *Blood flows from arteries -> capillaries -> veins* If flow is blocked at the venous side, blood would accumulate in the capillaries. Thus, hydrostatic pressure would build up in the capillaries, causing a net increase in fluid flow into the interstitial spaces.

Although Creutzfeld-Jakob Disease is perhaps the best-known prion disease, recent hypotheses suggest that neurodegenerative diseases such as Alzheimer's disease and Parkinson's disease may have prion components. Which of the following, if true, would best refute the claim that prions contribute to the pathology of Parkinson's disease?

A: Neurons from Parkinson's patients show absolutely no increase in expression of heat shock proteins or cellular chaperones - Prions tend to induce misfolding and aggregation of endogenous cellular proteins, forming highly stable amyloid fibers. The natural cellular response to the presence of misfolded proteins is the production of heat shock proteins. The absence of heat shock activity suggests that the cell is not experiencing any problems with protein aggregation.

Is there a correlation or linkage between wing characteristics and eye color?

A: No, because each trait is sorted independently from a genetic perspective Mendel's law of independent assortment states that separate genes which encode separate traits are passed down from parents to offspring independently of each other, This is because the genetic alleles on chromosomes are sorted separately due to the separation of homologous chromosomes during the meiosis stage of the reproductive process, when gametes are created which will eventually recombine and grow into offspring. This independent assortment is a consequence of genetic structure and mixing

From SDS-PAGE, The researcher concludes that a silent mutation differentiates the two proteins (labeled A and B). Is this conclusion supported by the data shown?

A: No, this conclusion is not supported because SDS-PAGE can't distinguish proteins that differ only due to silent mutations Silent mutations DON'T change the size of structure of the resulting protein; thus, there is no way to differentiate the variants on a gel. - The most reliable way to confirm the presence of a silent mutation is by analyzing the sequence of the associated gene (DNA sequencing).

A student postulated that the sodium pump directly causes action potential along neurons. Is this hypothesis reasonable?

A: No; action potentials result in an increased permeability of the plasma membrane to sodium The massive influx of sodium through voltage gated ion channels causes the depolarization of the neuron that occurs during the action potential. - Movement of Na+ does occur down the axon, but this is independent of the sodium pump and does not cause action potentials - The sodium pump does use energy released by the hydrolysis of ATP to move sodium and potassium ions across cell membranes; however, this energy doesn't cause the influx of sodium responsible for causing action potential - 3 Na+ molecules are extruded for each ATP molecule utilized by the sodium pump

Would an increase in the level of plasma aldosterone be expected to follow ingestion of excess quantities of NaCl?

A: No; aldosterone causes Na+ reabsorption by kidney tubules, which decrease Na+ levels in the urine. Because ingestion of excessive NaCl triggers Na+ secretion into the urine, plasma-aldosterone levels wouldn't increase. Rather, the body relies on homeostatic mechanisms excreting the excess Na+ which explains why there wouldn't be ingestion of large NaCl upon increase in plasma aldosterone

A certain protein release factor functions to specifically recognize stop codons and terminate translation. How many tRNA molecules bind to the same codons as this factor?

A: None To ensure the proper termination of translation, stop codons are only recognized by PROTEIN RELEASE FACTORS. In other words, no tRNA molecules bind to these codons, and they do not correspond to any amino acid residues. If tRNAs were able to recognize such codons, the translational machinery would be able to bypass them and produce inappropriately long protein chains.

Which of these organelles or structures are paternally inherited by the fetus?

A: None of the above (mitochondria, golgi apparatus, lysosomes) Virtually all of the cellular machinery and organelles are maternally inherited. Paternal organelles are destroyed during fertilization.

If the concentration of amino acid transport protein is increased, the transport affinity of Kt of L-alanine will

A: Not change Transport affinity for a given substrate (alanine) is a property of the amino acid transport protein itself; thus, Kt won't change as a result of there being more transport protein.

What mechanism of gene transfer is most likely to be responsible for the transfer of genes from bacteria to human hosts? A. Transfection B. Transduction C. Nuclear-phage recombination D. Transposable elements

A: Nuclear-phage recombination The term "phage" gives away this answer, which would be the case if a virus acquired bacterial genetic material and transferred it to a human host. - Transfection and transduction are means of horizontal gene transfer amongst prokaryotes - Transposable elements are responsible for moving genes around in the eukaryotic genome

Progressive multifocal leukoencephalopathy (PML) is a viral disease that causes inflammation of white matter in the brain and demyelination of the neurons. Symptoms include clumsiness, visual and speech abnormalities, and, occasionally, personality changes. From this information, PML most likely triggers the death of:

A: Oligodendrocytes Oligodendrocytes produce the myelin sheath of CNS neurons. - In contrast, schwann cells form the myelin sheath in the peripheral nervous system

Cells under stress sometimes use gene amplification to increase the copies of specific transcripts. However, duplication of certain genes may result in tumorigenesis - a clonal population of cells with uncontrollable cell division. The amplification of which of the following genes would most likely contribute to tumorigenesis?

A: Oncogene Oncogenes are genes involved in cell growth and have the potential to cause cancer. Proto-oncogens don't exhibit either a greater/lesser rate of transcription in cancer patients. It's the activated form of a proto-oncogene, an oncogene, that is related to tumor growth and metastasis. (proton-oncogen -> oncogen due to mutation or over-expressed proteins) Acetylation promotes gene expression while methylation discourages it (metylation lowers its proclivity to undergo transcription)

The finches observed by Darwin on the Galapagos islands are an example of adaptive radiation. To set up conditions producing adaptive radiation, it's necessary to place members of

A: One species in several different environments (I picked: "one species in one rapidly changing environments") Adaptive radiation involves the divergence of one species into multiple species over time, which occurs when subgroups of the original species are separated or isolated in different environments so they evolve independently of one another. My answer is incorrect because the environment change would enable the species to evolve into a single new species adapted to the new environment

In the first trial of the PCR procedure, only one primer was added to the mixture. What was the most likely outcome during this trial?

A: Only one strand of the DNA was linearly amplified When performing PCR, if only one primer is added to the mixture, then that primer will bind to one of the strands and initiate replication of that strand to produce the complementary strand. For one double-stranded template DNA, after one cycle, we are left with two of the complementary strand while still having only one of the strands where the primer binds. Only the complementary strands will be replicated. As you can see here, one of the strands is being replicated linearly (1 → 2 → 3 →...n), while the other strand is not replicated at all. After 30 cycles, for example, assuming that we began with one copy of each strand, we would have 30 copies of one strand and only one copy of the other. This differs from PCR that includes two primers (which is typical), where we would expect exponential amplification (230 copies of each strand).

Which of these statements accurately describes the function of a type of bone cell?

A: Osteoclasts are responsible for the reabsorption of pre-existing bone Osteoblasts create new bone tissue by secreting a mineral-dense matrix. In contrast, osteoclasts perform bone REABSOPRTION and recycling. When stimulated, osteoclasts facilitate the removal of calcium from the bone into circulation - PTH is responsible for activating osteoclasts that remove calcium from the bone to increase plasma Ca2+ concentrations. - Osteoblasts build bone tissue while reducing plasma calcium levels

Which participant in the electron transport chain has the greatest attraction for electrons?

A: Oxygen Electrons move to a slightly more ELECTRONEGATIVE carrier as the electrons pass through each step in the electron transport chain. Thus, the final electron acceptor of this chain, oxygen, has the greatest attraction for electrons

Under certain conditions, PKA and GSK-3 have been shown to autophosphorylate. The control group used in experiment 2 (lane 1) was designed to account for this possibility. Given this, the control group most likely contained

A: PKA and GSK-3 with ATP but no CREB327 (I picked: "PKA and GSK-3 with no ATP or CREB327") The best control against the variable of enzyme autophosphorylation would be the enzymes alone without substrate. ATP must be included so that autophosphorylation would be possible. The substrate should be excluded so that it's clear that any phosphorylation detected is due to autophosphorylation and not phosphorylation of the substrate

If an artery that supplies blood to a lung lobe was blocked but ventilation to the lobe was unaffected, how would alveolar gas partial pressure change?

A: PO2 would increase while PCO2 decreases If the blood flow to an alveolus were blocked, there would be no flow of hemoglobin-rich red blood cells to take away O2 and no influx of CO2 from the blood. Therefore, the air in the alveolus would be more like the atmosphere where it requires a higher PO2 and lower PCO2

The autonomic nerve fibers that directly innervate the heart to cause cardiac slowing are

A: Parasympathetic motor fibers Sympathetic motor fibers increases heart rate. Sensory fibers carry information from the heart to the central nervous system.

Production of which of the following hormones will be inhibited by the administration of dietary calcium to prevent osteoporosis?

A: Parathyroid hormone Calcium levels in the blood need to be kept constant. Parathyroid hormone and calcitonin regulate blood levels of calcium. Calcitonin takes calcium out of the blood and into the bone while preventing the loss of calcium from bone into the blood. Parathyroid hormone is inhibited by high levels of calcium.

The passage of IgG antibodies from mother to fetus illustrates:

A: Passive immunity Passive immunity is the TRANSFER of active humoral immunity in the form of ready-made antibodies, from one individual to another - Natural immunity is immunity that's present in the individual at BIRTH, prior to exposure to a pathogen or antigen which includes intact skin, salivary enzymes, neutrophils, NK cells, and complement - The innate (non-specific) immune system includes anatomical barriers, secretory molecules, and cellular components like skin, internal epithelial cells, intestines, and as such

Postmenopausal women receiving estrogen and progesterone therapy will most likely experience which of the following side effects?

A: Periodic menstruation will resume Estrogen and progesterone are actively secreted by the ovaries of pre-menopausal women and act to maintain the uterine cycle. With advancing age, the ovary becomes less responsive to pituitary gonadotropins and cyclical changes in the endometrium of the uterus disappear. The menstrual cycle can be re-established by administration of estrogen and progesterone in a regimen that approximate the rise and fall of hormone levels in pre-menopausal women.

The cellular function of PTEN is most similar to which class of enzyme?

A: Phosphatases The passage shows that PTEN acts to inhibit the signaling of P13K. It also tells us that dephosphorylation generally inhibits signaling. *Phosphatases are responsible for dephosphorylation*. - ATPase functions to transport substances across the cell membrane using ATP hydrolysis as a source of energy - KINASE is an enzyme that phosphorylates other proteins

Of the following, which correctly describes the fluid mosaic model?

A: Plasma membranes act as two-dimensional fluids that allow for the free diffusion of proteins and lipids within the leaflet

What type of control does siRNA exert on G6P expression?

A: Post-transcriptional control Due to its structure, siRNA can only bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred; specifically, siRNA prevents the translation of mRNA corresponding to the target protein

One characteristic common to arteries, veins, and capillaries is the

A: Presence of a layer of endothelial cells All three vessels have an inner layer of endothelial cells. - Only veins have valves - Only certain types of arteries dilate/constrict to regulate blood flow - The exchange of nutrients with the surrounding tissues occurs only in capillaries.

Which type of enzyme is responsible for activating subunit A?

A: Protease (I picked "endonuclease") It's protease because the subunit A is part of a polypeptide chain that is subject to hydrolytic cleavage, which is most likely catalyzed by a protease. Endonuclease is enzyme that cleaves the phosphodiester bond within a polynucleotide chain.

Which of the following is NOT a function of the blood-brain barrier?

A: Protection of the brain from carbon dioxide poisoning - The passage states that the BBB protects the brain from harmful agents that are large or polar. Carbon dioxide is both small and nonpolar, making it highly lipid soluble and freely pass through the BBB. - Hormones are large and won't enter the brain. - Glucose is the SOLE fuel for the human brain except during prolonged starvation. The brain lacks fuel stores and requires a CONTINUOUS supply of glucose. It accounts for ~60% of the utilization of glucose by the whole body in the resting state.

The pancreas produces which of the following substances for the digestive system?

A: Proteolytic enzymes The pancreas produces several proteolytic enzymes that are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase

Skin cancer of melanocytes (melanoma) does not cause apparent symptoms beyond pigmentation. If left untreated, metastasis of the cancerous cells to other sites in the body is highly likely, and patient mortality significantly increases. Which of the following best explains the high rate of metastasis seen in melanoma?

A: Proximity of melanocytes to blood vessels Melanocytes are located between the dermis and epidermis in the skin and are found in close proximity to blood vessels. Upon induction of oncogenes, melanocytes can migrate into the blood vessels and disseminate to other sites in the body (metastasis)

Which metabolic reaction is most likely affected by treatment of cells with C75? The reaction that converts

A: Pyruvate to acetyl-CoA The passage states that mtKAS is involved in the synthesis of lipoic acid, which is inhibited by C75. Thus, treatment of cells with C75 results in lower cellular level of lipoic acid, a cofactor for the enzyme pyruvate dehydrogenase which catalyzes the conversion of pyruvate to acetyl-CoA

Based on the FSH_pep sequence, which amino acid substitution in the FSHR binding domain is most likely to have the greatest effect on reducing bone density loss in the presence of high levels of FSH?

A: R8D FSHR activation increases bone density loss. Thus, to decrease loss, the substitution should disrupt binding. The R8D substitution replaces a positively charged residue with a negatively charged residue, disrupting binding of FSH to the FSHR and stimulation of osteoclast activity.

Which of the following enzymes must be present within the viral capsid of a (-)RNA virus for successful infection and replication to take place?

A: RNA-dependent RNA polymerase Negative-sense RNA can't be directly translated by host ribosomes to produce functional viral proteins. The (-) RNA must be used as a template strand for the production of (+) RNA by a viral RNA polymerase carried within the capsid. This enzyme is RNA-dependent RNA polymerase

When the environment temperature is 33 celsius, vasodilation of cutaneous blood vessels helps to regulate the body temperature of a human by

A: Radiating excess body heat into the environment When vasodilation occurs, the walls of blood vessels relax, allowing more blood. Increased blood allows heat to escape from the surface into the environment. Vasodilation typically functions to facilitate (not to prevent) heat loss.

To be an effective therapy, an antisense gene that's incorporated into a genome that contains the target gene must be

A: Regulated in a similar manner as the target gene To provide effective therapy, this antisense gene needs to be regulated in a manner similar to the manner where the target gene is regulated so that the antisense RNA is produced at the same time the sense mRNA is produced; this would ensure that the antisense RNA is available to bind with the sense mRNA, preventing its subsequent translation. - An effective method for the deliver of an antisense gene could be infection of an embryo by a virus modified to carry the gene. For an antisense gene to work, it needs to be incorporated into the cell that performs its job so its product is available to hybridize with the sense mRNA that needs to be blocked. The appropriate virus could be incorporated into the genome of the embryonic cells, causing all cells derived from these embryonic cells to contain the antisense gene.

The production of the identified immunoglobulin would most directly involve which eukaryotic structure?

A: Ribosome Translation (production of proteins) occur at the ribosomes - Immunoglobulin is same as antibodies; Y-shaped protein used to neutralize pathogens by recognizing and binding antigen at antigen-binding site - Nucleolus is the nuclear subdomain that assembles ribosomal subunits

Which of the following sequences identifies the cellular locations of the wild-type AAT protein product after its release from the ribosome?

A: Rough endoplasmic reticulum -> golgi apparatus -> plasma membrane -> extracellular space When a protein is translated by an ER-bound ribosome, it enters the ER for processing, then is sent to the Golgi where it is packed for export through the plasma membrane

Two known competitive inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then lysozyme was also added into the reaction mixture; where would the resulting enzyme kinetics curve for this experiment fall?

A: Same as before Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH)-. Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would be affected and the curve would fall at the same line.

A drug that disrupts hydrogen bonding would most likely affect what level of protein structure?

A: Secondary Secondary structure includes the formation of alpha helices and beta-pleated sheets; both structures are defined by patterns of hydrogen bond. - Primary structure related to the linear sequence of AA held together by peptide bonds - Tertiary structure is mainly by hydrophobic interactions - Quaternary structure is stabilized by disulfide bonds and other noncovalent interactions

Increased blood viscosity is known contributor to the development of chronic hypertension. What of the following increases blood viscosity?

A: Sickle-cell anemia Sickle cells are jagged and aggregate readily, which causes increased viscosity. This is why sickle-cell patients commonly have hypertension.

Absent any mutagenic effects, homologous recombination (HR)-mediated repair is likely to have the greatest accuracy when the exchange of identical genetic information occurs between which molecules?

A: Sister chromatids during G2 Sister chromatids are two identical copies of a chromosome attached at a centromere. They are formed during the S phase of interphase during the cell cycle. Barring any mutation, exchange of identical genetic material between adjacent segments of sister chromatids shouldn't result in genetic recombination of the resultant chromatids (in contrast to homologous chromosomes which are a pair of chromosomes inherited separately from one's parents that carry potentially different copies of genes on each chromosome). Exchange of genetic material between them (either during mitosis or meiosis or during repair processes) could result in genetic recombination.

According to hypothesis A, enhanced activity of which of the following basic muscle types would be most likely to cause hypertension?

A: Smooth Enhanced activity of smooth muscles in blood vessels would cause vasoconstriction, which is a major cause of hypertension

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

A: Spermatozoon The mature ovum is the female gamete that has completed meiosis and contains the haploid number of maternally derived chromosomes, analogous to spermatozoa.

A complex system of immunological tolerance permits the developing fetus to live without eliciting a hostile response. hCG (Human chronic gonadotropin) is thought to play a role by causing uterine endometrial cells to inactivate maternal lymphocytes. Insufficient hcG results in

A: Spontaneous abortion during early pregnancy hCG is required to permit embryonic implantation and continued embryonic growth. If hCG levels are too low, the embryo is attacked by maternal lymphocytes at the site of attachment and ultimately expelled from the uterus.

In a certain population of wild coyotes, brown fur (B) is dominant over spotted fur (b) and approximately 80% of a given breeding population has brown fur. Over the course of two decades, a genetic change occurs such that nearly the entire population expresses the spotted fur phenotype. In this population:

A: Spotted fur is the wild type he term "wild type" refers to the traits an animal typically possesses when found in nature. This usually refers to a dominant trait, but not always. If nearly all of the coyotes now have spotted fur, then spotted fur is the wild type.

A newly discovered biological factor is found to bind to a cytosolic nuclear receptor after entering the cell. Based on this observation, this molecule is likely which type of hormone?

A: Steroid hormone The key is that this hormone enters the cell, meaning that it is able to diffuse through the plasma membrane. Steroid hormones, which are derived from cholesterol, are small and nonpolar. These features allow them to traverse the membrane without a channel. - Peptide hormones are large and relatively polar (explains why it's on the membrane). As such, they are unable to enter a cell through its plasma membrane. - While some tyrosine-based hormones are lipophilic, the catecholamines (epinephrine and norepinephrine) must bind to receptors on the surface of the cell membrane. - aseous hormones, such as NO, can freely pass through the plasma membrane. However, these signaling molecules are fast-acting and not strongly associated with nuclear receptors.

A lack of ATP supply to muscle cells would most likely result in which of the following observations?

A: Stiffening and reduced flexibility of muscle tissue This happens because the breaking of cross-bridges between thick and thin filaments requires fresh ATP to replace the ADP bound to the myosin ATPase enzyme; Without ATP, the cross-bridges remain intact indefinitely

Researchers discover another cytokine (X) that is upregulated even more than Z as a result of exercise. Cytokine X binds to receptors in skeletal muscle to stimulate angiogenesis. After strenuous exercise, local [X] would be LEAST similar to post-exercise [Z] in:

A: Subjects in group A After you exercise, your body repairs/replaces damaged muscle fibers by fusing muscle fibers together to form new muscle protein strands (myofibrils). These repaired myofibrils increase in thickness and number to create muscle hypertrophy. If cytosine X acts to promote angiogenesis in the new muscle tissue, this means we would expect large increases in the local concentration of X. - Cytokines are important signaling molecules (certain cytokines are produced more/up-regulated during exercise)

Cancer cells frequently exist in an immunologically privileged environment within the body. This is accomplished by production of high levels of cytokines, including transforming growth factor (TGF). Of the following, the cells that are most likely activated by TGF are

A: Suppressor T cells Activation of suppressor T cells triggers production of a cytokine cocktail that inactivates other immune cells including the cytotoxic T cells and NK cells all vital for tumor elimination. - NK cells DON'T directly attack pathogens; they destroy infected HOST cells to stop the spread of an infection. Infected host cells signal NK cells for destruction through antigen presentation/expression of receptors

The outer layers of human skin are composed of dead cells impregnated with keratin and oil, which make the epidermis relatively impermeable to water, yet humans sweat freely in hot temperatures. This occurs because

A: Sweat glands have special channels through the skin The sweat glands secrete onto the surface of the skin through channels continuous with the most superficial layer of the skin (epidermis). These channels prevent water loss by isolating the water-permeable, sweat-secreting cells from dry surface air. The openings of the sweat glands on the surface of the epidermis are pores

In an isolated cave, two bat species are discovered and found to share a distant ancestor. The ears of the two species seem to have adapted to find different sizes of prey using echolocation. When crossed, individuals of the distinct species are unable to produce a virile litter. This situation exemplifies:

A: Sympatric speciation Sympatric speciation is that which occurs without a physical barrier. A population that diverges into two separate species in a single cave certainly falls under this form of speciation. - Parapatric speciation occurs when segments of two distinct populations overlap. Due to environmental differences, these segments may develop into two species, but individuals in the overlapping areas can typically still interbreed. - Allopatric speciation occurs when populations, or parts of the same population, are separated by a physical barrier. Peripatric speciation is a subtype of this concept that occurs specifically when one of the two populations is much smaller than the other.

The diagram shows the size and position of the exons (numbers) and introns (lines) of a gene that codes for hypothetical protein X, which can exist as two isoforms (either 16 or 17 amino acid residues long): -- (24 bp) - (9bp)--- (6bp) - (18bp) -- Which technique can be used to determine if a sample of cells expresses both isoforms?

A: Synthesize cDNA from protein X mRNA using primers overlapping exons 1 and 4, followed by electrophoresis and band visualization Isoforms are produced through alternative splicing of pre-mRNA. Based upon the lengths of the two isoforms, each isoform must contain at least exon 1 and 4. From that, it can be determined that one consists of exons 1, 2, and 4 (17 residues) and the other of exons 1, 3, and 4 (16 residues). Generating cDNA from mRNA will generate a sequence for each isoform, as splicing will have occurred. Since both isoforms must contain exons 1 and 4, primers overlapping exons 1 and 4 will produce cDNA for both of the isoforms. The band pattern of the gel will identify whether both or only one of the isoforms is expressed. - In electrophoresis, "DNA ladder" is predetermined set of DNA fragments with known sizes to compare with other DNA fragments

Mitochondria-initiated apoptosis is one way the human body fights viral infection. Which immune cell is also likely to be directly involved with inducing host cell death?

A: T lymphocytes Cytotoxic T cells (a form of T lymphocytes) have proteins on their cell surface which allow them to recognize virally-infected cells. On contact with an infected cell, T cells release cytotoxic factors, which can initiate apoptosis

ω-Conotoxin binds irreversibly to calcium channels in myocytes, locking the calcium channels open and preventing calcium from being isolated. Exposure to this toxin would most likely cause:

A: Tetanic muscle contraction (I picked: "inability of the muscle to contract") High calcium concentrations allow muscle contraction to take place. Calcium binds to troponin, which causes tropomyosin to stop blocking the binding site on the thin filaments for myosin. If the calcium channels are locked open, the muscle will not be able to stop contracting. A state of constant contraction is tetany

The enzyme pepsin, which catalyzes the hydrolysis of proteins in the stomach, has a pH optimum of 1.5. Under conditions of excess stomach acidity (pH of 1 of less), pepsin catalysis occurs very slowly. The most likely reason for this is that below a pH of 10

A: The 3D structure of pepsin is changed\ Proteins are long 1D strings of amino acids. For a protein to function properly, it must have a specific 3D structure, which is stabilized by covalent and noncovalent interactions between different regions of the linear peptide. This 3D structure can be disrupted by heating/changing the pH (denaturation).

The two primary factors that normally determine the level of blood pressure are

A: The cardiac output and the resistance to blood flow Cardiac output (stroke volume * heart rate) determines the amount of blood pumped into the system by the heart per unit time.

Cyclin and cyclin-dependent kinases (Cdks) play an essential role in the regulation of the cell cycle. If a cell from a mutated line displays permanently reduced levels of both cyclin and Cdks, which consequence is most likely?

A: The cell cycle would be prevented from occurring Cyclin and Cdks are crucial cell cycle regulators. Depending on the phase of the cycle, certain Cdks will be expressed at higher concentrations than others; this signals the cell to prepare for the next transition. In other words, the levels of these proteins are dynamic throughout the cell cycle. If this cell has perpetually low concentrations of cyclin and Cdks, it lacks the signaling molecules required to initiate and prep the cell for upcoming phases.

What component can best be expected to experience the same type of heavy staining as occurs in petri dish A to the nuclei peripheries and chromosome ends?

A: The center of each chromosome pair, since chromosomal centromeres are made of heterochromatin Heterochromatin stains more heavily than euchromatin does, and centromeres are made of heterochromatin. Most histones are located on the arms (made of euchromatin) and don't stain as heavily as would heterochromatin. Supercoiling of DNA is mediated by euchromatin protein that causes sufficient euchromatin to be consolidated to be microscopically observable

Cells with the longest type C polymers are likely obtained from:

A: The cerebral cortex Type C has the largest diameter, which is characteristic of microtubules. Microtubules are known for their rigidity. As some of the longest cells in the body, neurons rely on microtubules to transport neurotransmitter-containing vesicles and macromolecules along their axons.

If there were a level 6 of myelomeningocele, what would the DQ likely be (given a linear graph with value ~82)?

A: The correlation is not strong enough to predict this with any statistical confidence An R^2 value of 1 indicates that the regression lone is a perfect fit of the data while that of 0 indicates that there is no correlation among the data. With an R^2 of 0.002 in this analysis, the correlation is virtually nonexistent so we can't have statistical confidence in a calculation

The figure below (halted growth then sigmoidal) shows a population growth curve for a bacterial colony before and after the addition of a polysaccharide. Which of the flooding most likely explains why the bacterial colony did not grow immediately after the polysaccharide was added?

A: The digestive enzyme for the polysaccharide had to be transcribed and translated (I picked "The bacteria that were unable to digest the polysaccharide died") Regulation of gene expression is one method by which bacteria respond to changes in their environment (in this case the addition of a new metabolite polysaccharide). When polysaccharide is absent, the enzymes necessary to digest it are not expressed. However upon addition of polysaccharide, in order for it to be used as an energy source, expression of the enzymes becomes necessary. Thus, the lag in colony growth represents the period in which the expression of the enzyme is occurring

A chemist looking to enhance parenteral nutrition attempts to create amino acids and sugars in the lab that are produced naturally in the body. He finds that his workups consistently produce racemic mixtures of the amino acids that the body normally synthesizes and uses in only a single enantiomeric form. The most likely reason for this distinction is that:

A: The enzymes the body uses are typically stereospecific and only produce one enantiomer (I picked: "The solvents used for workup in the lab are less selective than the enzymes in the body") Many enzymes used in body are stereospecific; they will only react with one particular stereoisomer form of a molecule and catalyze a reaction that produces a particular isomer of the product. Often drugs are administered as a racemic mixture because the workup in the lab even though only one of the isomers has any biological effect.

During most of the duration of an action potential, another stimulus, regardless of its strength, cannot cause the neuron to fire again. This is termed the absolute refractory period, and it can be attributed to:

A: The fact that sodium channels are inactivated Many voltage-gated channels have two gates (an activation and an inactivation gate). When either is closed, transmission through the channel is blocked; however, inactivation gates are generally closed after a channel has been open, coinciding with the duration of an absolute refractory period- from immediately after sodium channels have opened through the repolarization

Muscles with striated fibers are the primary muscle type in

A: The heart Skeletal and cardiac muscles contain striated muscle fibers while smooth muscles do not. The heart is made of cardiac muscle and has striated muscle fibers

The name and function of the circular tube (notochord), indicated by the arrow in the figure above, is:

A: The notochord that develops into the vertebral column The notochord provides the primitive axis of the developing embryo and, in vertebrates, develops into the vertebral column - The neural tube is above the notochord that develops into the CNS

In which organelle of a eukaryotic cell is the pyrimidine uracil, as part of uridine triphosphate (UTP), incorporated into nucleic acid?

A: The nucleus The nitrogenous base, uracil, combined with the sugar ribose and phosphate makes up the nucleotide uridine. It's found in RNA but not in DNA. RNA is manufactured in the nucleus from a DNA template.

The initial source of energy replacement in the liver of newborn infants is formed by glycosidic bonds between glucose molecules through

A: a(1->4) linkage linearly and a(1->6) linkage at branch points The glucose polymer in liver (glycogen) is formed by glycosidic bonds between glucose molecules through a(1->4) linkage linearly and a(1->6) linkage at branch point

In humans, the quadriceps muscle is a skeletal leg muscle under voluntary control. As such, it contracts in response to signals from the motor cortex in the cerebrum. If such a muscle contracts involuntarily, that indicates:

A: The operation of a reflex arc (I picked: "an appropriate modulation of Ca2+ ions in the muscle tissue") Voluntary muscles may contract involuntarily due to a reflex arc (i.e patellar tendon reflex). - If the Ca2+ ions are modulated, the muscle would simply contract whenever it's normally supposed to (whether such contractions was voluntary or not)

Due to a sudden mutation, an important enzyme gains an unusually high affinity for numerous substrates in addition to its intended one. If these new substrates bind at an allosteric site, how will this most likely affect the enzyme's original reaction?

A: The original reaction will no longer occur due to conformational changes of the enzyme. Allosteric inhibition involves the binding of a molecule at a site other than the active site. This binding causes a conformational change in the enzyme, rendering it unable to bind its original substrates. If the enzyme in the question stem is suddenly able to allosterically bind several new molecules, it is possible that this will occur

An ulcer that penetrated the wall of the intestine would allow the contents of the gastrointestinal tract to enter

A: The peritoneal cavity If an ulcer penetrated the walls of the intestine, it would allow the contents of the gastrointestinal tract to enter the peritoneal cavity. Membranes surround this cavity, further transporting the gastrointestinal contents through the rest of the body. An ulcer in the small intestine wouldn't allow the contents to enter the lumen because this is the normal place where the contents are found

Eukaryotes perform aerobic respiration in the mitochondria by pumping protons into the intermembrane space. What membrane do prokaryotes use to perform this function?

A: The plasma membrane (proton gradient is established in the plasma membrane)

A cell is perfectly normal except for a deficiency in the proper production of centromeres. Which cellular function would be most inhibited in this case?

A: The presence of attachment points on the chromosomes for microtubules during mitosis Centromeres are heterochromatin-based regions of DNA that assist in the connection of sister chromatids. Kinetochore is situated on the centromere and attaches to microtubules from the spindle apparatus. - Centrioles are protein-based structures from which microtubules can emanate during mitosis and meiosis (~ "organizational center" where the spindle apparatus can extend)

One quality of eukaryotic, but not prokaryotic, cells is A.the presence of DNA-binding proteins. B.the presence of a cell wall. C.the presence of ribosomes that are 70S in overall size. D.the presence of multiple linear chromosomes.

A: The presence of multiple linear chromosomes Only eukaryotes possess linear chromosomes in a true nucleus. For example, human somatic cells contain 46 such chromosomes. In contrast, prokaryotic cells contain circular DNA that lacks a membrane-bound nucleus

The true weight of the p53 gene is 43.7 kDa. What effect would best explain this discrepancy from its measured weight, as mentioned in the passage?

A: The protein migrated a smaller distance because it contains many positively charged residues This asks to determine a reason for the 9.3 kDa discrepancy. SDS-PAGE is used to grant a UNIFORM NEGATIVE CHARGE to all proteins in an assay. However, if the protein has enough charged residues on its own, it causes the measurements to be less accurate. If there are enough positive charge, the negative charge of the protein won't be as great as anticipated and the molecule will travel a smaller distance; the mass of the protein is greater and would lead to the discrepancy.

The tertiary structure of myoglobin is virtually identical to that of a single hemoglobin subunit. Hemoglobin and myoglobin differ in that:

A: The two molecules have different tissue distribution Myoglobin is found in the muscle while hemoglobin is found only in the blood. - Both molecules use a heme cofactor with an iron center (not just hemoglobin) - Myoglobin has a higher oxygen affinity, allowing it to draw oxygen out of the blood effectively. However, myoglobin has only one oxygen-binding site. It doesn't bind oxygen cooperatively whereas hemoglobin utilizes cooperative binding. - The binding affinity of hemoglobin for carbon monoxide (CO) is over 250 times greater than its affinity for oxygen. CO is a competitive inhibitor that binds hemoglobin directly at the heme site (i.e heavy smokers have ~20% of oxygen binding sites blocked by CO) - 2,3-bisphosphoglycerate (2,3-BPG) increases at high altitudes to enable more oxygen to be delivered to the tissues.

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, the nitrogenous bases form hydrogen bonds with each other in a 1:1 ration: guanine-cytosine and adenine-thymine. This implies that the amount of guanine and cytosine would be the same while the amount of adenine and thymine are also the same

A clinical trial for an experimental form of birth control involves injecting patients with progestin, or synthetic progesterone. If a woman receives this injection during the early part of the follicular phase, and if its effects persist for approximately three weeks, what can be expected to occur?

A: The woman won't experience ovulation at the usual time Usually, estrogen and progesterone negatively feed back on GnRH. Therefore, high levels of these lead to reduced release of LH and FSH. Because SURGE IN LH is required to stimulate ovulation, persistently high progesterone levels near the beginning of the menstrual cycle prevents this woman from ovulating.

If the neomycin resistance gene on the pC27-53 and pC27-53X plasmids was abnormally susceptible to mutation, the most likely effect on the results from experiment 1 would be that:

A: There would be a decrease in the number of colonies in all trials (I picked: "there would be an increase in the number of colonies in all trials") If the neomycin resistance gene was unusually susceptible to mutation, then there would be an increased likelihood of a mutation rendering the resistance gene non-functional. Without that resistance, fewer colonies would grow in the gentamicin solution, resulting in fewer colonies in all trials - An absence of resistance DECREASE COLONIES (a mutated non-functional resistance reduce growth in all colonies)

What ideal solution exhibits the greatest osmotic pressure? A. 0.1 M MgCl2 B. 0.2M NaCl C. 0.2M CaCl2 D. 0.5M Glucose

Answer: C The answer is C because the solution listed has the greatest concentration of solute particles. CaCl2 dissociates into three ions, making the solution listed 0.6M in solute particles

A bacterium has a faulty lac operon in which there is a structural defect in the operator. In this bacterium:

A: There's a structural problem with a segment of DNA that binds a repressor (I picked: "there will be no proteins available capable of digesting lactose") In the lac operon, the operator is the segment of DNA that binds to the repressor. In the absence of functional repressor/operator binding, the cell constitutively produce the proteins needed for lactose metabolism. Although this may not be fatal to the cell, it will waste energy if the surroundings lack lactose - The promoter is a segment of DNA that binds to RNA polymerase - The operator binds to the repressor. The repressor is coded elsewhere - If lactose is not utilized by epithelial cells, they stay in the intestinal limen and acts as a solute drawing water out of lumenal epithelial cells and cause problems like Diarrhea and gas production

In comparison with the wall of the right ventricle of the heart, the left ventricular wall is

A: Thicker and generates a higher pressure when it contracts The peak pressure in the left ventricle is 120mmHg while that of right ventricle is only 25mmhg. The right ventricle pumps blood through the lungs when the left ventricle pumps blood through the entire rest of the body. The organs through which the left ventricle pumps blood are farther away from the heart than the lungs and resistance in a tube is inversely proportional to the length of the tube, suggesting a thicker wall for the left ventricle produces greater pressure

A pharmacologist is testing the vasoconstrictive effects of caffeine on thermoregulation. He theorizes that, in cold weather, caffeine consumption can have a warming effect on the body by promoting the constriction of smooth muscle in arterioles.

A: This theory is accurate Arteries, arterioles, veins, and venules contain walls with a layer of smooth muscle. When this muscle constricts in vessels near the body's surface, blood is shunted away from the colder environment so that vasoconstriction helps to keep blood near the body's warmer core. A layer of smooth muscle in the walls allows for processes such as vasoconstriction and vasodilation. - Capillary walls are composed solely of a one-cell-thick layer of endothelium to facilitate gas exchange; thus, caffeine wouldn't promote vasoconstriction in the capillaries. - Thermoregulation: When it's hot, we use anterior hypothalamus and we use posterior hypothalamus when it's cold. Body maintains its core temperature using skin where skin acts as insulation. Vasodilation occurs when it's too hot to allow more blood flow to near the skin, allowing cooling off (heat lost). Skeletal muscles do NOTHING when it's hot. Respiratory system involved in thermoregulation as breathing causes you to lose heat by breathing out warm, moist air

How many molecules of reduced electron carrier are generated during conversion of a-ketoglutarate to oxaloacetate in the citric acid cycle?

A: Three During conversion of a-ketoglutarate to oxaloacetate in the citric acid cycle 2 molecules of NADH and one FADH2 are generated

Which of the structures listed below is found in the blood-brain barrier?

A: Tight junctions The blood-brain barrier is composed of endothelial cells with tight junctions that prevent the movement of most solutes

Which of the following reasons does NOT describe why FSH_pep was included in the ELISA experiment?

A: To determine the amount of FSH-Ab that is most effective as a therapeutic treatment The use of FSH_pep would provide no useful information about the drug's therapeutic use as the peptide does not exist in vivo; further studies using FSH would need to be performed to accomplish this

Shortly after engulfing bacteria, macrophages undergo a "respiratory burst" characterized by substantial ATP production in matter of milliseconds. This metabolic activity is so intense that macrophages usually die after engulfing 40-50 bacteria. The purpose of the respiratory burst is most likely:

A: To provide energy for the rapid acidification of phagosomes via proton pumps Bacteria must be promptly exposed to acidic conditions following phagocytosis. Otherwise they could simply replicate within the phagosome and cause lysis of the immune cells. Macrophages can acidify their phagosomes to pH ~1.5.

A cellular abnormality results in the rapid production of sense RNA that is complementary to a large portion of a certain gene. What will most likely happen to the transcription of that gene?

A: Transcription of the gene will largely be unaffected mRNA is transcribed directly from the antisense strand (template), giving it the same base sequence as the sense strand. Therefore, the production of abnormal sense RNA won't affect the antisense strand, which is the strand acting as a template for transcription

Most viral proteins are produced directly by

A: Translation of viral nucleic acid Transcription refers to taking DNA and making an RNA copy while translation refers to taking RNA and putting it into protein language. Some viruses have a genome consisting of RNA that can be directly translated by the ribosome. Others use DNA as the genetic material and require transcription. Both make proteins by translating their specific RNAs into protein

Although eukaryotes differ from prokaryotes in many significant ways, their replication, transcription, and translation mechanisms are fairly similar. However, notable distinctions can be seen. Which of these aspects of translation is uniquely prokaryotic?

A: Translation requires multiple release factors (RFs) Prokaryotic translation does rely on the presence of several release factors. In contrast, eukaryotes need only one such factor: eukaryotic translation termination factor 1 (eRF1).

A researcher is investigating the absorption of manganese ions by epithelial cells. After observation, he concluded that manganese is moved into a cell until it reaches a concentration equilibrium, at which point transport stops. Which of the following can be inferred?

A: Transport is passive and manganese uses a protein channel to pass through the membrane If transport stops when manganese reached equilibrium, the ion must not be moving against its concentration gradient, making this a form of passive transport. Since manganese is charged, it must use a protein channel to enter the cell. - Ions CAN'T diffuse through the plasma membrane

Between trisomy 21 and trisomy 14, which is likely to have the most deleterious effect and why?

A: Trisomy 14, because chromosome 14 has more genes and thus is more likely to contain a gene that is lethal at three equivalents The primary reason certain chromosomal disorders are deadly is because the genes on the chromosome that are lacked or in excess impact fetal survival. The more genes a chromosome has, the more likely that an aneuploidy of that chromosome will be fatal to the fetus. Trisomy 21 is survivable and results in Down's syndrome, while trisomy 14 is incompatible with life.

A student begins with a sequence of dsDNA. He then adds the enzymes necessary for DNA replication but substitutes artificial bases G', C', T', and A' for the normal bases. He allows the DNA to replicate once, then removes the artificial bases and replaces them with bases G'', C'', T'', and A''. Finally, the DNA is allowed to replicate two more times. At the end of the experiment, how can the DNA molecules present be described with regard to their composition? (In the notation below, the / symbol separates the two strands of the double-stranded DNA.)

A: Two GCTA / G''C''T''A'' molecules, two G'C'T'A' / G''C''T''A'' molecules, four G''C''T''A'' / G''C''T''A'' molecules DNA replicates in a semiconservative manner, meaning that each replicating strand of double-stranded DNA (dsDNA) produces two daughter molecules (each of which contains one parental and one new strand). Here, we begin with one molecule of dsDNA that is GCTA / GCTA. After the first round of replication, we have two molecules: GCTA / G'C'T'A' and GCTA / G'C'T'A'. After the next round, we now have two GCTA / G''C''T''A'' and two G'C'T'A' / G''C''T''A'' molecules. In the final round of replication, we are left with two GCTA / G''C''T''A'', two G'C'T'A' / G''C''T''A'', and four G''C''T''A'' / G''C''T''A'' molecules.

Tests performed on the M.tuberculosis strain infecting the patient's coworker indicated that the strain was susceptible to both ampicillin and kanamycin, and the coworker was successfully treated. The M.tuberculosis most likely survived in the patient because it had

A: Undergone conjugation with cells of resistant E.coli The strain of M.tuberculosis in the coworkers was killed by both ampicillin and kanamycin, indicating that this strain did not carry a plasmid gene that made it resistant to two antibiotics. However, the antibiotic didn't induce mutations. The resistant cells flourish in the absence of the nonresistant bacteria. The M.tuberculosis didn't adapt to its new environment by modifying its metabolism; rather, there was a strain with a metabolic capability that wasn't compromised by the antibiotic. It's most likely that these bacteria underwent conjugation with resistant cells.

Would the Cl- concentration of the RBCs be expected to be greater in the systemic veins or the systemic arteries?

A: Veins, because the HCO3- concentration is higher in veins than in arteries Systemic veins and arteries carry high and low carbon dioxide levels respectively. The passage states that Cl- ions enter the RBCs to equalize the HCO3- ions that leave the RBCs (and HCO3- are produced by a reaction involving carbon dioxide)

Which of the following conclusions regarding virion biology is supported by information given in the passage?

A: Virions are obligate parasites One of the characteristics of viruses is that they are obligate intracellular parasites. Thy are UNABLE to reproduce without the aid of the host cell's metabolic machinery. - Virus structure can't be visualized with a light microscope; therefore, they are smaller than all known eukaryotic cells

A single point mutation in a gene results in a nonfunctional protein. Individual heterozygous for this mutation were identified using a southern blot. Which pair of wild-type (WT) and mutant alleles most likely contains this mutation?

A: WT 5'-TAGTCGAAGCTTAGGCATCT-3' Mutant 5'-TAGTCGATGCTTAGGCATCT-3' A southern blot uses a restriction digest to different between mutant and wild-type alleles. In order for a Southern blot to be useful, the mutation should either create or eliminate a restriction site, most of which are *palindromes and 4-6 base pairs long*. The mutation shown in this option is the only one that disrupts a palindromic sequence (AAGCTT; read same backward & front), which is the recognition sequence for Hind - Palindromic sequence: A sequence made up of double helix of DNA and RNA that is the same when read from 5' -> 3' on one strand and 5' -> 3' on the other; Cleaved by restriction enzyme

The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins?

A: Western blot and RT-PCR Rapidly dividing cells undergo mitosis under the influence of specific signaling molecules; these molecules are expressed when their genes are transcribed and then are translated into proteins. To understand how a signaling protein's levels are regulated, both the protein and mRNA levels are need to be studied. Western blotting gives us information about the amount of protein expressed in a cell while RT-PCR gives us information about the amount of RNA expressed.

Under what condition would the level of calcitonin tend to increase

A: When the level of calcium in the plasma is high Calcitonin reduces bone resorption which occurs when the level of calcium in the blood plasma is low; however, resorption is not needed when the level of calcium is high. Therefore, resorption is reduced by calcitonin under conditions where the level of calcium in the plasma is high

Carbonic anhydrase catalyzes the conversion of water and carbon dioxide to carbonic acid, which is crucial for the maintenance of a buffered blood pH. Which metal cofactor is necessary for carbonic anhydrase to function?

A: Zinc Carbonic anhydrase is a metalloenzyme that catalyzes the reversible reaction between carbon dioxide and water. The active site of this enzyme contains a zinc molecule.

In normal human extracellular fluid, the concentration of Na+ is approximately 140 mEq/L. A beaker of water is filled with a solution with exactly that sodium concentration. If a water-permeable synthetic cell is dropped in the beaker, what NaCl concentration can it have and still lose water to its environment?

A: [NaCl] < 70 mEq/L For a water-permeable (not solute-permeable) cell to lose water via osmosis, its contents must be hypotonic relative to its environment (the extracellular fluid must include less solute). Here, the extracellular fluid has a sodium ion concentration of 140 mEq/L, meaning that any cell with a smaller ion concentration should leak water. However, NaCl is virtually completely soluble in water, so 70 mEq/L of NaCl will dissociate into 140 mEq/L Na + and Cl- ions.

A researcher's agar plate becomes partially contaminated with an unknown contaminant. When growing a culture of gram-negative bacteria on the plate, only the area with the contaminant is clear the next morning. The contaminant is most likely:

A: a phage in the lytic phase (I picked "B- a nutrient broth lacking an essential nutrient for the bacteria") Whatever is contaminated the plate is killing the bacteria, preventing them from growing into the contaminated region. A virus in the lytic phase will infect and kill bacteria. - B is not trie because the agar itself has the necessary nutrients since the bacteria aren't growing just where the contaminant is. Thus, adding some extra unnecessary nutrients wouldn't stop growth - In the lysogenic phase, a virus incorporates into the bacteria and allows the bacteria to grow and reproduce - Distilled water would just diffuse across the plate and is unlikely to be toxic as to prevent all growth

When viewing an X-ray of the bones of a leg, a doctor can tell if the patient is a growing child, because the X-ray shows

A: cartilaginous areas in the long bones We need to identify the characteristic that differentiates growing, developing long bones from adult bones. Long bones grow via endochondral ossification that requires cartilaginous growth plates at the ends of long bones that thicken as cartilage and become ossified. - Dividing bone cells and haversion canals can be present in fully ossified adult bones.

Compared to the T11 residue, the T3 residue of amino acid is

A: closer to the N-terminus, and that terminus is likely to be positively charged in vivo The amino acid of a protein is conventionally written from the *N-terminus to the C-terminus*, therefore, T3 residue (the third residue in the protein) must be closer to N-terminus than T11. In vivo, ("in the living"), we are talking about the conditions found inside the living human body with physiological pH ranging from 7.2 to 7.4; at this pH, the N terminus tends to exist in the (+) charged NH3+ group

The chemical gramicidin inserts into membranes and creates an artificial pathway for proton movement. Based on figure 1, if mitochondria are treated with gramicidin, the rate of ATP synthesis will most likely:

A: decrease, because the proton gradient will rapidly reach equilibrium The provision of a channel for proton flow across the membrane allows hydrogen ions to flow across the membrane until equilibrium had been achieved between the concentrations on each side of the membrane. Because ATP synthesis is driven by a flow of hydrogen ions down a concentration gradient, ATP production will decrease and eventually stop as equilibrium is established.

Unlike the cells from which human organs are composed, the cell of a unicellular organism such as algae:

A: has a genome where nearly all material codes for protein Nearly 95% of the human genome doesn't code for proteins/RNA. In contrast, the genomes of both prokaryotes and unicellular eukaryotes largely lack introns. In these, most genetic material does code for protein products.

A mother gives birth to fraternal twin girls. Both appear healthy, behave similarly, and develop normally during the first few weeks of life. During a well-baby checkup, the physician administers a biochemical test of the girls' saliva. This test reveals that one of the girls has an elevated level of a certain solute, indicating that she has inherited a recessive trait that runs in the father's family. These girls:

A: have different genotypes and phenotypes Fraternal twins are the result of the fertilization of two different ova by two different sperm. Thus fraternal twins share only 50% of their genes, making them no more closely genetically related than any two siblings. Since 50% of their genes are different, the two girls have different genotypes. One of the girls exhibits a different result in a biochemical test, meaning they must also have different phenotypes.

Pulmonary arterial blood differs from the aortic blood because it has

A: less O2, more CO2, and lower pH This is because the blood in the pulmonary artery, unlike blood in other arteries, is essentially the same as venous blood. It passes through the tissues of the body where it's given up oxygen and taken on CO2. It has been pumped through the right side of the heart from the veins and headed for the lung where it is oxygenated and dispose its CO2 load. *Venous blood is lower in pH than arterial blood due primarily to the CO2 it carries*

In comparison to a neural cell, a gastrointestinal cell would likely spend:

A: less time in G0 G0 is a non-growing state that accounts for the observed differences in length in cell cycle. Intestinal cells divide twice per day while neuronal cells DO NOT divide following initial differentiation and remain permanently in G0 - It would spend less time in S and M phase - Cell differentiation is mediated primarily by gene expression level

One characteristic of prokaryotic, but not eukaryotic, transcription is that

A: mRNA doesn't undergo post-transcriptional modification Eukaryotic mRNA is produced in the nucleus, but must be exported to ribosomes in the cytoplasm for translation. To prepare a transcript for this process, it is modified by the addition of a 5' cap and a 3' poly(A) tail; additionally, its introns are spliced out by cellular machinery. Prokaryotic mRNA transcripts are not modified in this manner.

Tuberculosis begins with colonization of the lungs by the bacterium M. tuberculosis. The lungs contain billions of alveolar macrophages that normally control the spread of airborne pathogens. M. tuberculosis lives within these macrophages, but is unable to enter alveolar epithelial cells. Consequently, a rapid and robust macrophage-mediated response accelerates the spread of the infection. In vitro, the bacterium would be most infectious to:

A: neutrophils If the bacterium can enter macrophages but not other somatic cells, it capitalizes on the natural entry mechanism offered by phagocytosis. Although T cells and endothelial cells can phagocytize under certain circumstances, neutrophils are professional phagocytes that are rapidly infected by the bacterium - Phagocyte eats pathogens; it has receptors that bind pathogen. Once bound, it's engulfed and enters the cell in a vesicle called "phagosome" which contains things that break phagosome (i.e lysosome)

Genotypic analysis of another colorectal cancer cell line, CRC200, indicates that the cancerous cells are expressing a single nonsense mutation in one of their copies of p53. Given that information, which of the following would most likely represent the number of colonies observed after an identical transfection assay?

A: pC27-53: 647, pC27-53X: 648 (Picked 16, 56) Human cells are diploid; to remove the function of a tumor suppressor protein like p53, both copies of the allele would have to be mutated (this is in contrast to mutations in oncogenes where a single gain of function mutation can lead to an overactive protein product). Despite having one mutated copy of p53, therefore, CRC200 would still be expressing WT p53 proteins and as a result transfection with an additional copy of WT p53 would not significantly alter cell growth. Of the options.

Polyadenylation, or the attachment of a repeating sequence of adenine nucleotides, can

A: prevents enzymatic degradation of the transcript - Nuclear export is done by the 5' cap

Following the LH surge, progesterone remains elevated for the latter half of the menstrual cycle. Treatment with which of the following would most likely place a pregnant female in danger of miscarriage?

A: progesterone receptor antagonist Progesterone must remain elevated to maintain the uterine endometrium and prevent contractions. Antagonizing its receptor would prevent progesterone from binding, likely leading to spontaneous abortion. - FSH remains low during the luteal phase and is unlikely to play a significant role in maintaining pregnancy

Which type of RNA is most active during post-transcriptional processing?

A: snRNA snRNAs combine with proteins to form the spliceosome, which splices introns out of the pre-mRNA transcript. The exons then are rejoined to form the mature mRNA. We use snRNA (small nuclear RNA) and snRTP (small nuclear ribonucleoproteins) to form a splicesome complex to excise the intron - a small RNA binds to an mRNA and directs degradation of the mRNA or prevents translation of the mRNA

A sample of non-coding RNA is isolated from a cell's nucleolus. The sample most likely contains

A: snoRNA Small nucleolar RNA (snoRNAs) are involved in the modification of rRNA like methylation and pseudouridylation; thus, they are located in the nucleolus where ribosomes are assembled. - snRNA is in nucleus aiding in the splicing of pre-mRNA - hnRNA is alternative name for pre-mRNA - ncRNA (non-coding RNA): 1) MicroRNA (miRNA) & Small nuclear RNA (snRNA): help regulate gene expression (both transcriptional and post-transcriptional) 2) Small nucleolar RNA (snoRNA): Covalent modifications of RNA (i.e methylation and pseudouridylation) 3) Ribosomal RNA (rRNA): Catalytic- help make up ribosome 4) Transfer RNA (tRNA): Carries amino acids to ribosome for polypeptides

In cardiac muscle fibers, action potentials display a plateau after their initial depolarization. This allows the duration of an action potential to better align with muscle contraction and to help maintain a steady heartbeat. The described effect is due to

A: the action of slow-opening, voltage-gated calcium channels. Voltage-gated Ca2+ channels allow calcium cations to flow into the cardiomyocytes, delaying repolarization of these cells. This produces a long plateau at a relatively depolarized voltage.

Of these combinations, which could result in a successful transfer of genetic material through conjugation? A. An F+ bacterium with another F+ bacterium B. An Hfr bacterium with an F- cell C. An F+ bacterium with an Hfr cell D. An F- bacterium with another F- bacterium

Answer- B: An Hfr bacterium with an F- The F (fertility) factor is a piece of genetic material coding for the construction of a sex pilus. The bridge allows a bacterium to transfer the factor to another cell, promoting genetic diversity. ONLY bacteria with the factor can initiate this transfer and ONLY to cells that lack it. Usually the F factor is held outside the main genome in the form of a small plasmid. However, in Hfr, the F plasmid has become incorporated into the organism's circular chromosome. As they do contain the F factor, they can initiate conjugate to F-. But they often transfer genomic material along with part of all of the factor

Arachidonic acid, released during AEA hydrolysis, is NOT a precursor for the synthesis of what class of molecules? A.Prostaglandins B. Catecholamines C. Thromboxanes D. Phospholipids

Answer- B: Catecholamines In the passage, it's given that AEA is degraded to ethanolamine and arachidonic acid. Ethanolamine is included as a polar head group in the phospholipid phosphatidylethanolamine while arachidonic acid is important as a precursor for the biosynthesis of the eicosanoid signaling molecules: prostaglandins, thromboxanes, and leukotrienes. Catecholamines is a class of molecules derived from tyrosine.

Which of the following molecules is/are most likely to have selective proteins in the blood-brain barrier (BBB) to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea

Answer- C: Amino acid The correct answer will be a molecule/substance essential to brain function. Amino acids are necessary for the production of proteins that are essential for the function of any cell. - In general, antibodies are too large to cross the BBB - While select monosaccharides (i.e glucose and fructose) can cross the BBB, large polysaccharides like starch are unable to cross - Urea is a waste product that is unlikely to have specialized proteins for entry into the brain

Scientists who wished to study the metabolic function of cells with balanced translocations while preventing cell replication would be best served by arresting the cells during which phase of the cell cycle? A.Anaphase B. Metaphase C. Interphase D. Prophase

Answer- C: Interphase If the scientists wanted to prevent cellular respiration, they would need to halt cell division (mitosis); interphase is the stage of the cell cycle that occurs between rounds of divisions. A,B,D are all phases of mitosis; if cells were arrested at any of these stages, they would already have begun to replicate

All of the following are true of genetic drift EXCEPT: A. it can lead to loss of alleles from the population. B. it can lead to alleles becoming fixed in a population. C. it is the result of random allele segregation from parents. D. it can increase the genetic diversity of the population.

Answer- D Genetic drift is simply the change in allele frequencies due to random processes. Specifically, random chance plays a role in determining which alleles are inherited by offspring from their parents. This can cause some alleles not to be passed down at all, leaving others "fixed" as the only alleles present for that locus. However, genetic drift does not relate to the introduction of new alleles (as mutation does), and cannot increase a population's genetic diversity.

All of the following are differences between the genetic material in a parasitic bacterium and that of its eukaryotic hosts EXCEPT: A. supercoiling of DNA. B. presence of introns. C. size. D. number of origins of replication.

Answer: A Both eukaryotic and prokaryotic DNA are supercoiled to save space. Eukaryotes have an additional level of supercoiling achieved through histones to squeeze more DNA into their nuclei. - Eukaryotic genomes and chromosomes are much larger than prokaryotic chromosomes - Bacterial chromosomes, owing to their shape and size, have only one origin of replication. Each eukaryotic chromosome has multiple origins of replication to speed up the process of copying all that DNA

Which event is NOT a likely outcome of glucagon binding to its receptor? increase in A. GDP binding Ga subunit of the G protein B. Adenylate cyclase activity C. protein kinase A activity D. cAMP generation

Answer: A Following glucagon binding to its receptor and activation of its coupled G protein, activities of the adenylate cyclase and the protein kinase A, and level of cAMP are all increased. In contrast activation of the G protein promotes the dissociation of bound GDP and its exchange for GTP on the a-subunit.

In cases of azoospermia, the lack of sperm generation most likely results directly from exogenous suppression of which hormone? A. FSH B. LH C. GnRH D. Somatotropin

Answer: A In both males and families, FSH directly stimulates the maturation of germ cells. In females, it does so by stimulating the recruitment and growth of the immature ovarian follicle. In males, FSH stimulates primary spermatocytes to proceed through meiosis. - A rapid rise in level of luteinizing hormone (LH) triggers ovulation and the development of the corpus luteum in women. In males, LH stimulates production of testosterone by Leydig cells - GnRH is a hormone released by the hypothalamus in order to regulate the release of FSH and LH from the anterior pituitary - Somatotropin is growth hormone that is responsible for cell growth and regeneration and plays a major role in maintaining the health of the brain and other organs.

Damage to which of the following cell structures would NOT cause the mitotic rates measured in the experiment to be reduced? A. Nucleolus B. Kinetochore C. Centrosome D. Aster

Answer: A The best answer will be a structure uninvolved with the direct actions of mitosis. The nucleolus is the nuclear subdomain that assembles ribosomal subunits in eukaryotes. It contains the genes for pre-ribosomal ribonucleic acid (rRNA), which serves as the foundation for nucleolar structure. The nucleolus disassembles at the beginning of mitosis and reassembly occurs during telophase and early G1 phase; thus, it's unlikely to cause a significant decrease in mitotic ability over the six hours. Asters are radial microtubule arrays found in animal cells. These star-shaped structures form around each pair of centrioles during mitosis. Asters help to manipulate chromosomes during cell division to ensure that each daughter cell has the appropriate complement of chromosomes

Which of these statements exemplifies a characteristic that makes mitochondria unique among organelles found in animal cells? A. The mitochondrion likely originated as an independent prokaryotic organism that was engulfed by an early eukaryote. B. Mitochondria have their own set of DNA that is entirely separate from the nuclear genome. C. Mitochondria are able to replicate in a fashion that is independent of cell replication. D. All of the above are correct.

Answer: All of the above are correct

All of the following are functions of cholecystokinin EXCEPT: A. suppressing hunger. B. inhibiting somatostatin. C. inhibiting gastric emptying. D. stimulating pancreatic acinar cells

Answer: B Cholecystokinin (CCK) acts in the small intestine upon the entry of food into the duodenum from the stomach. This peptide hormone functions to aid a series of processes involved in digestion. Among these are stimulating pancreatic acinar cells to release digestive enzymes, stimulating feelings of satiety (fullness) to suppress hunger, inhibiting stomach emptying, and lowering gastric acid secretion. Mainly, it contracts the gall bladder and release the bile. - Somatostatin (= growth hormone inhibiting hormone) is the hormone that inhibits the release of CCK

In mammals, which of the following events occurs during mitosis but does NOT during meiosis I? A: Synapsis B. The splitting of centromeres C. The paring of homologous chromosomes D. The breaking down of the nuclear membrane

Answer: B One of the key differences between mitosis and meiosis occurs during anaphase. During anaphase, 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

All of the following are functions performed by macrophages EXCEPT: A. antigen presentation. B. opsonization. C. phagocytosis. D. cytokine release.

Answer: B Opsonization is the process by which antibodies bind to and recognize antigens on the surface of a pathogen. The antibodies then attract macrophages to phagocytose the invader

Which type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein? A. Disulfide bond B. Phosphodiester bond C. Hydrogen bond D. Salt bridge

Answer: B Phosphodiester bonds link adjacent nucleotides in DNA; they do not contribute to the stabilization of protein structure

It has been shown that pheromones induce the body to secrete sex hormones more readily. All of the following are expected effects of sex hormones EXCEPT: A. estrogen causes the endometrial lining to thicken. B. testosterone causes the Leydig cells in the seminiferous tubules to produce sperm. C. estrogen inhibits bone resorption. D. testosterone causes the development of secondary sex characteristics in boys.

Answer: B Sperm is produced by the Sertoli cells of the seminiferous tubules, not the Leydig cells - A is true; estrogen and progesterone together cause the endometrium to thicken. Estrogen produced from the developing follicles stimulate endometrial growth, and progesterone is responsible for converting the estrogen-primed endometrium into a receptive state. - C is true; this is why post-menopausal women experience significant declines in bone density - D is true; testosterone causes secondary sex characteristics to develop during puberty in boys

All of the following will typically affect transcription or translation EXCEPT: A. a mutation in the promoter region of the gene of interest. B. a mutation in the third position of a codon of the gene of interest. C. a mutation in one exon of an allosteric regulator of the gene of interest. D. a mutation causing a frameshift in the gene's coding region.

Answer: B The third position of a codon is sometimes called the "wobble base." This name comes from the observation that mutations to this base often have no effect at all, since multiple codons code for the same amino acid. For example, the codon CUA is associated with the amino acid leucine, while CUU, CUC, and CUG all code for leucine as well. - A mutation in the promoter region will make the initiation of transcription much less likely - Mutation in allosteric regulator will likely slow/speed up transcription

Which enzyme was least likely used in cloning of WT-PRR cDNA? A. DNA polymerase B. RNA polymerase C. DNA ligase D. Reverse transcriptase

Answer: B While DNA polymerase (in DNA amplification), DNA ligase (in ligation of the cDNA to DNA) and reverse transcriptase (reverse transcription of RNA to cDNA) are used during cDNA cloning, RNA polymerase is not used in cDNA cloning (already made with post-transcription modifications).

Following puberty, the testes begin producing large amounts of testosterone. After production, the testosterone: A. is stored in secretory granules until needed to be secreted in response to declining plasma testosterone concentrations. B. diffuses into the circulatory system and is transported around the body while bound to a plasma protein. C. is sequestered in the nucleus, where it upregulates genes associated with secondary sex characteristics in males. D. is transported to target tissues, where it binds to cell surface receptors and triggers a second messenger cascade.

Answer: B (I picked C) Testosterone is a steroid hormone that can freely diffuse through cell membranes. As a hormone, testosterone is transported throughout the body in the circulatory system, not confined to the nuclei of the tests. Testoserone must bind to a transport protein since it's not hydrophilic and can't simply dissolve in the blood plasma directly. - A and D are describing characteristics of peptide hormones - As a steroid, testosterone can't be sequestered in any membrane-bound organelle

The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders? A. Cholesterol B. Glycoproteins C. Glycolipids D. Phospholipids

Answer: B- Glycoproteins Membrane transport is most likely to be affect if the disruption occurs in components spanning the membrane. Transmembrane proteins (many of them are glycoproteins) are the only component that pass all the way though the cell membrane and facilitate membrane trnasport made of unbroken stretches of hydrophobic amino acids. - Cholesterol does not facilitate membrane traffic nor extend across the membrane. - Phospholipids and glycolipids are located on the surface of cell membranes and typically do not extend through the entire bilayer. Glycolipids act to provide energy and serve as markers for cellular recognition & phospholipids are a structural component of the membrane not involved in traffic/transport.

#4. All of the following are examples of endosymbiosis EXCEPT: A. mitochondria in eukaryotic cells. B. nitrogen-fixing bacteria in the roots of legumes. C. ribosomes in bacterial cells. D. Dengue fever (a virus) in A.aegypti

Answer: C Endosymbiosis occurs when one organism lives inside another one. Ribosomes are assemblages of RNA and protein that are responsible for translation. All cells contain ribosomes; thus, they are not an example of endosymbiosis - A: It's agreed that certain organelles of the eukaryotic cell (especially mitochondria and plastids such as chloroplasts) originated as bacterial endosymbionts - D: Dengue fever is a virus that uses A.aegypti as an intermediate host and vector. Since the virus lives inside the mosquito, it's an example of endosymbiosis

Which of the following statements is NOT true with regard to the replication of the Srebp1 gene in humans by the eukaryotic DNA polymerases involved? A. The strand that is being copied in the direction of the advancing replication fork is synthesized continuously. B. The strand that is being copied in the direction away from the replication fork is synthesized discontinuously. C. DNA polymerases responsible for the elongation of the complementary DNA strand read the parental nucleotide sequence in the 5' → 3' direction only. D. Newly synthesized stretches of nucleotide chains are elongated in opposite directions relative to the position of the replication fork.

Answer: C Eukaryotic DNA polymerases ε and δ are principally responsible for the elongation of the leading and lagging strands, respectively, during DNA replication. Both polymerases read the parental nucleotide template in the 3'->5' direction, adding nucleotides to the growing strand in the 5'->3' (antiparallel) direction. - A is true; synthesis of the leading strand (copied in the direction of the advancing replication fork) by DNA polymerase advances continuously. The lagging strand is synthesized discontinuously via the addition of short RNA primers by the primase activity of DNA polymerase α (Answer choice B). D is true because the leading and lagging strands are elongated in opposite directions.

Which of the following would be LEAST useful in cellular movement? A. Flagella B. Actin polymerization C. Microtubule depolymerization D. Cilia

Answer: C For cells to travel, they need to migrate. Microtubule DEpolymerization is responsible for separating chromosomes during anaphase of mitosis/meiosis. - Flagella and cilia are structures that allow cells to propel themselves or nutrients in environment. Bacterial flagella made of flagellin (powered by proton/sodium gradient) <-> eukaryotic flagella made of microtubules (powered by ATP) - Rapid actin polymerization at the edge of the cellular membrane is responsible for cellular motility in complex EUKARYOTIC cells

A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT: A. Electrons are passed from carriers with lower reduction potential to those with higher reduction potential. B. The first electron carrier is also a proton pump. C. All electron carriers are mobile and hydrophobic. D. The electron carriers can transport a maximum of 2 electrons.

Answer: C In the ETC, carriers travel inside the inner mitochondrial membrane, passing electrons from one to another and pumping protons across the inner mitochondrial membrane. They are mobile! Cytochrome c is a highly water-soluble proteins. - A: Electrons are passed from carriers with lower reduction potential to those with higher reduction potential - D: Each member of the electron transport chain can carry one to two electrons at a time

An experimental drug designed to prevent capsular contracture around medical implants works by sequestering a cofactor required for collagen synthesis. Which cofactor is the most likely target of this drug? A. Vitamin A B. Vitamin B1 C. Vitamin C D. Vitamin D

Answer: C Of all choices, only vitamin C acts as a cofactor in collagen synthesis. It's required for the hydroxylation of the collagen precursor. A deficiency of vitamin C can result in impaired collagen synthesis and the development of the disease known as scurvy - Vitamin A refers to a group of unsaturated compounds including retinal, retinol, retinoid acid, and beta-carotene. It's involved in proper immune function, normal growth and development (especially of epithelial cells), and vision - Vitamin B1 (Thiamine) is one of the B complex vitamins. Its phosphate derivatives including thiamine pyrophosphate (TPP) act as coenzymes in the catabolism of sugars and amino acids. Its deficiency can lead to memory dysfunction (Korsakoff's), visual disturbance (optic neuropathy), and cardiovascular/neurological problems (beriberi). - Vitamin D refers to a group of fat-soluble steroid hormones responsible for enhancing intestinal absorption of calcium and phosphate. Synthesis of vitamin D in the skin is the principal natural source of this vitamin and depends on exposure to UVB radiation, most typically from sun exposure. Dietary or dermally-synthesized vitamin D must be converted to its active form, calcitriol, by sequential hydroxylation in the liver and kidneys. Deficiency results in impaired bone mineralization and bone damage.

Lack of bile acids could lead to deficiency in which of the following? A. Cholesterol B. Glucose C. Vitamin K D. Glycine

Answer: C This question is asking us to determine which of the four options is fat-soluble and taken up from the small intestine with the help of bile. We know that vitamin K (along with A, D, and E) is a fat-soluble vitamin and its absorption thus depends on fat's absorption from the small intestine.

A student researcher plans to study the effects of removing the endoplasmic reticulum from a cell. He discusses this idea with the professor overseeing his research, who informs him that the cells would almost immediately die. Why might the professor believe this? A. Without the ER, the cell would be unable to synthesize a variety of essential proteins. B. After removal of the ER, the cell would be unable to convert cholesterol into a vital steroid hormone precursor. C. Removal of the ER might disrupt the outer leaflet of the nuclear envelope, potentially releasing the contents of the nucleus into the cell and triggering apoptosis. D. The ER is directly involved in the aerobic production of ATP, so its removal would immediately render the cell energy-deficient.

Answer: C (Picked A) The membrane of the endoplasmic reticulum is an extension of the outer leaflet of the nuclear envelope. Its removal would greatly diminish overall nuclear integrity (causing that organelle to lyse), triggering apoptosis. - A is false. ER does function mainly in the production of proteins. However, the question references NEAR-immediate cell death. Even if the synthesis of vital proteins were to cease, this would cause a gradual death than the professor implies.

Which of the following is NOT considered an organic acid? A. Folic acid B. Carbonic acid C. Ascorbic acid D. Citric acid

Answer: Carbonic acid An organic compound must contain a COVALENT bond between a carbon and hydrogen. Organic acids are WEAK acids that generally have formulas of RCO2H with the acidic hydrogen bonded to an oxygen. Carbonic acid does not have a hydrogen atom bound to CARBON; thus it is not organic

Of the following tissues, which is NOT derived from embryonic mesoderm? A. Circulatory B. Bone C. Dermal D. Nerve

Answer: D Nervous tissue arises developmentally from ectoderm

All of the following are glia in the CNS EXCEPT: A. oligodendrocytes. B. astrocytes. C. ependymal cells. D. Schwann cells.

Answer: D Schwann cells make and maintain the myelin sheaths of peripheral nerves; they exist in the CNS. Oligodendrocytes form and maintain myelin in the CNS while astrocytes are a key component of the integrity of the blood-brain barrier. Astrocytes are star-shaped glial cell (surround neuron and support for insulation) that's most common in the CNS (especially in brain); they have structural, metabolic, regulatory, and repair functions. Ependymal cells make cerebrospinal fluid. Ependymal cells line the cerebrospinal fluid (fluid cushioning the brain and spinal cord that are kept separate from blood and lymph fluid) with main functions of 1) forming barrier and 2) secreting CSF.

Hepatic scarring that develops as a result of cirrhosis can lead to progressive loss of liver function. All of the following processes may be impaired due to cirrhosis EXCEPT: A. cholesterol production. B. blood clotting. C. bilirubin excretion. D. digestive enzyme synthesis.

Answer: D Since the question asks about liver complications, we deduce that the disease interfere with hepatic cholesterol and clotting factor synthesis as well as bilirubin (the breakdown product of heme) conjugation and excretion. NO digestive enzymes are synthesized in the liver. Bile, which is synthesized in the liver and is involved in lipid digestion, is an emulsifier not a digestive enzyme.

Sickle-cell disease is a rare blood disorder inherited in an autosomal recessive pattern. Which of the following is true about this condition? A. All affected children always have at least one affected parent. B. The condition exhibits a vertical pattern of inheritance. C. Half of the children of affected parents will be affected. D. If both parents are carriers, the probability of having an affected child is 0.25.

Answer: D The probability that a carrier parent passes along the mutated allele is 1/2. Therefore, the probability that both carrier parents pass along their mutated alleles is 1/2 * 1/2 = 1/4 - A is false; since the trait is recessive, an affected individual may have two parents who are carriers. In the event that both parents pass along their mutated allele to the offspring, the child will be affected. - B is false; vertical patterns of inheritance are typical in dominant traits where the phenotype is very like to show up in every generation. A vertical pattern of inheritance for a recessive trait is possible if the trait is common in the population. However, the question states that the disease is rare. Thus, it's reasonable to conclude that the pattern of inheritance is horizontal - C is false, affected parents should have two copies of the mutated allele. Therefore, all their children will inevitably be affected.

Viruses are unique from most biological organisms in that they: A. are able to reproduce inside a host. B. only use RNA as genetic material. C. are obligate parasites. D. are pathogenic entities which have been described as not being "living."

Answer: D Viruses are unique because they occupy a gray area between living and non-living - A: Other organisms (i.e bacteria and fungi) are able to reproduce inside other organisms - B: Some viruses use DNA as genetic material - C: Other infectious agents are obligate parasites (i.e tapeworms)

Which of the following compounds is NOT a gluconeogenic precursor or substrate? A. Lactate B. Glycerol C. Oxaloacetate D. Phosphogluconate

Answer: D (I picked "B") Phosphogluconate is involved in the pentose pathway and is not a precursor/substrate in gluconeogenesis.

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

Answer: D (Picked C) 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

Which of the following is NOT a function of the Golgi apparatus A.Packaging of proteins in vesicles for transport to other parts of the cell B. Targeting of proteins for excretion C.Production of lysosomes D.Production of the enzymes found within peroxisomes

Answer: D- Production of the enzymes found within peroxisomes The golgi apparatus functions like a "post ofice". It receives proteins from the ER and packages them into vesicles; these membrane-bound sacs can then travel to specific locations within the cell or to the plasma membrane for excretion. Lysosomes are formed from vesicles that bud off the Golgi'a larger structure. - Since nearly all enzymes are protein-based, it is the ribosomes- whether free-floating or attached to the RER that synthesize them

How do the chemical modifications described int he passage differ from eukaryotic post transcriptional modifications? A.Post-transcriptional modifications are made to mRNA, whereas post-translational modifications only affect enzymes. B. Post-transcriptional modifications are not catalyzed by enzymes. C. Post-transcriptional modifications do not involve the addition or removal of parts of the substrate. D. Post-transcriptional modifications are carried out entirely within the nucleus

Answer: D- post-transcriptional modifications are carried out entirely within the nucleus Protein modifications must be post-translational modifications that can take place in a variety of locations within the cell like ER and cytoplasm. In contrast, post-transcriptional modifications occur in the nucleus. A- post-translational modifications can be made to enzymes, but they can also be made to proteins that lack enzymatic function. B- post-transcriptional do involve catalysis by enzymes (i.e poly(A) tail is catalyzed by a poly(A) polymerase) C- This is false; both the poly(A) tail and the 5' cap require addition to the original pre-mRNA substrate and splicing involves the removal of introns

In an adult, which of the following cell types is LEAST likely to enter a programmed G0 phase of the cell cycle? A. Liver cells B. Kidney cells C. Epithelial cells D. Neurons

Answer: Epithelial cells G0 phase is the state that a cell will enter if it DOESN'T NEED TO DIVIDE. Epithelial cells are those that divide the most. - Liver and kidney cells don't divide as much - Neurons in adults DO NOT divide and almost ALWAYS in G0

Of these pairings, which accurately group(s) an organelle with one of its functions? I. Rough ER - synthesis of transmembrane proteins II. Nucleolus - rRNA production and ribosome assembly III. Peroxisome - breakdown of proteins

Answer: I and II The rough ER is dotted with bound ribosomes that serve to produce a wide variety of proteins including transmembrane proteins. Nucleolus is vital for the synthesis of ribosomes from protein and rRNA. - Peroxisomes are small membrane-bound organelles that function mainly to break down LIPIDS <-> lysosomes facilitate the enzymatic catabolism of protein

Which of these components is / are involved in prokaryotic transcription and its regulation? I. Repressor proteins II. Promoter DNA sequences III. Activator proteins IV. Enhancer DNA sequences

Answer: I, II, III Prokaryotic and eukaryotic transcriptional regulation requires transcription factors, including repressor and activator proteins. Repressors inhibit gene expression by blocking specific regions of the DNA or by preventing RNA polymerase from attaching to the promoter. In contrast, activators upregulate transcription by binding to either operators (in prokaryotes) or enhancers (in eukaryotes). Finally, promoters are essential in both types of organisms, since they function as the DNA sequences to which RNA polymerase initially binds. Enhancers are regulatory DNA sequences found only in eukaryotes. In prokaryotes, the operator serves as the regulatory DNA region instead.

Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration B. Low blood potassium C. Low blood sodium D. Hypotension

Answer: Low blood potassium Aldosterone increases water and sodium reabsorption from the kidney while EXCHANGING K+ ions for Na+ ions - Hypersecretion of aldosterone negatively feedback and inhibit renin production, results in high blood sodium and pressure

When an individual quickly moves from a warm to a cold environment, the body will cope in all of the following ways EXCEPT: A.sweat glands will decrease, or cease entirely, their secretion of fluid onto the exterior of the skin. B.adipose tissue in the hypodermal layer will serve to insulate the body from the cold ambient air. C.capillaries in the skin will constrict to reduce the flow of blood near the cold environment. D.arterioles leading to the skin will narrow to keep more blood near the body's warmer core.

Answer: capillaries in the skin will constrict to reduce the flow of blood near the cold environment. Constriction is performed by arteries and arterioles not capillaries. Capillary walls consist only of a single endothelial layer. Since they lack smooth muscle entirely, they cannot possibly constrict or expand.

Assuming that the vertebrates were all of comparable size, which of the following vertebrates would be expected to have the strongest and heaviest bones? A. land-dwelling mammal B. Water-dwelling mammal C. Flying bird D. Amphibian

Answer: land-dwelling mammal Of the animals listed, a land-dwelling mammal has the heaviest bones because enhanced bone density would be required to withstand the load bearing activity that results from the impact of gravity on land-dwelling animals. Less dense bones would be present in water-dwelling creatures and amphibians (impact of gravity is ameliorated by life in an aquatic environment) and flying object.

Prokaryotic cell

Bacteria do not have spindles and asters that make up the eukaryotic mitotic apparatus. Instead, the prokaryotic cytoskeleton helps pull the replicated DNA apart - Regulation at the translation level: some mRNA gets translated more. In prokaryotes, mRNAs with better Shine-Dalgarno sequence are translated more. In eukaryotes, post-transcription modifications. Therefore, prokaryotes regulate gene expression predominantly at the transcription level. Transcription-Translation coupling: in prokaryotes, translation occurs as the mRNA is being transcribes (NO RNA processing in prokaryotes) In a coupled transcription-translation system, regulation by attenuation can occur for the Trp gene 1) When cell is full of Trp, translation occurs fast because of abundant Trp amino acid; this fast ribosome movement across the transcribing mRNA causes the Trp mRNA transcription to terminate as Trp is not needed 2) When starved of Trp, transcription occurs slow, and this slow ribosome movement across the transcribing mRNA causes the Trp mRNA to be made to its completion

Red blood cells (erythrocytes)

DON'T contain DNA or a membrane-bound nucleus , which allows them to have more space for oxygen-carrying hemoglobin molecules. - Under normal conditions, hemoglobin is carried in RBC and is NOT found in the plasma; thus, the plasma hemoglobin can work as an operational measure of the extent of hemolysis (rupture of RBC) i.e) A typical eosinophil would contain a membrane-bound nucleus while an erythrocyte doesn't

In prokaryotes, genes exists as operons transcribed into a polycistronic mRNA; in eukaryotes, transcripts exist only as monocistronic mRNA. What is responsible for this distinction?

In eukaryotes, each gene has their own transcription initiation site - Operon is a segment of DNA that consists of a regulatory and coding section

Insulin

In general, insulin causes the body to build up large molecules to store energy (glycogen, lipids) and stop the body from breaking down large molecules to provide energy; therefore, it's responsible for: 1) increased glycogen synthesis (store sugar), 2) increased lipid synthesis, 3) increased esterification of fatty acid, and 4) decreased gluconeogenesis

Which of the following choices accurately identifies a property of deoxyribonucleic acid (DNA)?

I. DNA carries the instructions necessary for all cellular functions II. DNA can be found in a supercoiled structure when it's not being actively transcribed or replicated

Arrange the following in increasing order of compactness. I. Nucleosome II. Heterochromatin III. Euchromatin IV. DNA helix

IV, III, II, I A DNA helix is the least compact structure given. Nucleosomes form the fundamental repeating units of eukaryotic chromatin, meaning that they are more compact than either chromatin structures. Chromatin is primarily the combination of DNA and histones, the proteins responsible for packaging DNA into smaller volumes so that they can fit inside the cell. There are two varieties of chromatin where heterochromatin is formed from compactly coiled regions while euchromatin is loosely coiled

Hill coefficient

If an enzyme has a hill coefficient greater than 1, it means that the enzyme exhibits cooperativity. (i.e if hill coefficient = 1, the enzyme doesn't have cooperativity)

The Dicer enzyme

Involved in producing siRNA

What is arranging nuclear DNA in preparation for transmission to offspring?

It's the creation of gametes through meiosis, which is a noncyclical process. It happens once and terminates in gamete creation (either sperm cells or ova)

At day 14..

LH surges, estrogen threshold reached, follicle ruptures (ovulation) into fallopian tube, and corpus luteum forms Corpus luteum = makes estrogen and progesterone, which maintains endometrium - Female check point: 1) primary oocyte at prophase I & 2) secondary oocyte at metaphase II (where secondary oocyte arrested; comes out of arrest if fertilization occurs)

Fertilization vs. no fertilization

No fertilization -> LH falls -> corpus luteum dies -> estrogen and progesterone fall -> endometrium dies (menses) -> cycle begins anew with FSH and LH re-rising Fertilization occurs -> implanted embryo releases hCG -> hCG mimics LH to maintain corpus lute -> estrogen and progesterone maintained by corpus lute -> placenta takes over the responsibility of making estrogen/progesterone later on

Fibroblast

Structure embedded in the extracellular matrix that produce the fibers comprising connective tissues such as collagen - Fibronectin is the basement membrane connected to an epithelial layer and endothelium of the capillary. - Connexin is a gap junction protein - Cadherins from cell-cell junctions - Occludin comprises tight cell-cell junction

RER vs. SER

RER (site of ribosomes): The ribosomes attach to the outside of rough ER and synthesis protein into the lumen. SER (= Makes lipids of the plasma membrane) <-> RER (= Makes transmembrane proteins, carries them on its membrane. RER forms vesicles and bud off, fuses with the plasma membrane, transmembrane proteins ow on the plasma membrane)

Transcription in prokaryote

Regulation of transcription in prokaryotes: Transcription factors (proteins) bind to enhances or silences (DNA) to affect transcription. Enhancers and silencers in prokaryotes are close to the core promoter and is part of the extended promoter. Operons are groups of genes whose transcription can be regulated by binding of either repressors (decrease transcription)/inducers(increase transcription) onto the a stretch of DNA on the operon called the operators. Sometimes you come across the term co-repressors and co-inducers. When a co-repressor binds to its target, the resulting complex becomes either an active repressor or an inactive inducer. When a co-inducer binds to its target, the resulting complex becomes either an active inducer or an inactive repressor. Alpha factors are how phages control transcription inside their bacterial host. By making α factors at different times, the phage can control the correct transcription sequence of early, middle, and late genes. Transcription attenuation works in the trp (tryptophan) operon. When tryptophan is scarce and needed, transcription occurs normally. However, if there's already a lot of tryptophan present, then transcription terminates prematurely. Operons in prokaryotes consist of groups of genes that work together and are all controlled by one promoter, which is controlled by an operator region (all the genes in the operon are turned on and off together).

phylogenetic tree

Represents the evolutionary relationships among a set of organisms. We want the tree to require the fewest "events" (i.e mutation, transfer, etc) to produce the genetic picture.

Afferent motor neurons

They use chemical synapses that are eventually ignored by the body (i.e desensitized or turned out) and are less likely to be continuous (they may be physiologically continuous, but the brain begins to disregard them). - In contrast, electrical synapses are faster and continuous in their activity; their speed is the primary reason why they are preferred by the body over the traditional chemical synapse. - Chemical synapses are slower than electrical synapses because chemical synapses require the movement of agents through intercellular space (i.e at gap junctions, the cells approach within ~2-4nm of each other, a much shorter distance than the 20-40nm intracellular space that separates cells at a chemical synapse). Because the chemicals diffuse over larger distances, they take more time.

In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are heterozygotes for the polydactyly allele?

We can use the Hardy-Weinberg equation to solve this question. Remember that the total number of alleles in the population has to add up to 1: A + a = 1 And the total number of genotypes in the population must also add up to 1: AA + 2Aa + aa = 1 We're told that a = 0.012. (Note that the question gave the frequency of the recessive autosomal allele, not the frequency of individuals in the population that are homozygous recessive!) By the first equation above, A = 0.988. The carriers are the *heterozygotes* with the genotype Aa. Their frequency is: 2Aa = 2 x 0.988 x 0.012 = 0.988 x 0.024 = 2.37%

SOS response in E.Coli

during replication, when there's too much DNA damage for normal repair to handle, the SOS repair system comes alone. Instead of correcting any DNA damages during replication, the polymerase replicates over the damaged DNA as if it were normal. By using the damaged DNA as a template error rates are high, but still better than not replicating at all.