BIOCHEM I MCAT SELF PREP plus KA notes from Biochem passages

SSB proteins

-bind to open DNA to make sure two strands don't reanneal

transversion mutation

A point mutation in which a pyrimidine is substitued for a purine, or vice versa.

parabiosed

Parabiosis, meaning "living beside", is a technical term in various contexts in fields of study related to ecology and physiology.

What is the function of reverse transcriptase? Choice A,Transcribe mRNA from a pre-mRNA template( Choice B, Transcribe DNA from a RNA template( Choice C, Transcribe RNA from a DNA template( Choice D, Transcribe pre-mRNA from a mRNA template

Reverse transcriptase is found in retroviruses. Retro"/ "reverse" = backwards he normal pattern of genome processing is DNA→ transcription→ RNA. The function of reverse transcriptase is to transcribe DNA from RNA.

Enzyme X is activated by the presence of acetate ions. How would you expect the presence of sarin to affect the rate of enzyme X? Choice A, Sarin exposure would lead to more acetylcholine breakdown, causing an increase in the amount of available acetate, and an increase in the rate of Enzyme X.( Choice B,Sarin exposure would lead to less acetylcholine breakdown, causing a decrease in the amount of available acetate, and an increase in the rate of Enzyme X. (Choice C, Sarin exposure would lead to more acetylcholine breakdown, causing an increase in the amount of available acetate, and a decrease in the rate of Enzyme X.( Choice D, Sarin exposure would lead to less acetylcholine breakdown, causing a decrease in the amount of available acetate, and a decrease in the rate of Enzyme X.

Where in the passage is acetate mentioned? Sarin would prevent acetylcholine metabolism, and would lower the amount of acetate production. Since Enzyme X is activated by acetylcholine, a decrease in available acetylcholine would decrease the rate of Enzyme X.

Neuron

a specialized cell transmitting nerve impulses; a nerve cell.

How are peptide bonds broken?

hydrolysis non specific (through acid hydrolysis and heat) specific (through proteolysis, specific protease) ex. trypsin cleaves at the c terminal of arg and lysine

How does the breaking of a disulfide bond change typhoid toxin? (Choice A, The newly created terminal sulfides have different valences for binding and although will not bind exactly to cell surface receptors, they will remain mildly toxic to some cells. (Choice B, A broken disulfide bond alters the tertiary structure of the protein and inhibits normal binding to cell surface receptors, thus eliminating the toxicity.( Choice C,The breaking of the disulfide bond decouples the toxic protein from the typhoid protein complex, yet the toxic protein itself retains its toxicity and causes cellular arrest.( Choice D,The disulfide bond is the main active site of the protein, so the protein is unable to make toxins that will bind to cell surface receptors therefore there is no toxicity.

(Choice B, CORRECT A broken disulfide bond alters the tertiary structure of the protein and inhibits normal binding to cell surface receptors, thus eliminating the toxicity

A hypothetical mouse mutation is identified that results in the absence of any circulating leptin in mice harboring the mutation: which of the following statements describe a cellular mechanism that could explain this lack of circulating leptin? Assume that the mutation is a loss of function mutation. I. The mutation occurs in the leptin gene. II. The mutation occurs in the gene that codes for a chaperone protein involved in post-translational modification of leptin. III. The mutation occurs in the gene that codes for a transcription factor that acts as a co-repressor of the leptin gene. Choice A)AIII only (Choice B, I and II only (Choice C, I and III only (Choice D)DI, II, and III

A mutation in a gene coding for a co-repressor of another gene could render the co-repressor inactive, which could reasonably be assumed to result in increased production of the latter gene's protein product. Problems in post-translational modification of a protein could result in a failure by cells to produce the protein. The mutation could occur in the leptin gene, or in a gene that codes for a chaperon protein involved in the post-translational modification of leptin.

transition mutation

A point mutation in which a pyrimidine is susbstituted for a pyrimidine, or a purine is substituted for a purine.

How would you expect the mutation associated with PKU to affect the rate of the reaction catalyzed by phenylalanine transaminase? Choice A, The reaction rate will increase, due to an increase in the concentration of phenylalanine.( Choice B, The reaction rate will decrease, due to an increase in the concentration of phenylalanine.( Choice C The reaction rate will decrease, due to the mutation associated with the disease.( Choice D, The reaction rate will remain unchanged, as Km and Vmax have not changed.

Consider phenylalanine transaminase's role in phenylalanine metabolism. PKU results in an increase in phenylalanine concentration. As a result of the decreased rate of phenylalanine hydroxylase, the concentration of phenylalanine, which would be [S] in this case, would be abnormally high, leading to an increase in the rate of phenylalanine transaminase's reaction.

Which of the following compensatory changes should one expect to see as a result of the increased concentration of plasma glucose in hyperglycemic ob/ob mice? Choice A,Increased secretion of glucagon by pancreatic \alphaαalpha-cells.( Choice B, Increased secretion of insulin by pancreatic \betaβbeta-cells.( Choice C, Increased secretion of growth hormone by somatotropic cells of the anterior pituitary. Choice D, Increased secretion of cortisol by cells in the zone fasiculata of the adrenal cortex.

Cortisol is a glucocorticoid hormone. One of its biological effects is to increase plasma glucose. One of the biological effects of growth hormone is to increase plasma glucose. Glucagon is secreted by pancreatic alpha cells in response to decreased plasma glucose levels. Increased secretion of insulin by pancreatic beta cells would be a compensatory change in response to hyperglycemia of ob/ob mice.

Recombinant DNA

DNA produced by combining DNA from different sources

Why was DNase needed in this experiment? Choice A, DNase destroyed the E. coli DNA so no new transcripts were produced( Choice B, DNase activated mRNA transcription so the radioactive amino acids would incorporate into the protein( Choice C, DNase activated translation of the poly-uracil transcript Choice D, DNase was used as a control in the experiment

DNase is an enzyme that breaks apart DNA Since the researchers only wanted to measure protein output from the added poly-uracil, they needed to remove any other DNA that could make a transcript. Therefore, DNase was added to destroy the E. coli DNA so no new transcripts were produced, and the main radioactivity measured was from the protein synthesis from the added poly-uracil RNA.

The pure, isolated EBOV genome is transcribed immediately after entering the host cell. Considering the products of this transcription, the pure, isolated EBOV genome is analogous to what molecule? Choice A, Antisense DNA strand (Choice B,Plasmid DNA (Choice C, tRNA strand( Choice D, mRNA

EBOV is a negative sense RNA virus, which means it must first transcribe it's genome before protein synthesis can begin. Negative sense RNA viruses have a genome in a 3'→5' orientation, thus a 5'→3' strand is produced from its transcription. The antisense DNA strand is read and transcribed by RNA polymerase to produce mRNA. mRNA is produced in a 3'→5' fashion, but the template strand is read in a 5'→3' fashion. mRNA is analogous to a positive-sense viral genome, and can be read for translation in a 5'→3' fashion.

The rate of the uncatalyzed hydrolysis of sialic acid is 107 times slower than the rate of the same reaction with the neuraminidase catalyst. What accounts for this difference? Choice A,Neuraminidase decreases the activation energy of the reaction, thereby making the reaction more thermodynamically favorable. (Choice B,Neuraminidase increases the activation energy of the reaction, thereby making the reaction more thermodynamically favorable. (Choice C, Neuraminidase increases the activation energy of the reaction, thereby making the reaction more kinetically favorable.( Choice D, Neuraminidase decreases the activation energy of the reaction, thereby making the reaction more kinetically favorable.

Enzymes do not change any thermodynamic properties of a reaction. Reactions with lower activation energies have faster rates. By decreasing the activation energy of the reaction, neuraminidase makes the reaction more kinetically favorable.

Where is the most likely place where surface glycoproteins are glycolated? Choice A, In endosomes BIn lysosomes( Choice C, On the cell surface( Choice D, In the endoplasmic reticulum

Glycoproteins are mostly located on the outer plasma membrane surface of the cell. For a protein to be located on the cell surface it had to undergo exocytosis via a vesicle. Vesicles for surface proteins are formed on the rough endoplasmic reticulum. All modifications to proteins will typically begin in the endoplasmic reticulum and continue in the Golgi body. Modifications are not made in the vesicle during exocytosis

In the normal cellular response to Hormone A, the efflux of Cl- should immediately stop after which step? Choice A, Synthesis of Protein X is terminated( Choice B, The complex of Hormone A and Protein X undergoes endocytosis( Choice C, Hormone A no longer binds to Protein X( Choice DProtein X is dephosphorylated

Hormone A binds to Protein X to induce a response, which leads to the efflux of Cl-. There is no response whenever Hormone A is not bound to Protein X. The efflux of Cl- will stop when Hormone A is no longer bound to Protein X.

If each member of a couple is heterozygous for homocystinuria, and they have a child, what is the probability that the child will have at least one normal allele? Choice A, 100%( Choice B,75% (Choice C, 50%( Choice D, 0%

If both parents are heterozygous, then they have each have a 50% chance of passing on the abnormal allele. A child that is either homozygous for the normal allele or heterozygous will have at least one normal allele. There is a 25% chance (Remember, 50% x 50% = 25%) that a child will receive both abnormal alleles from the parents, leaving a 75% chance that the child will have at least one normal allele. Alternatively, you could do a punnett square analysis to come with the same answer.

Where in the cell would LPS likely be found? Choice A, Attached to the plasma membrane( Choice B, Free-floating in the cytoplasm( Choice C, Inside vesicles ready for exocytosis( Choice D, Attached to the nucleolus

LPS has long hydrocarbon tails that will be hydrophobic, so the hydrocarbon tails of LPS will not exist in a polar environment like the cytoplasm. LPS also has polar groups, so the polar part of the molecule will be able to exist in a polar environment. LPS is located on the exterior plasma membrane. The structure and shape of LPS is similar to a phospholipid in that there is a polar region and a non-polar region. The fatty acid tails insert themselves into the phospholipid bilayer while the polar glycosylated regions are located outside the cell.

How many nucleotides long is the gene that codes for the hormone leptin? Choice A)Exactly (167×3)+6 (Choice B)BExactly (167×3)+3 Choice C,Exactly (167×3) (Choice D)DGreater than (167×3)+3

Leptin is a 167 amino acid protein. According to the central dogma of molecular biology, each amino acid in a polypeptide is coded for by a three nucleotide sequence in the gene that codes for the polypeptide. Eukaryotic genes contain a start codon, which codes for an amino acid (methionine), and a stop codon, which does not. Eukaryotic genes contain introns, sequences of nucleotides within genes that do not correspond to amino acids in the functional protein. The leptin gene is greater than [(167*3)+3] nucleotides long.

Which of the following describes the solubility of methionine? Choice A,High in both water and lipids.( Choice B, Low in water and high in lipids.( Choice C, High in water and low in lipids.( Choice D,Low in both water and lipids.

Methionine is a non-polar, hydrophobic amino acid. Water is a polar solvent, and lipids are nonpolar. Thus, methionine will have low solubility in water and high solubility in lipids.

Like all types of cancer, GBM is best characterized by which of the following cellular pathologies? Choice A,Gain of function mutations in genes controlling apoptosis( Choice B,Gain of function mutations in tumor suppressor genes( Choice C,Changes in basal cellular protein levels leading to loss of cell cycle control (Choice D, Loss of function mutations in oncogenes

Oncogenes are genes that have the potential, if rendered constitutively active or overexpressed by genetic mutation, to cause the uncontrolled growth of cells. If an oncogene is rendered nonfunctional by a loss of function genetic mutation, it loses its ability to function in a way that can lead to uncontrolled cell growth. Tumor suppressor genes are genes that code for proteins that prevent the uncontrolled growth of cells. Thus, if a tumor suppressor gene is up-regulated or rendered constitutively active by genetic mutation, it will most likely not lead to the uncontrolled growth of cells. Apoptosis is the process of programmed cell death. A gain of function mutation in a gene controlling apoptosis would lead to the opposite of uncontrolled cell growth. GBM is best characterized by changes in basal cellular protein levels leading to loss of cell cycle control.

At what point during the course of the neuraminidase catalyzed reaction with sialic acid, does the substrate most closely resemble the Oseltamivir analogue? Choice A, When the energy of the enzyme-substrate complex is at its highest point.( Choice B, When the energy of the enzyme-substrate complex most closely resembles the energy level of the resulting products. (Choice C,When the energy of the enzyme-substrate complex most closely resembles the energy level of the original reactants.(Choice D, When the energy of the enzyme-substrate complex is at its lowest point.

Oseltamivir is an analogue of sialic acid's transition state structure. Transition states are the most energetically unfavorable state during a reaction. Sialic acid will most closely resemble Oseltamivir when the energy of the enzyme-substrate complex is at its highest point - at the time of the reaction's transition state.

Using the results from the SDS PAGE experiment presented in Figure 2, what is the approximate size in kDa of the enzyme PGM/PMM that is found in Pseudomonas aeruginosa? Choice A)50 kDa( Choice B, 30 kDa (Choice C, 15 kDa (Choice D, 300 kDa

PAGE stands for PolyAcrylamide Gel Electrophoresis. In this process there are small pores that molecules travel through. It is more challenging for large molecules to transverse the small pores. Longer fragments will travel through pores at a slower rate. Length of polypeptides is proportional to the molecular mass measured in kDa. The distance traveled by most polypeptides is linearly proportional to the logarithm of their mass. 550 kDa is the most reasonable answer, and it is actually the real measurement made by the research team. The fragment must be larger than 45 kDa because it traveled less distance. However, the fragment will not be an order of magnitude different in size because there is a linear relationship between the distance traveled and the length of the fragment.

Most serum homocysteine exists in a form bound to other thiol amino acids and proteins in the form of disulfides (e.g. cystine-homocystine and homocystine-homocystine). Disulfide bonds play a role in which levels of protein structure? Choice A, Tertiary structure only( Choice B, Tertiary and quaternary structure (Choice C, Quaternary structure only (Choice D, Secondary structure only

Primary structure is determined by covalent peptide bonds. Secondary structure is determined by hydrogen bonding between distance side chains of amino acids in the same polypeptide. Disulfide bonds play a role in determining tertiary and quaternary structure.

Based on the information in the passage, what is another name for Protein L in EBOV? Choice A, DNA helicase( Choice B, Reverse transcriptase( Choice C, RNA polymerase( Choice D, Telomerase

Protein L catalyzes the transcription of the viral genome to produce a strand capable of being translated by host machinery. According to the passage, this transcription product is analogous to a positive-sense virus's genome, which is analogous to mRNA. mRNA is a transcription product.Transcription producing mRNA is catalyzed by RNA polymerase. Reverse transcriptase forms DNA through the transcription of RNA. Protein L in EBOV is also called RNA polymerase.

If electrophoresis was performed without SDS, what would be a reasonable location for the enzyme PGM/PMM? Choice A Near letter C in Figure 2( Choice B, Near letter B in Figure 2( Choice C, Near letter D in Figure 2( Choice D, Near letter A in Figure 2

SDS (sodium dodecyl sulfate) inserts itself into an amino acid chain and has a net negative charge in buffer solution. SDS also interrupts the tertiary structure of the protein, making a straighter chain of the polypeptide. The net negative charge is what causes a force of attraction between the anode (+) terminal of the electrophoresis machine and the negative charge on the polypeptide. Without a net negative charge and straight shape, the amino acid chain will not migrate quickly through the gel. The molecule will appear significantly larger than it actually is. Therefore the most reasonable answer near letter A in Figure 2.

Based on Table 1, which amino acid did poly-uracil code for? Choice A, Phenylalanine( Choice B, Leucine( Choice C, Phenylalanine and Leucine (Choice D, Alanine

Table 1 shows the levels of radioactivity with or without poly-uracil RNA. Exogenous poly-uracil RNA was added in large quantities when compared to the rest of the cell RNA. Phenylalanine increased from 68 to 38,300, therefore poly-uracil encodes for Phenylalanine. Leucine only increased a small amount over the control levels.

What can Mg2+ be considered in the hexokinase reaction? Choice A,Cofactor( Choice B,Reactant( Choice C, Product( Choice Denzyme

The concentration of Mg2+ is the same before and after the reaction, so it is neither used nor produced. It is neither a reactant or a product. Enzymes are by definition organic protein catalysts. Mg2+ is an inorganic ion. Mg2+ is not an enzyme, but it is clearly required for the proper function of the enzyme. Therefore, it is an inorganic cofactor. (As it turns out, it associates with the negatively charged triphosphate group of ATP to help hexokinase facilitate the reaction)

From the results in Figure 3, what can be concluded about EBOV's mechanism of cell entry, and why? Choice A,EBOV probably enters the cell via clathrin-mediated endocytosis because the EBOV group treated with drug X shows a decreased percentage of infected cells (Choice B, EBOV probably does not enter the cell via clathrin-mediated endocytosis because no significant change in the number of infected cells was seen between the control group and the group receiving the drug (Choice C, EBOV probably does not enter the cell via clathrin-mediated endocytosis because the number of infected cells was significantly higher in the EBOV control group than in the poliovirus group( Choice D,EBOV probably enters the cell via clathrin-mediated endocytosis because no change in the number of infected cells was seen between the control group and the group receiving the drug

The high percentage of infected cells in the EBOV control group simply suggests that EBOV is more infectious relative to poliovirus. Poliovirus is known to enter the cell by a process dependent on dynamin. Inhibiting dynamin with drug X greatly decreased the infectivity of poliovirus, presumably due to inhibiting cell entry. Although the drug X treated EBOV rats show a mildly decreased percent of infected cells, the difference between the control and treated group falls within the standard error, and is not statistically significant. EBOV probably does not enter the cell via clathrin-mediated endocytosis because no significant change in the number of infected cells was seen between the control group and the group receiving the drug.

a large margin of error means

The margin of error is a statistic expressing the amount of random sampling error in the results of a survey. The larger the margin of error, the less confidence one should have that a poll result would reflect the result of a survey of the entire population

Which of the following adaptations of the human body functions in the most similar manner as the adaptation of setae into highly-branched structures with a large number of densely-packed Choice A, The large number of microbes in the colon. (Choice B, The high density of ridges in a fingerprint.( Choice C,The high density of cilia in the trachea.( Choice D, The large number of microvilli in the small intestines.

The passage states that the purpose of the setae branching into spatulae is to provide "enough surface area in contact with the surface to create a strong adhesive force." In other words, this adaptation of the setae functions to increase the surface area of the setae. Microvilli are small projections from the wall of the small intestine that serve to increase the surface area of the small intestine in contact with food that is being digested. Therefore, "the large number of microvilli in the small intestines" functions in the same way as the branching of the setae in a large number of densely-packed spatulae.

Why do patients suffering from Addison's disease display a phenotype associated with hypersecretion of melanin by melanocytes without increased plasma levels of \alphaαalpha-MSH, the primary melanocyte stimulating hormone in humans? Choice A, IAddison's disease is caused by a genetic mutation that renders melanocytes constitutively active. (Choice B, The similarity in the amino acid sequence of ACTH and \alphaαalpha-MSH allows ACTH to bind \alphaαalpha-MSH receptors under certain conditions.( Choice C,Addison's disease is associated with increased levels of glucocorticoids, which can be interconverted to \alphaαalpha-MSH by prohormoneconvertases secreted by melanocytes. (Choice D, Addison's disease is associated with decreased levels of glucocorticoids, which stimulate the production of ACTH via a positive feedback loop.

The passage states the Addison's disease is associated with decreased levels of glucocorticoids. The passage does not state that Addison's disease is in any way related to a genetic mutation leading to constitutive activity of melanocytes. The sequence of amino acids in a protein determines the function of the protein, thus two proteins that have similar or identical sequences of amino acids may be able to function in similar ways by binding similar receptors. The similarity in the amino acid sequence of ACTH and \alphaαalpha-MSH allows ACTH to bind \alphaαalpha-MSH receptors under certain conditions.

What is the difference in the number of potential RNA combinations that can exist with a single-base RNA code and a three-base RNA code? Choice A,4 Choice B, 16 (Choice C)60 Choice D)64

There are four different bases used in RNA. If the code used only one base to code for an amino acid, then only four different amino acids could be coded (4 x 1 = 4). A three-base system can code for 64 different amino acids (4 x 4 x 4 = 64). The question is asking for the difference between the two scenarios, which is 60 (64 - 4 = 60).

Transcription vs. Translation

Transcription : DNA -> MRNA (nucleic acid to nucleic acid) Translation : MRNA -> TRNA -> Protein (nucleic acid aka. RNA to amino acid aka. protein)

isomers

Two different molecules that have the same chemical formula

Given that Sarin causes irreversible damage to a cell's acetylcholinesterase enzymes, which medication would be most appropriate for an individual who has already been exposed to sarin for a significant amount of time prior to receiving treatment. Choice A,Pralidoxime, since it would remove the Sarin molecule from the acetylcholinesterase enzyme.( Choice B, Pralidoxime, since it would remove sarin from affected cells.( Choice C,Atropine, since it would reactivate inactive acetylcholinesterase enzymes.( Choice D, Atropine, since it would counteract the effects of the irremediable acetylcholine surge.

When sarin irreversibly binds acetylcholinesterase, it cannot be removed and that enzyme will be permanently inhibited after a short period. Atropine serves to block the excessive activation of the acetylcholine receptor It would be "too late" for pralidoxime treatment, making Atropine the best choice as it would counteract the effects of the acetylcholine surge.

The following base sequence is given for the DNA coding strand: 5' ACTGTTACATTG 3'. Insertion of the base thymine (T) between the 8th and 9th base may affect how many amino acids? Choice A. 0 (Choice B, 1 (Choice C, 2( Choice D,3

Without the insertion of the extra base, the sequence generates four amino acids from four tRNA anti-codons - UGA, CAA, UGU, AAC. With the extra base inserted, the third tRNA anti-codon changes from UGU (cys) → UGA (stop), and the fourth tRNA anti-codon changes from AAC (asn) → UAA (stop). The first two amino acids are unaffected, but the third and fourth amino acids are affected. This is a particularly important insertion since the third codon is a stop codon that will terminate the protein.

RNase H

aka DNA polymerase I the enzyme that helps to remove the RNA primer during DNA replication

margin of error

an amount (usually small) that is allowed for in case of miscalculation or change of circumstances.

The bond holding the sugar to the base is called and formed where

it is a beta glycosidic bond and is formed from the 1' carbon to the N of the nitrogenous base

How are peptide bonds formed?

nucleophillic addition elimination rxn condensation reaction, dehydration synthesis

Backbone of DNA

remains constant and does not change sugar and phosphate

describe the polarity of nucleic acids

the 5' end has a phosphate group and the 3' end has a OH group. It reads from 5' to 3'

Recombination

the genetic process by which one chromosome breaks off and attaches to another chromosome during reproductive cell division

the O- instead of th OH from the phosphate backbone prevent

the phosphate of the DNA molecule from binding with nucleophiles renders it more stable. the neg charge will help the phosphate group to interact with other polar molecules in aqueous environment to stabilize the structure of DNA

What is the most likely explanation for how short hairpin RNAs (shRNA) reduced the toxicity of typhoid toxin? Choice A, The shRNA molecules bind with the mRNA for PODXL receptors. (Choice B,the shRNA molecules bind with the "A" subunit of the typhoid toxin.( Choice C,The shRNA molecules are inhibitors for PODXL gene transcription.( Choice D, The shRNA molecules block PODXL receptors on the cell surface.

shRNA are RNA strands produced inside of the cell. After a shRNA molecule is produced, the loop of the hairpin is cleaved and the two short strands of RNA can separate. Once separated, the two short strands of RNA can hybridize to other strands of nucleic acid sequences that are in the cytosol. The readily available strands of nucleic acid inside the cell are mRNAs. mRNA is abundant and also single stranded, so it will bind quickly to the short single strands of the shRNA molecule. Once the short strands of shRNA have hybridized to mRNA, the translation of the hybridized mRNA sequence into a peptide chain is blocked. It is most likely that the shRNA contains a complementary sequence for the PODXL gene and therefore blocks PODXL receptors from ever being made on the rough ER. Without PODXL receptors, cells will not bind with the typhoid toxin.

Transposons

(jumping genes) short strands of DNA capable of moving from one location to another within a cell's genetic material (are found in DNA and RNA)

How would you expect the mutation associated with PKU to affect the catalytic efficiency of phenylalanine transaminase? Choice A)Kcat and Km will decrease, leading to an indeterminate effect on the enzyme's catalytic efficiency. (Choice B, Neither Kcat nor Km will change, leading to an unchanged catalytic efficiency.( Choice C)Kcat will increase and Km will decrease, leading to an increased catalytic efficiency.( Choice D, Kcat will decrease, and Km will increase, leading to a decreased catalytic efficiency.

Catalytic Efficiency is calculated by dividing Kcat by Km. The mutation in PKU patients results in an increase in the substrate concentration (phenyalanine). Km and Kcat measure kinetic parameters intrinsic to the enzymes, and are independent of the substrate concentration. Changing the concentration of phenyalanine will only affect [S]. Kcat and Km would be unaffected leading to an unchanged catalytic efficiency.

If a 5 mL sample of 0.1 mM neuraminidase is added to a flask containing equal parts Oseltamivir and sialic acid, which of the two substrates is more likely to be bound to the neuraminidase? Choice A, Oseltamivir, as it has a greater affinity for neuraminidase, as it can more readily bind the enzyme's transition state structure.( Choice B,Oseltamivir, as it has a greater affinity for neuraminidase, as it can more readily bind the enzyme's resting structure.( Choice C, Sialic acid, as it has a greater affinity for neuraminidase as it can more readily bind the enzyme's transition state structure.( Choice D, Sialic acid, as it has a greater affinity for neuraminidase, as it can more readily bind the enzyme's resting structure.

Enzymes and substrates bind most tightly at the corresponding reaction's transition state. Oseltamivir's structure most closely resembles the structure of sialic acid when at the transition state of the neuraminidase catalyzed reaction. Oseltamivir will have a greater affinity for neuraminidase as transition state analogues are often the most tightly bound substrates for any enzyme.

Which fundamental principle surrounding enzymes explains why transition state analogues are able to act as potent inhibitors of naturally occurring enzymes? Choice A, Allosteric Inhibition( Choice B, Proximity and Orientation Effects( Choice C, Enzyme Specificity( Choice D, The Induced Fit Theory

Enzymes and substrates do not fit together like puzzle pieces immediately upon binding. Enzymes and substrates will change their structure slightly upon binding to increase the affinity between the two during the reaction's transition state. The induced fit theory states that binding between enzymes and substrates is strongest at the corresponding reaction's transition state.

Why is the reverse reaction of endoprotease catalyzed hydrolysis of POMC (the endoprotease catalyzed condensation of ACTH with the remaining polypeptide fragment of POMC) generally not observed under physiological conditions? Choice A, The endoprotease responsible for the hydrolysis reaction does not lower the activation energy for the reverse condensation reaction. (Choice B, The endoprotease in question lowers the activation energy for the condensation reaction, but not enough for it to occur under physiological conditions.( Choice C,The endoprotease responsible for the reaction undergoes a permanent alteration to its active site during the hydrolysis reaction.( Choice D, The endoprotease responsible for the hydrolysis reaction lowers the Gibbs free energy of ACTH but not POMC.

Enzymes catalyze chemical reactions by lowering the activation energy required for the reaction to occur. The energy required to drive biochemical reactions is obtained from the surrounding environment. All chemical reactions are reversible, at least in principle. The endoprotease in question lowers the activation energy for the condensation reaction, but not enough for it to occur under physiological conditions.

Given a deficiency in tetrahydrobiopterin, how might one go about calculating the kinetics of phenylalanine hydroxylase using the Michaelis-Menten equation? (Choice A)No modifications would be needed as there is no term in the Michaelis-Menten equation corresponding to cofactor concentration. Choice B, Since cofactors are also substrates, the term representing substrate concentration could be used to account for cofactor concentration.( Choice C)The effects of the substrate would be incorporated into the Kcat value.( Choice D This reaction would exhibit non-Michaelis-Menten kinetics, so the Michaelis-Menten equation could not be used in this case as it stands.

Examine the role of tetrahydrobiopterin in this reaction. The presence of a cofactor means that it is possible for the enzyme to exist in forms aside from "E" and "ES" The presence of a cofactor violates one of the assumptions made during the derivation of the Michaelis-Menten equation, thereby giving the reaction non-Michaelis-Menten kinetics.

An experiment is designed in which synthetic leptin is administered intravenously to db/db mice and ob/ob mice: which strain would most likely experience a cessation of the symptoms associated with their respective genetic mutations? Choice A,Db/db mice only.( Choice B, Ob/ob mice only.( Choice C, Both strains.( Choice D,Neither strain.

Experiment three of table one shows that, when parabiosed with a lean mice with normal levels of circulating leptin, ob/ob mice experience a cessation of symptoms associated with their mutation. This suggests that the ob mutation prevents ob/ob mice from producing leptin. Experiment one of table one shows that, when parabiosed with lean mice with normal levels of circulating leptin, no change occurs in db/db mice, while the lean mice experience symptoms associated with overexposure to leptin. This suggests that db/db mice produce large amounts of leptin, but are unable to respond properly to the presence of the hormone in the circulatory system. Only ob/ob mice would be likely to experience a cessation of the symptoms associated with their genetic mutation upon intravenous administration of synthetic leptin

Individuals heterozygous for the PKU mutation have very slightly increased levels of phenylalanine. How would you expect the phenylalanine hydroxylase reaction rate of an individual heterozygous for the PKU mutation to be different in than an individual with no mutated genes? Choice A, The reaction rate would decrease due to an increase in the reaction's Km( Choice B, The reaction rate would increase due to a decrease in the reaction's Km( Choice C,The reaction rate would increase due to an increase in the reaction's Vmax( Choice D, The reaction rate would decrease due to a decrease in the reaction's Vmax

Having a heterozygous PKU mutation would mean that a portion of the expressed phenylalanine hydroxylase will be mutated. This would lower the effective [E]_\T, . A reaction's Vmax is dependent on enzyme concentration, while an enzyme's Km is not.

The protein product of the lacI gene is best classified as what type of molecule? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTProsthetic group(Choice B, Checked, Correct)CORRECT (SELECTED)Transcription factor(Choice C, Incorrect)INCORRECTZwitterion(Choice D, Incorrect)INCORRECTZymogen

Hint #11 / 3A zwitterion is molecule with positive and negative charges but no net charge. A zymogen is an inactive form of an enzyme. A prosthetic group is a non-protein molecule that binds to an enzyme. Hint #22 / 3A transcription factor is a protein that controls gene transcription by binding a DNA sequence in close physical proximity to the gene. Hint #33 / 3The protein product of the lacI gene is best classified as a transcription factor.

With respect to DM1, how would you expect the size of the mutant protein, coded by the DMPK gene, to compare to the size of the non-mutated protein? Choose 1 answer:Choose 1 answer:(Choice A, Checked, Correct)CORRECT (SELECTED)The protein would be larger, due to the insertion of at least 333 nucleotides in the DMPK gene.(Choice B, Incorrect)INCORRECTThe protein would be larger, since the nucleotide addition alters the reading frame of the DMPK gene.(Choice C, Incorrect)INCORRECTThe protein would be smaller, since the nucleotide addition does not alter the reading frame of the DMPK gene.(Choice D, Incorrect)INCORRECTThere is not enough information to answer this question

Hint #11 / 3An insertion of 333 nucleotides would not alter the DMPK reading frame. Hint #22 / 3An insertion of 333 nucleotides would add an additional amino acid to the protein. Hint #33 / 3Since at least one codon is being inserted into the gene, there would be at least one additional amino acid present in the DMPK protein, making it larger than the original.

A population of the Tangarana Ant lives in one of the trees being cut down to make room for the highway. The ants are forced to migrate underground, where they are exposed to a widely different living environment, causing the destruction of over 99\%99%99, percent of the ant population. Two years after the tree was cut down, a concerned Tangarana Ant sympathizer collects a huge population of ants he finds in an underground tunnel, near to where the tree once stood, and moves them to a tree deeper in the forest that is similar to the original tree they once inhabited. Was this action necessarily helpful to the Tangarana Ant population? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTNo, as the new tree would not likely resemble the previous tree due to the passage of time.(Choice B, Incorrect)INCORRECTYes, as Tangarana Ants are always more suited to living in trees than they are to living underground.(Choice C, Incorrect)INCORRECTYes, as this will put an end to the environmental stress caused by the tree removal, thus allowing the ants to grow their population back to its original size.(Choice D, Checked, Correct)CORRECT (SELECTED)No, as the remaining ants are likely to be more suited to life underground than they are to life in a tree.

Hint #11 / 3Are the Tangarana Ants the sympathizer finds underground the same as the ants that were once in the tree? Hint #22 / 3What happens when a population survives a bottleneck? Hint #33 / 3The population of ants that survived the destruction of the original tree have likely grown more suited to life underground as the population's gene pool changed along with the living environment.

Based solely off of the information in the passage, which of the following types of mutations is not likely to cause Crohn's Disease? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTDeletion(Choice B, Incorrect)INCORRECTTransition(Choice C, Incorrect)INCORRECTInsertion(Choice D, Checked, Correct)CORRECT (SELECTED)Translocation

Hint #11 / 3Crohn's disease results from either a point mutation or frame shift mutation. Hint #22 / 3Insertion and deletion can result in a frame-shift mutation, translocation can result in large scale chromosomal changes, transition can result in point-mutations. Hint #33 / 3Translocation is the only answer choice that cannot lead to either a frame-shift mutation, or a point mutation.

In the lac operon, the repressor protein exerts a negative effect on the expression of the structural genes in the absence of the inducer molecule. In the case of the lac operon, what is the inducer molecule? Choose 1 answer:Choose 1 answer:(Choice A, Checked, Correct)CORRECT (SELECTED)Allolactose(Choice B, Incorrect)INCORRECTRNA polymerase(Choice C, Incorrect)INCORRECTGlucose(Choice D, Incorrect)INCORRECTLactose

Hint #11 / 3Glucose is the usual source of energy for the cell, and prevents the expression of the lac operon structural genes. Hint #22 / 3Although lactose ultimately turns on the expression of the lac operon, it does not directly regulate the transcription of the operon. Hint #33 / 3Allolactose, a metabolite of lactose, is the inducer molecule in the lac operon.

It turns out that Marfan syndrome has a penetrance of 80\%80%80, percent and Sandy's sister actually has the FBN1 mutation. What are the chances of the sister's next child having Marfan syndrome? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECT808080%(Choice B, Incorrect)INCORRECT505050%(Choice C, Checked, Correct)CORRECT (SELECTED)404040%(Choice D, Incorrect)INCORRECT000%

Hint #11 / 3Penetrance is the percentage of individuals with the allele for a trait who express that trait. Hint #22 / 3The sister must be heterozygous for the fibrillin-1 mutation. Hint #33 / 3The odds of the next child having Marfan syndrome is the odds of the mother passing on the mutation multiplied by the odds of expressing that phenotype. This is \dfrac{1}{2}21​start fraction, 1, divided by, 2, end fraction \cdot⋅dot 80\%80%80, percent = 0.40.40, point, 4, or 40\%40%40, percent.

Why does E.coli have polycistronic messages? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTTranslation happens faster with polycistronic messages, so proteins are produced more quickly once transcription is initiated.(Choice B, Checked, Correct)CORRECT (SELECTED)Polycistronic messages allow the bacteria to regulate several functionally related genes at the same time.(Choice C, Incorrect)INCORRECTPolycistronic messages recruit RNA polymerase more efficiently.(Choice D, Incorrect)INCORRECTPolycistronic messages are activated by an inducer, allowing tighter regulation of the production of operon protein products.

Hint #11 / 3Polycistronic messages are mRNAs that encode for more than one gene. Hint #22 / 3Polycistronic messages do not recruit RNA polymerase or ribosomes more efficiently. They may or may not be activated by an inducer. Hint #33 / 3Polycistronic messages often contain genes involved in the same pathway, therefore it allows E. coli to regulate several functionally related genes at the same time.

Given the information in the passage, hypothesize why DM1 presents with more significant symptoms than DM2. Choose 1 answer:Choose 1 answer:(Choice A, Checked, Correct)CORRECT (SELECTED)The DMPK gene likely codes for a protein that is more integral to the muscular system than the ZNF9 gene.(Choice B, Incorrect)INCORRECTThere must be more nucleotide repeats present in the DMPK gene than the ZNF9 gene.(Choice C, Incorrect)INCORRECTMutations that result in an insertion of a greater number of nucleotides must be more significant.(Choice D, Incorrect)INCORRECT444 nucleotide inserts cause frame-shift mutations, which can effect proteins more severely than333 nucleotide inserts.

Hint #11 / 3The DM1 and DM2 mutations affect two different proteins. Hint #22 / 3There is no information in the passage about the number of nucleotide repeats present in each type of Myotonic dystrophy, therefore, based on the information in the passage, the number of nucleotide repeats cannot account for the differences in severity. Hint #33 / 3Based on the information in the passage, one would hypothesize that the DMPK gene likely codes for a protein that is more integral to the muscular system than the ZNF9 gene, and therefore the phenotype is more severe when DMPK is mutated.

A population of the Tangarana Ant lives in one of the trees being cut down to make room for the highway. The ants are forced to migrate underground, where they are exposed to a widely different living environment, causing the destruction of over 99\%99%99, percent of the ant population. This is an example of: Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTInbreeding, as the ants are now forced to reproduce with a smaller subset of the larger original population of ants.(Choice B, Checked, Correct)CORRECT (SELECTED)The bottleneck effect, as only a few ants were able to survive the change in habitat.(Choice C, Incorrect)INCORRECTExtinction following a significant environmental stress, as the ants are unable to survive outside of their original habitat.(Choice D, Incorrect)INCORRECTHabitat isolation, as the ants are unable to access their original habitat.

Hint #11 / 3The destruction of the tree and the living conditions underground represent significant environmental stresses that the ants face. Hint #22 / 3Only a tiny fraction of the ant population was able to survive. Hint #33 / 3A bottleneck occurs when the majority, but not all, of a population is killed by an environmental stress.

What percentage of biologically active STAT3 homodimers would one expect to find in cells of AD-HIES patients as compared to healthy individuals? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTCells of AD-HIES patients should have 100\%100%100, percent as many biologically active STAT3 homodimers as cells of healthy individuals.(Choice B, Incorrect)INCORRECTCells of AD-HIES patients should have 75\%75%75, percent as many biologically active STAT3 homodimers as cells of healthy individuals.(Choice C, Incorrect)INCORRECTCells of AD-HIES patients should have 50\%50%50, percent as many biologically active STAT3 homodimers as cells of healthy individuals.(Choice D, Checked, Correct)CORRECT (SELECTED)Cells of AD-HIES patients should have 25\%25%25, percent as many biologically active STAT3 homodimers as cells of healthy individuals.

Hint #11 / 3The mutation associated with AD-HIES is heterozygous, and data in figure one shows that AD-HIES cells have about 50\%50%50, percent as much non-mutated STAT3 mRNA as control cells. Hint #22 / 3Since each individual STAT3 protein has to dimerize with another STAT3 protein to have the intended biological effect, reducing STAT3 expression levels by 50\%50%50, percent will reduce biologically effective STAT3 homodimers by 75\%75%75, percent. Hint #33 / 3Cells of AD-HIES patients should have 25\%25%25, percent as many biologically active STAT3 homodimers as cells of healthy individuals.

What is the correct sequence of proteins found in the lac operon? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTBeta-galactosidase, transacetylase, permease(Choice B, Incorrect)INCORRECTTransacetylase, permease, beta-galactosidase(Choice C, Checked, Correct)CORRECT (SELECTED)Beta-galactosidase, permease, transacetylase(Choice D, Incorrect)INCORRECTPermease, beta-galactosidase, transacetylase

Hint #11 / 3The three enzymes are always copied together in sequence. Hint #22 / 3The first enzyme is produced before the second, and the second is produced before the third. Hint #33 / 3When gene 1 is blocked, no enzymes are produced. When gene 2 is blocked, only beta-galactosidase (gene 1) is produced. When gene 3 is blocked, only the transacetylase (gene 3) is absent. Thus, the sequence of enzymes is beta-galactosidase, permease, and transacetylase.

Which of the following could explain the variable expressivity shown in Sandy's family? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTEach individual inherits different mutations of FBN1 gene(Choice B, Incorrect)INCORRECTThe FBN1 allele exhibits incomplete penetrance(Choice C, Checked, Correct)CORRECT (SELECTED)Each individual has different genetic and environmental factors that contribute to the disease's phenotype(Choice D, Incorrect)INCORRECTThe FBN1 allele exhibits incomplete dominance

Hint #11 / 3Variable expressivity refers to phenomenon of different phenotypes coming from the same genotype. Hint #22 / 3Penetrance refers to whether the disorder manifests at all. Hint #33 / 3Phenotypes are multifactorial, and the differing genetic environments found in every member of the family would contribute to the variety of expression.

DM1 usually results from an increase in the number of CTG triplet repeats in the DMPK gene, which encodes for a protein kinase, with a larger number of inserted repeats leading to more significant symptoms. DM2, on the other hand, is a result of an increased number of CCTG repeats in the zinc finger ZNF9 gene. Myotonic dystrophy can be difficult to diagnose as it presents in a similar way to other defects in the muscular system, and is also difficult to treat due to its multi-systemic effects. Which mutation classification for the mutation seen in DM1 is most applicable? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTPoint mutation(Choice B, Incorrect)INCORRECTFrame-shift mutation(Choice C, Incorrect)INCORRECTMissense mutation(Choice D, Checked, Correct)CORRECT (SELECTED)None of these classifications are appropriate

Hint #11 / 3What does the passage say about the nature of the DNA change seen in DM1? Hint #22 / 3Will the addition of three nucleotides cause a shift in the DNA reading frame? Hint #33 / 3If a triplet nucleotide is inserted, it will result in an additional amino acid and not a point mutation, frame-shift mutation or missense mutation. Therefore, none of the classifications listed are appropriate.

Which of the following scenarios occurs when lactose is present as the sole energy source? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTProduction of the repressor protein is increased.(Choice B, Checked, Correct)CORRECT (SELECTED)The enzymes for lactose metabolism are produced.(Choice C, Incorrect)INCORRECTThe repressor protein binds to the operator.(Choice D, Incorrect)INCORRECTThe lac operon genes are not transcribed.

Hint #11 / 3When lactose is present, allolactose (derivative of lactose) binds to the repressor protein. Hint #22 / 3Binding of the repressor protein by allolactose causes the repressor protein to be released from the operator. Hint #33 / 3When the repressor protein is gone, RNA polymerase is free to bind to the promoter and transcribe the enzymes for lactose metabolism.

Duchenne Muscular Dystrophy (DMD) is an X-linked recessive degenerative disorder of the muscle. The normal gene codes for a protein named dystrophin, which is approximately 500500500 kDa in size. It is a component of muscle, present in rather low amounts. In DMD patients, there is a deletion mutation in the dystrophin gene that leads to either a defective form of dystrophin or a lack of dystrophin completely. Utilizing an experiment where DNA probes, made from the normal dystrophin sequence, were hybridized to DNA from normal individuals and to DNA from DMD patients, scientists were able to identify sequences that correspond to the region of DNA that contains the deletions characteristic of DMD. The deletions occur at different locations in each patient, suggesting that they occur de novo. The most severe and common deletion is located in the center of the locus. The sequence of the deletion in the coding strand was determined to be: 555'—GCCATAGAGCGA—333'. What type of mutation characterizes the most common form of DMD deletion? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTFrameshift mutation(Choice B, Incorrect)INCORRECTGain-of-function mutation(Choice C, Checked, Correct)CORRECT (SELECTED)Loss-of-function mutation(Choice D, Incorrect)INCORRECTSilent mutation

Hint #11 / 4The deletion identified in DMD patients was a segment of DNA that was a multiple of three, suggesting that there was not a frameshift mutation, which would cause a shift in the reading frame of the coding sequence. Hint #22 / 4A silent mutation does not change the resulting amino acid sequence, which is not the case in DMD patients where a portion of the sequence is deleted. Hint #33 / 4In gain-of-function mutations, the mutant form of the protein also interferes with the function of the normal protein, thus resulting in a dominant phenotype. Since DMD is a recessive phenotype, this mutation is not a gain-of-function mutation. Hint #44 / 4Loss-of-function mutations render the mutated allele non-functional, however the wild-type allele can still function normally. In the case of DMD patients, the deletion results in a loss-of-function mutation.

Assuming that the IDH mutation discussed in the passage confers a loss of function on the protein associated with the gene, which of the following would be the most likely immediate effect in cancer cells harboring the IDH mutation? Choose 1 answer:Choose 1 answer:(Choice A, Checked, Correct)CORRECT (SELECTED)Decreased concentration of electron-donating molecules.(Choice B, Incorrect)INCORRECTDecreased concentration of electron-accepting molecules.(Choice C, Incorrect)INCORRECTDecreased concentration of proton-donating molecules.(Choice D, Incorrect)INCORRECTDecreased concentration of proton-accepting molecules.

Hint #11 / 4The normal function of the IDH enzyme is to reduce NADP^{+}+start superscript, plus, end superscript to NADPH. Hint #22 / 4In a oxidation-reduction reaction, the molecule that is reduced gains electrons, while the molecule that is oxidized loses electrons. Hint #33 / 4A loss of function mutation in a gene impacts the normal biological function of the protein associated with the gene. A loss of function mutation in the IDH gene would result in decreased ability of IDH to reduce NADP^{+}+start superscript, plus, end superscript. Hint #44 / 4The most likely immediate effect of an IDH mutation in cancer cells would be a decreased concentration of electron-donating molecules.

What is the benefit of the repressor being constitutively produced? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTThe repressor provides a feedback loop preventing the lac operon from being transcribed in the absence of glucose.(Choice B, Checked, Correct)CORRECT (SELECTED)The bacteria does not want to expend energy creating lactose metabolizing enzymes if lactose is not available.(Choice C, Incorrect)INCORRECTThe repressor acts as a lactose sensor, and recruits RNA polymerase only in the presence of lactose.(Choice D, Incorrect)INCORRECTThe operator needs either a constitutively expressed activator or repressor binding to it to regulate transcription.

Hint #11 / 4The repressor negatively regulates the lac operon, preventing lactose metabolizing enzymes from being transcribed. Hint #22 / 4The repressor occupies the operator in the absence of lactose. Hint #33 / 4Operators do not need to necessarily have a repressor or activator present. Hint #44 / 4E. coli does not want to expend energy creating lactose metabolizing enzymes if lactose is not available, therefore the operon is turned off unless lactose is present.

Assuming all are approximately 888 nucleotides in length, which of the following would be effective as primers during the reverse transcription step of the experiment described above? A multitude of random, scrambled primers end text Primers specific to the first intron of POMC A string of thymine nucleotides The same set of primers as used in the PCR amplification step How many of these are correct

Hint #11 / 5Reverse transcription involves making cDNA from an mRNA template. After genes are transcribed, introns are spliced out of the nascent mRNA. Therefore, primers specific to the first intron of POMC would not target mRNA. Hint #22 / 5Besides splicing, another important post-transcriptional modification is poly-adenylation. This process results in a "poly-A tail," or a string of adenines. Thymine and uracil base pair with adenine. Hint #33 / 5The reverse transcription step of this experiment requires that at least one POMC mRNA yield a complementary cDNA. In other words, this step does not require specificity. PCR amplification will exponentially and specifically multiply any POMC cDNA generated. Therefore, with enough random 666 - 888 nucleotide primers, one would expect to amplify at least a small fraction of POMC mRNA. Hint #44 / 5Option \text{IV}IVstart text, I, V, end text uses primers that are specific for the cDNA, which will also produce the desired RT-PCR product. Hint #55 / 5Options \text{I}Istart text, I, end text, \text{III}IIIstart text, I, I, I, end text, and \text{IV}IVstart text, I, V, end text would all be suitable primers, and genetic researchers often use these very techniques.

If light microscopy of a muscle biopsy from a patient affected with DMD revealed that some dystrophin was present in their cells, but that it appeared misshapen, what type of mutation is LEAST (I misread this) likely to have caused their disease? Choose 1 answer:Choose 1 answer:(Choice A, Incorrect)INCORRECTFrameshift mutation(Choice B, Incorrect)INCORRECTMissense mutation(Choice C, Checked, Correct)CORRECT (SELECTED)Nonsense mutation(Choice D, Incorrect)INCORRECTSingle base deletion

Hint #11 / 6Note that the question states that although it is misshapen, some of the protein is still apparent. The mutation must have been one which did not entirely interfere with transcription and translation. Hint #22 / 6A missense mutation is a single nucleotide substitution which results in a codon which codes for a different amino acid. Mutations such as this still allow for normal transcription and translation, but may interfere with normal protein structure or function. Hint #33 / 6A frameshift mutation is a deletion or insertion of a number of nucleotides that is not divisible by three. As such, all downstream nucleotides will be 'shifted' and therefore misread. This is may still be transcribed/translated, but will usually result in a truncated, misshapen, and/or nonfunctional protein. Hint #44 / 6A single base deletion is a type of frameshift mutation. Frameshift mutations are associated with DMD. Hint #55 / 6A nonsense mutation is a substitution of a nucleotide which causes the mutated codon to be read as a "Stop" codon. No protein product will be produced. Hint #66 / 6Because some dystrophin is visualized, the least likely mutation in this case would be a nonsense mutation.

Since the laboratory analysis of homocysteine is complicated by the fact that it exists bound to other thiol amino acids via disulfide bonds, it is first necessary to subject these disulfide bonds to reduction prior to analysis. It is necessary to reduce these disulfide bonds in order to get a true value of the quantity of homocysteine in a plasma sample. Why is this so? Coice A The true quantity of homocysteine would be artificially high if the reduction step were not carried out prior to analysis. (Choice B, It is necessary because disulfide bound homocysteine would otherwise precipitate out of the serum before analysis could be conducted. (Choice C, The true quantity of homocysteine would be artificially low if the reduction step were not carried out prior to analysis. (Choice D, It is necessary in order to prevent homocysteine from degrading in the sample.

Homocysteines that are bound via disulfide bonds do not register as isolated homocysteine molecules. Reduction of disulfide bonds separates homocysteine molecules that are bound to other amino acids. If the reduction step were skipped, then the true quantity of homocysteine would be artificially low.

What is the most likely function of phosphorylated Protein Y? Choice A, Phosphatase (Choice B, Membrane channel( Choice C, Antibody( Choice D, Hormone receptor

Hormone A binds to Protein X to induce a response. The response is that Protein X phosphorylates Protein Y. Phosphorylated Protein Y results in the efflux of Cl- from the cell, thus it is likely a membrane channel protein.

Based on the passage, what does the mutation in the gene for affected males from Family 2 most likely result in? (Choice A)The introduction of a premature stop codon in Hormone A B. The replacement of a single amino acid in Protein Y. (Choice C,The introduction of a premature stop codon in Protein X. (Choice D)The replacement of a single amino acid in Protein X.

Hormone A was bound to protein X in all of the isolates, so neither of those biomolecules is affected. Protein X was normal in Family 2, so there was no premature stop codon in Protein X for that family. Protein X could not phosphorylate Protein Y in the males in Family 2, thus there was likely a mutation resulting in the replacement of a single amino acid in Protein Y.

Enzyme Y is deactivated by downstream effector molecules that are synthesized following acetylcholine receptor activation. How would you expect the presence of sarin to affect the rate of Enzyme Y? Choice A, Sarin exposure would increase the rate of Enzyme Y by decreasing the level of Acetylcholine receptor activation (Choice B, Sarin exposure would decrease the rate of Enzyme Y by increasing the level of Acetylcholine receptor activation( Choice C, Sarin exposure would increase the rate of Enzyme Y by increasing the level of Acetylcholine receptor activation.( Choice D,Sarin exposure would decrease the rate of Enzyme Y by decreasing the level of Acetylcholine receptor activation

How does Sarin affect activation of the acetylcholine receptors? By limiting acetylcholine metabolism, sarin would increase acetylcholine receptor activation. It is stated that "Enzyme Y is deactivated by downstream effector molecules that are synthesized following acetylcholine receptor activation." Enzyme Y's rate would therefore decrease in the presence of sarin due to increased acetylcholine receptor activation.

Which of the following statements represents the best explanation for why db/db mice produce abnormally high levels of leptin? Choice A, The db gene codes for a transcription factor bound by leptin in the nucleus of adipocytes, and this transcription factor is constitutively active in db/db mice, resulting in leptin resistance in leptin-sensitive cells and hypersecretion of leptin by leptin-producing cells.( Choice B)BThe db gene codes for a transmembrane receptor on the surface of leptin sensitive cells, and this transmembrane receptor is constitutively active in db/db mice, resulting in leptin resistance in leptin-sensitive cells and hypersecretion of leptin by leptin-producing cells. (Choice C)CThe db gene codes for a transmembrane receptor on the surface of leptin sensitive cells, and this transmembrane receptor is rendered inactive in db/db mice, resulting in leptin resistance in leptin-sensitive cells and hypersecretion of leptin by leptin-producing cells.( Choice D)DThe db gene codes for a transcription factor bound by leptin in the nucleus of adipocytes, and this transcription factor is inactivated in db/db mice, resulting in leptin resistance in leptin-sensitive cells and hypersecretion of leptin by leptin-producing cells.

Leptin is a polypeptide hormone. Protein hormones function via interaction with transmembrane receptors. Molecules secreted by leptin-sensitive cells participate in a negative feedback loop that modulates the production of leptin. The db gene codes for a transmembrane receptor on the surface of leptin sensitive cells. This transmembrane receptor is rendered inactive in db/db mutants, resulting in leptin resistance in leptin-sensitive cells and hypersecretion of leptin by leptin-producing cells.

Which of the following players in an enzymatic reaction may you find pyridostigmine bound to in the case of acetylcholinesterase? I. Free Enzyme II. Enzyme substrate complex III. Free Substrate Choice A, I only( Choice B, II only( Choice C, I & II Choice D, I, II, & III

Pyridostigmine is a competitive inhibitor of acetylcholinesterase Competitive Inhibitors can bind to enzymes only when they are not bound to substrate Competitive inhibitors will bind to free enzyme, but not to enzyme substrate complex, or free substrate.

How would you expect Pyridostigmine to affect to the apparent Km and Vmax values of the acetylcholinesterase enzyme? Choice A, Km would increase and Vmax would remain unchanged( Choice B Km would decrease and Vmax would remain unchanged.(Choice C, Km and Vmax would both decrease( Choice D, Km and Vmax would both increase

Pyridostigmine is a competitive inhibitor of acetylcholinesterase The Km value is used to measure an enzyme's efficacy at low substrate concentrations, while the Vmax value is used to measure an enzyme's efficacy at high substrate concentrations. Competitive inhibitors only hurt an enzyme's catalytic ability by increasing Km, at high concentrations of substrate where the reaction velocity approaches Vmax, the competitive inhibitors can be overcome.

Why does nucleic acid electrophoresis not require the use of SDS? Choice A, Under physiological conditions, DNA exists in its lowest oxidation state. Choice B, The pH of a solution will not alter the net charge on DNA molecules.( Choice C, DNA molecules already have a net positive charge proportional to their length.( Choice D, DNA molecules already have a net negative charge proportional to their length.

SDS is used to ensure that polypeptides have a negative charge proportional to the length of the molecule. Altering the pH of a solution can increase or decrease the concentration of proton donors and/or acceptors, which can effect the net charge on DNA molecules. Phosphate groups, which participate in the phosphodiester bonds linking nucleotides in DNA, are negatively charged. DNA molecules already have a net negative charge proportional to their length

If SDS-page is performed with a sample containing ACTH and \alphaαalpha-MSH, how far will \alphaαalpha-MSH migrate as compared to the distance ACTH migrates? Choice A)AThe distance \alphaαalpha-MSH migrates is equal to one-third the distance ACTH migrates.( Choice B, The distance \alphaαalpha-MSH migrates is equal to the natural logarithm of the distance ACTH migrates.( Choice C)The distance \alphaαalpha-MSH migrates is greater than the distance ACTH migrates. (Choice D)More information is needed, since the distance of migration in an SDS-page gel also depends on the conformational shape of the polypeptides.

SDS, BME, and heat disrupt the molecular forces that confer conformational shape on polypeptides. Distance of migration depends on molecular weight. ACTH is heavier than \alphaαalpha-MSH, since, according to table one, ACTH has 26 more amino acids than \alphaαalpha-MSH. The distance \alphaαalpha-MSH migrates is greater than the distance ACTH migrates.

Why is SDS-page carried out in a pH buffered solution? Choice A, Using a pH buffered solution disrupts protein-protein interactions that occur under physiological conditions, which are not pH buffered, allowing for complete separation across the gel.(Choice B, Using a pH buffered solution ensures that SDS-bound polypeptides maintain the desired electrical charge.( Choice C, Using a pH buffered solution maintains the native conformations of the proteins being separated.( Choice D, Using a pH buffered solution prevents the autoionization of water, inhibiting the production of reactive hydronium and hydroxide species.

SDS-page electrophoresis requires the disruption of the native conformations of the proteins of interest. SDS-page is not carried out under physiological conditions (BME and SDS are not present in living systems). SDS-page uses a voltage gradient to spatially separate electrically charged proteins. SDS-page is carried out in a pH buffered solution to ensure that the proteins being separated maintain the desired electrical charge.

Why are Km values for hexokinase IV not reported in the scientific literature? Choice A, HK IV [V]0 decreases with increasing [Glc].( Choice B, HK IV [V]0 remains constant at physiological [Glc].(Choice C,)HK IV Glc affinity changes as [Glc] changes.( Choice D, HK IV Glc affinity remains constant at physiological [Glc].

Since hexokinase IV is not included in Figure 2, the main source of information about it is in Figure 1. Notice that the curve for hexokinase IV is S-shaped, suggesting cooperativity. Cooperativity is when the binding of the substrate to the enzyme facilitates the binding of more substrate. In other words, as the concentration of glucose changes, the affinity of the enzyme for glucose increases. An enzyme that displays cooperativity does not follow Michaelis-Menten kinetics and does not have a Km value. Therefore the reasons why hexokinase IV does not have a Km value because "HK IV Glc affinity changes as [Glc] changes."

allosteric regulation

The binding of a molecule to a protein that affects the function of the protein at a different site.

In which direction(s) is the mRNA strand read during translation? Choice A, '3, '→'5, '( Choice B, 5, '→3, '( Choice C, Both 5, '→3, ' and 3, '→5, ' (Choice D, Either 5, '→3, ' or 3, '→'5, '

The entire mRNA strand does not have to be read to translate a single protein. herefore, protein synthesis may begin anywhere along the strand (where it sees a start codon). mRNA and proteins are produced in the order of 5'→3'.No matter where it begins, the mRNA strand is always read from 5'→3'.

To further validate the findings of experiment 1 and make more direct connections to the physiology of humans, researchers should perform which one of the following experiments? Choice A, Design a mouse model that has receptors for Samonella Typhi uptake along with engineeredSamonella that secrete modified toxins.( Choice B, Create computer 3-D models of the molecular interactions between typhoid toxin subunits and cell membrane receptors.( Choice C, Use immunoflorescent staining to observe typhoid toxin's intercellular interactions in mice and observe locations of binding.( Choice D, Inject mice with concentrated toxin and observe additional symptoms of typhoid fever beyond just weight loss.

The minor problem with experiment 1 is an observed correlation with weight loss does not necessarily imply typhoid fever. Moreover, mice do not produce typhoid toxin endogenously, so ideally to compare the physiology of mice to that of humans, scientists should use a model that produces typhoid toxin endogenously. The best solution to study typhoid toxicity in humans is to use an analogous mouse model. In 2012 a mouse model was developed that has human like receptors so that mice will produce the toxin endogenously.

The scientists responsible for the phylogenetic analysis in Figure 2 could have utilized any section of DNA found in geckos. Why does their inclusion of mitochondrial DNA, in addition to nuclear DNA, increase the accuracy of their results? (Choice A, Mitochondrial DNA and nuclear DNA are found in different locations in the cell.( Choice B, There are significantly more copies of mitochondrial DNA than nuclear DNA.( Choice C, Mitochondrial DNA and nuclear DNA undergo different modes of inheritance. (Choice D, There are fewer nucleotide base pairs in mitochondrial DNA than nuclear DNA.

The passage tells us that the scientists undertook a "phylogenetic analysis." Even if you don't know exactly what this is, you can tell from the passage that they compared sequences of DNA to infer evolutionary relationships. This is how they generated the evolutionary tree in Figure 2. We know evolution consists of the gradual accumulation of genetic changes in a population over time. We need to consider which statement would provide the most information about accumulated The amount of DNA present wouldn't necessarily tell us anything about genetic changes that have accumulated over time. Therefore, we can eliminate "There are fewer nucleotide base pairs in mitochondrial DNA than nuclear DNA" and "There are significantly more copies of mitochondrial DNA than nuclear DNA." The location of DNA wouldn't necessarily tell us anything about genetic changes in geckos, since all geckos have DNA in the nucleus and in the mitochondria. Therefore, we can eliminate "Mitochondrial DNA and nuclear DNA are found in different locations in the cell." Mitochondrial and nuclear DNA undergo different modes of inheritance--which means they accumulate genetic changes in different ways. Therefore, if both sources indicate the same evolutionary relationship, it is much more likely to be a true evolutionary relationship.

Studies have shown that 75% of 18F-FDG injected intravenously enters the cell and undergoes radioactive decay with a half-life of ~120 minutes. 25% does not enter the cell and is excreted with a half-life of ~15 minutes. What percent of 18F-labeled molecules originally injected are still radiolabeled and present in the body after 2 hours? Choice A, 25% (Choice B, 38%( Choice C, 50% (Choice D, 63%

This is a half-life problem with two distinct groups of molecules. The first group is the 18F-FDG that enters the cell and has a half-life of ~120 minutes. The second group is the 18F-FDG that remains in the bloodstream with a half-life of 15 minutes. In two hours, the 18F-FDG in the cell undergoes one half-life. Since you began with 75%, half of that will remain, or ~38%. (Alternatively, this is ¾ of the sample; half of that is ⅜ or ~38%). The remaining 25% undergoes 8 half-lives (there are 8 15-minute periods in 2 hours). If you began with 25% (¼), you can quickly see that much less than 1% will remain. Therefore, we can ignore this group of molecules because they are essentially all excreted. Since only the 18F-FDG in the cell remains to an appreciable amount, we can say that ~38% of the radiolabeled molecules are still in the body.

DNA ligase

an enzyme that eventually joins the sugar-phosphate backbones of the Okazaki fragments A linking enzyme essential for DNA replication; catalyzes the covalent bonding of the 3' end of a new DNA fragment to the 5' end of a growing chain.

the missing OH from the ribose on the DNA is helpful to the DNA in that

it makes it more stable than RNA and not able to undergo hydrolysis due to the missing OH on the 2' carbon.