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Application of latrunculin B to RMCA held at 120mmHg will most likely result in:

Decreased F-actin levels According to the passage: -During actin polymerization monomers of G-actin polymerize into F-actin -Latrunculin B, which stimulates actin depolymerization, should result in decreased F-actin levels.

Which amino acid residues can be phosphorylated?

All the BASIC and -OH residues Arginine, Lysine, Histidine Serine, Threonine, Tyrosine

Which amino acid has UNBRANCHED alkyl chain?


The stereochemical designators ALPHA and BETA distinguish between

EPIMERS at an ANOMERIC carbon atom Remember: alpha/beta are anomers of a ring sugar BUT D/L are stereoisomers in open chain form

Higher osmotic pressure – which solution?

The MOST concentrated one

Which event is directly mediated by a LIGAND GATED ion channel

Influx of Na+ across the motor end plate resulting in depolarization of muscle cell fiber membrane –Influx of Na+ across the motor end plate occurs when Na+ ion channels bind the LIGAND ACETYLCHOLINE

Southern Blot

Uses restriction digest to differentiate between mutant and wild type alleles In order for 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

Restriction Enzyme

Also called restriction endonuclease They cut the double strand DNA at palindrome sequences. The resulting fragments are called RESTRICTION FRAGMENTS


Palindromes are 4-6 base pairs long

Restriction enzyme – sticky ends

Some restriction enzymes cut to make sticky ends, which can HYBRIDIZE

Restriction enzyme – blunt ends

Some restriction enzymes cut to make blunt ends, which can NOT HYBRIDIZE

Blot mnemonic

SNOW DROP Southern = DNA Northern = RNA O=O Western = Protein

Which technique CANNOT be used to analyze gene expression?

Southern Blot Coz southern blot only detects DNA sequence during the

What bond is cleaved by IN during first reaction of integration?

P-O Although there is no physical phosphate at the very end of 3′ end of DNA, but when we are talking about cleavage, we are referring to the cleavage between the last nucleotide and second last nucleotide. Those two nucleotides are bonded together by a PHOSPHODIESTER BOND

Phosphodiester bond

Whenever DNA bonding/cleavage question – think Phosphodiester and hydrogen bonds

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

5′- GGCAACUGACUA -3′ (the same) –HIV is an RNA virus –It injects RNA into the host cells –With that package also comes REVERSE TRANSCRIPTASE, INTEGRASE and PROTEASE –First, reverse transcriptase makes DNA from that injected RNA –This DNA will, of course, be complementary to the injected RNA –Integrase then integrates the viral DNA into the host genome –Then protein synthesis occurs by the usual cell’s mechanism, namely that DNA is transcribed to mRNA which will be complementary to the integrated viral DNA and therefore be IDENTICAL to the initial injected viral RNA –HIV protease then clips that protein at specific spots to activate it

Competitive inhibition

Competitve inhibitions can be determined through rate experiments by applying the principles Michaelis-Menten equation. By keeping -enzyme concentration CONSTANT -VARYING substrate concentration -either including or excluding the inhibitor the effect of the inhibitor on the Vmax and apparent Km of the reaction can be determined

What is the average molecular weight of amino acid?

110 Dalton

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

According to the passage – IN is a 288 residue protein And we know that average molecular weight of an amino acid is 110 daltons So – 288 * 110 = 31680 daltons Since IN is a TETRAMER 31680 * 4 = 126, 720 daltons ~ 128 KDa

To determine if a small molecule acts like a LEDGIN with respect to IN both with and without the small molecule and then 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

III. only The intent is to confirm that a small molecule induces the formation of integrase tetramers from integrase dimers, it is necessary to visualize the proteins in their native state –Use of denaturing agent will disrupt the interaction between monomers –Use of a reducing agent only will disrupt any sulfide bonds

Independent and Dependent variable in a graph

USUALLY: –x-axis is the independent variable –y-axis is the dependent variable

What does Hill coefficient mean?

If: –Hill coefficient >1 means it exhibits COOPERATIVITY –Hill coefficient =1 NO COOPERATIVITY

#31 Which technique can be used to determine if a sample of cells expresses both isoforms?

3 bp are translated into 1 AA. That means Exon 1: 8 AA Exon 2: 3 AA Exon 3: 2 AA Exon 4: 6 AA The question says that the final protein products are either 16 or 17 AA long. The only way you can get to those numbers is either having a protein coded by exons 1+2+4 (17) or by exons 1+3+4 (16). Exons 1 and 4 flank the spliced region, and will be expressed regardless of which isoform is made. So, you make primers against 1 and 4, PCR amplify, run it on a gel, and two products of different sizes tell you that both isoforms are expressed.

#30. 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? (Note: Assume Mendelian inheritance patterns)

4/9 –> read explanation in image


Adenine Guanine — 2 rings

Guanine structure

Adenine structure

Pyrimidines & their structures

Cytosine Thymine Uracil –1 ring

Isoelectric focusing

separates proteins based on isoelectric point –To do this a stable pH gradient must be established in the gel

How can different protein isoforms be produced?

Different protein isoforms are synthesized from same gene during ALTERNATIVE SPLICING, during which sections of the full transcript (both introns and exons) are spliced. –Different combinations of exons can produce different protein isoforms

Which statement about CREB327 phosphorylation is supported by the data presented in table 1?

The WT shows 27% activation and the control is 1.6%. We know that PKA phosphorylates at the 119 residue. Therefore, seeing CREB327(119) means that the 119 residue is mutated and thus PKA cannot phosphorylate there so we have essentially the same % activated as the control. Now, CREB327(115) has the 115 residue mutated and the 119 intact thus this one can be phosphorylated by CREB but only to a certain degree (8.5%). This 8.5% is definitely more than the control or the CREB327(119) but it is not completely as much as the WT. Thus, we can deduce that the PKA is responsible for partial activation (up to 8.5%) of the CREB protein.

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:

PKA and GSK-3 with ATP but noCREB327 WT 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 is clear that any phosphorylation detected is due to autophosphorylation and not phosphorylation of the substrate.

Higher Km or Kd =

LOWER affinity REMEMBER it is inversely proprotional

Ovarian cells are


Osteoclasts are

CONNECTIVE tissue cells

Which statement best explains why the absorbance levels for FSH differs from those of FSH pep?

The tertiary structure of FSH limits FSH-Ab binding interactions Absorbance levels show that FSH-Ab binds FSH pep at higher levels than FSH. The difference between the two is that FSH is a fully folded protein whereas FSHpep is a peptide sequence. Thus, the most likely explanation for why FSH-Ab binds FSH at lower levels than FSHpep is because the tertiary structure of FSH interferes with FSH-Ab binding

Peptide Hormones

are hydrophilic and soluble in blood –Hormones that must bind transport proteins are steroid proteins which are hydrophobic and lipophilic

Two gel electrophoresis analyses are performed on a sample with unknown structure: SDS_PAGE(1 band appears) and SDS-PAGE under reducing conditions(2 bands appear). Which prediction about the protein is directly supported by these results? Protein:

Is composed of multiple subunits Reducing agent is used during SDS-PAGE to CLEAVE DISULFIDE BONDS The appearance of 1 band on SDS-PAGE without reducing agent and 2 bands on SDS-PAGE with a reducing agent, suggests that at least one disulfide bond is present in the protein and that the disulfide bond(s) hold(s) two separate subunits of different masses together

Phosphodiester bond found ONLY in

DNA!! They do not contribute to stabilization of protein structure

Which type of inhibitor does NOT alter the slope of Lineweaver-Burk plot (Km/Vmax)?


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

II and III only –Uncompetitive inhibitors bind their target enzymes ONLY when 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 –Additionally an increase in inhibitor concentration results in increased enzyme binding and inhibition

Nuclear protein requires

Nuclear Localization Sequence – for nuclear translocation DNA binding domain – for binding to regulatory regions of targeted genes

Signal Sequence domain

Are protein domains required for proteins that are directed towards SECRETORY pathways

Binding domain

is a protein domain which binds to a specific atom or a molecule, such as calcium or DNA.

Which amino acid substitution will most likely result in upregulation of leptin signaling?

Y985 within LEPRb The very last sentence of the passage states SOCS3 binds to Y985 within the LEPRb, thereby blocking recruitment of STAT3 to the LEPRb/JAK2 complex. After reading the passage you should understand that when leptin binds it activates a signal cascade that is then terminated by the binding of SOCS3, if you then modify the binding site of SOCS3 on the SOCS3 then the binding of SOCS3 will be blocked and the cascade will become unable to be terminated. This results in upregulation of signal cascade.

Consensus binding sites

are short nucleotide sequences that are found in different locations but are thought to have the same function –Thinking of this in terms of gene regulation it would be the similar nucleotide sequence that are found in transcription factors and promoters.

Which mechanism restricts the expression of leptin to adipocytes? Only adipocytes contain:

A. the ob gene
B. a promoter for the expression of the ob gene
C. enhancers for the expression of the ob gene
D. nuclear factors for the expression of the ob gene

D. Nuclear factors for gene expression of ob gene Nuclear factors are the only elements that vary in different cells and therefore can confer both temporal and spatial regulation of their target genes

Which amino acids are most likely present at the dimerization interface of STAT3 protiens?

Hydrophobic amino acids Because the polar and charged amino acids most likely interact with water molecules in cytosol and would not be involved in protein-protein interactions. In contrast the side chains of hydrophobic amino acids are free and most likely participate in dimerization of STAT3

SDS gel electrophoresis

SDS-PAGE (sodium dodecyl sulfate-polyacrylamide gel electrophoresis) SDS_PAGE is a variant of polyacrylamide gel electrophoresis, an analytical method in biochemistry for the separation of CHARGED molecules in mixtures by their MOLECULAR MASSES in an electrical field

Isoelectric Focusing

An electrophoresis technique used to separate protein based on their ISOELECTRIC POINT (pI) pI = pH at which a protein has no net charge and does not move in an electric field

Ion exchange chromatography

ion exchange chromatography can be split into cation and anion versions. –The cation one has negatively charged resin and causes cations to elute slower than anions. –Inversely, the anion version has positively charged resin and causes anions to elute slower than cations.

Affinity chromatography

Affinity chromatography is a method of separating biochemical mixtures based on a highly specific interaction between antigen and antibody, enzyme and substrate, receptor and ligand, or protein and nucleic acid.

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

Three During conversion of alpha-ketoglutarate to oxaloacetate in the citric acid cycle: 2 molecules of NADH and and 1 molecule of FADH2 are generated

Which enzyme is used in both gluconeogenesis and glycogenolysis?


Efficiency of an enzyme is measured by

kcat/Km ratio Vmax = kcat[E] –Higher the ratio, more efficient the enzyme

74. Based on the data presented in Table I, which amino acid residues of PRR are involved in binding of prorenin?

I. Residue 109
II. Residue 140
III. Residue 201
IV. Residue 269

I, III and IV only Amino acid substitutions at position 140, 201, and 269 result in changes to the Kd, while a non-conservative substitution at position 109 yields no change in the Kd

76. Based on the data presented in Fig 1, overexpression of PRR:

I. Increases BP in part through an angiotensin II-dependent pathway

II. Increases BP in part through an angiotensin II-independent pathway

III. Decreases BP in part through an angiotensin II-dependent pathway

I and II only Compared to control, the PRR overexpression alone increased ROS (an indicator of blood pressure) levels even in presence of Losartan (angiotensin II antagonist), indicating that PRR overexpression includes ROS formation in an angiotensin-II independent manner. Moreover, the addition of exogenous prorerin, a ligand for PRR, induced an increase in ROS formation which can be reversed by losartan, indicating the angiotensin-II dependent nature

Based on the data presented in table 1, which amino acid residue of prorerin most likely interacts with the residue at position 201 of PRR?

A. Alanine
B. Glutamate
C. Arginine

Arginine Substitution of aspartate (D) at position 201 with asparagine (N) results in lower PRR affinity (higher Kd) towards prorerin. Since both asparagine and aspartate have equivalent carbon lengths, the decrease in affinity is most likely due to elimination of an ion pair. Among the possible options, only arginine could form an ion pair with aspartate The question is asking what interacts with residue 201, not what else could replace it. Given the lowered efficacy of the D to N switch, we can assume the – charge is interacting with something, likely a positive.

Which enzymes are used in cDNA cloning?

DNA Polymerase – DNA amplification DNA ligase – in the ligation of the cDNA to DNA vector Reverse transcriptase – in reverse transcription of RNA to cDNA

Which enzyme is NOT used in cDNA cloning?

RNA polymerase

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

Elimination of a n-stacking interaction decreases stability of the complex Based on results presented in Table 1 – in W140L variant, the replacement of a tryptophan (a hydrophobic, aromatic amino acid) by leucine (a hydrophobic but non-aromatic amino acid) results in a decrease in stability of the PRR-prorenin complex as indicated by the increase in Kd. This is most likely due to elimination of the n-stacking interaction with the side chain of tryptophan NOTE: It’s crucial that you know both the one- and three-letter codes for the amino acids. And yeah – I guess it’s not entirely fair to assume that you know the notation for mutations, but it’s quite common in biology so you might have come across it before in some biology course you’ve taken. In any case, the notation always is AXXXB, where A is the amino acid being switched out and B is the replacement amino acid. The XXX represents the amino acid number (1, 2, 3, etc. from the N-terminus).


Metabolism of glycogen polymers occurred during fasting. The glycogen is broken down in the liver, kidney, or muscles into glucose or to glucose-6-phosphate for use in glycolysis pathway


Metabolic pathway that results in the generation of glucose from non-carbohydrate carbon substrates such as lactate, glycerol, and glucogenic amino acids

Glycolysis and gluconeogenesis pathways

Citric acid cycle

G-protein coupled receptor leads to

INCREASE in –Adenylate cyclase –Protein kinase A –level of cAMP G protein promotes the dissociation of bound GDP and its exchange for GTP on the alpha-subunit

Which enzyme catalyzes the rate limiting step in glycogenolysis(glycogen breakdown)

Glycogen phosphorylase

Bonds in glycogen

Glycogen (glucose polymer) is formed by GLYCOSIDIC BONDS between glucose molecules through ALPHA (1->4) linkage linearly and ALPHA(1->6) linkage at branch point

Pentose Phosphate pathway

The pentose phosphate pathway results in formation of :

NADPH, which is used as an OXIDATIVE AGENT in cellular respiratory processed NADPH = reduced form = oxidizing agent NADP+ = oxidized form = reducing agent

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

Southern blotting –because southern blotting is used to detect a particular sequence in a sample of DNA Remember SNOW DROP

Reverse transcription PCR

which involves exploiting the ability of reverse transcriptase to synthesize complementary DNA (cDNA) from mRNA transcripts, and then using PCR to amplify regions of interest

90 – MAYBE its done already – Double check

91. Which conclusion about the data shown in figure 1 is valid?

Changes in the MMP likely affect cell viabiltiy The passage changes in MMP are noticed at 2 hours post treatment, whereas changes in cell viability appear after 12 hours of treatment. This suggests that changes in MMP precede and likely affect cell viabiltiy

92. In a follow up experiment, researchers treated cells with 50µM of C75 in the presence or absence of free radical N-acetyl cysteine (NAC) and then measured cell viability. The most likely rationale for this experiment was to test whether:

reactive oxygen species play a role in mediating the effects of C75 Free radical scavengers such as NAC mitigate the effects of reactive oxygen species (ROS), thereby indicating if ROS were involved in mediating the effects of C75 –Anti-oxidants like NAC, glutathione, and vitamin E act as reducing agents to neutralize ROS from reacting with things like DNA and proteins.

N-acetyl cysteine

–Anti-oxidants like NAC act as reducing agents to neutralize ROS from reacting with things like DNA and proteins


–Anti-oxidants like Gluthathione act as reducing agents to neutralize ROS from reacting with things like DNA and proteins

93. Which control experiments should be included in the design of experiment 2 to validate the results shown in figure 2?

I. Evaluating protein levels of FASN and mtKAS following siRNA treatment

II. Evaluating the siRNAs specificity for FASN and mkKAS by assessing the cellular levels of unrelated proteins

III. Evaluating the siRNAs specificity by showing that a non-specific siRNA has no effects on the protein levels of FASN and mtKAS

I, II and III In order to connect the effects of siRNA to the observed results in figure 2 researchers must ascertain that: I -> siRNA treatment results in knocking down the protein levels of FASN and mtKAS II-> each SiRNA is specific to its target protein and does not interfere with expression of other unrelated proteins III-> A non-specific scrambled siRNA does not interfere with the expression of FASN ot mtKAS

Fatty Acid synthesis and Beta-oxidation

95. Based on the information in the passage, the deletion of mtKAS will most likely result in decreased level of:

Protons in the mitochondrial intermembrane space Passage states that deletion of the gene encoding of mtKAS results in dissipation of the electrochemical gradient generated by electron transport chain. The ETC uses the free energy from redox rxns to pump protons from mitochondrial matrix to the intermembrane space thereby generating electrochemical gradient across the inner mitochondrial membrane. Therefore any event that causes the dissipation of the electrochemical gradient across the inner mitochondrial membrane will result in decreased level of protons in mitochondrial intermembrane space

Citric acid cycle with structures

99. A partial DNA sequence of the coding strand of a gene shown:


Which sequence corresponds to mRNA of this DNA sequence?

5′-GACAUGGACUCGCUA-3′ — for any given gene, the nucleotide sequence of both coding strand and mRNA are complementary to the template strand

Is coding strand the same things as mRNA sequence?

YES! Both coding strand and mRNA strand are both complementary to the template strand, they will have the same sequence

which technique CANNOT be used to analyze gene expression?

SOUTHERN BLOTTING Southern blotting is a technique used to detect particular sequence in a sample of DNA

Are transmembrane proteins

Most of the transmembrane proteins are HYDROPHOBIC

9. Compared to the WT mice, which experimental group of mice is most likely to remain lean when fed a HFD?

Mice in which GPCR43 is overexpressed in WAT Increased GPCR3 activity would be expected to compensate for the increased fat intake and prevent fat accumulation

12. Compared to untreated WT mice, antibiotic treatment of WT mice is likely to result in:

Increased volume of adipocytes If SCFA producing microbes were removed by antibiotic treatment, then GPCR43 would not be activated to reduce glucose uptake into adipocytes

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