AAMC FL 2 BB

What event will most likely occur if Protein X is inserted into the inner membrane of mitochondria? A.The citric acid cycle will cease to function. B.The electron transport chain will cease to function. C.The proton gradient across the inner membrane will dissipate. D.The pH of the intermembrane space will decrease. passage: Addition of Protein X to the liposomes caused a gradual release of the fluorescent dye from the liposomes

"gradual release" Okay so if it did this in a liposome membrane, then we would need to assume that this also happens in the mitochondrial membrane as well. POE: A- Citric Acid cycle would have nothing to do with this B- The fact that protons leak across the membrane would not stop the ETC - it would just mean that the ETC would go into overdrive to maintain the proton gradient. D- if there were holes in the membrane as the passage suggests, then the pH of the membrane would actually decrease instead of increase because the protons would be leaving. C is the correct answer. this indicates that the gradient would go away. Therefore the pump would have to work super super hard to keep going. also: B has a strong word in it. It has the word "cease"

In a female mouse born with two wild-type Foxp3 alleles, what is the minimum number of inactivating, recessive mutations that might be sufficient in the Foxp3 alleles in mammary epithelium to significantly increase the likelihood that this mouse will develop mammary tumors? the passage indicates that approximately 60% of female Foxp3sf/+ mice, which carry one inactivating scurfin (sf) allele and one wild-type (+) allele, spontaneously develop mammary tumors.

1 because the minimum number of inactivating mutations to increase the likelihood of tumor formation is one.

Osmotic pressure Π is given by the relation: Π = iMRT where i is the van't Hoff factor, M is the concentration of solute, R is the gas constant, and T is the temperature. The osmotic pressure of sea water is approximately 24 atm at 25°C. What is the approximate concentration of salt in sea water (approximated by NaCl with i = 2)? (Note: Use R = 0.08 L•atm/mol•K.) A.0.25 M B.0.50 M C.0.75 M D.1.0 M

24= (2)(x)(25+273)(8e-2) 24=2(298)(8e-2) 24= 2(300)(8e-2) 24= 2(3e2)(8e-2) 24=2x3x8 (X) 24= 48 (x) 24/48=x .5M

From Table 1, GnRH + NPY = 7.03 and GnRH alone = 4.74. NPY amplified pituitary responses to GnRH by: A.27%. B.48%. C.67%. D.83%.

7.03/4.74 7/5= 1.48 so NPY amplified pituitary responses to GnRH by 48%.

Given that the tumorigenicity of a certain mouse mammary tumor cell line is dependent on ErbB2-mediated intracellular signaling, mice injected with variants of this cell line that have which of the following modifications would be most likely to survive the longest? A. The Foxp3 genes deleted B. One of the ErbB2 genes amplified C. A Foxp3-expressing plasmid introduced D. The Foxp3 binding sites deleted in the promoter of one of the ErbB2 genes

C because according to the passage, cancerous tumors are associated with the inactivating scurfin allele, not the wild-type Foxp3 allele. Because Foxp3 regulates ErbB2 transcription and high levels of ErbB2 mRNA are found in cancerous epithelium, the best modification to a cell line would be Foxp3-expressing plasmid, to repress transcription of ErbB2.

Where in the human male reproductive system do the gametes become motile and capable of fertilization? A. Testis B. Urethra C. Epididymis D. Prostate gland

C because sperm, produced in the seminiferous tubules of the testes, completes maturation and becomes motile in the epididymis.

Based on the passage, does optimal VSV-EGP infection in vitro require CatB, CatL, or both? Optimal infection: A. requires CatB only. B. requires CatL only. C. requires both CatB and CatL. D. does not require CatB or CatL.

C because the data presented in Figure 1 indicate that CatB―/―, CatL−/− mouse cells expressing either CatB or CatL have low levels of viral infection. In contrast CatB−/−, CatL−/− mouse cells expressing both CatB and CatL exhibit several folds higher viral infection levels.

Based on the passage, is CatL expression sufficient for VSV-EGP infection of the mouse cell lines presented in Figure 1? (Note: In these experiments, assume that values of <3% of total cells infected are too low to be measured accurately.) A.Yes, because VSV-EGP infects cells expressing CatL better than it infects cells not expressing CatL B.Yes, because VSV-EGP infects cells expressing CatB better than it infects cells not expressing CatB C.No, because VSV-EGP does not infect CatB-/- cells expressing CatL better than it infects CatB-/- cells not expressing CatL D.No, because VSV-EGP infects cells expressing both CatB and CatL better than it infects cells expressing CatB but not CatL

C because the data presented in Figure 1 indicate that compared with the control mouse cell line (bacterial; column 5), introduction of only the CatL gene into the CatB−/−, CatL−/− cell line (column 7) does not affect cell infectivity and therefore is not sufficient for VSV-EGP infection of the mouse cell lines.

The purpose of the study described in the passage was: A.to determine whether LH can modulate GnRH or NPY secretion. B.to determine whether GnRH can modulate NPY secretion. C.to determine whether NPY can modulate LH secretion. D.to determine whether NPY can modulate GnRH secretion. passage: Shown in Table 1 are results from a study in which the pituitary effects of the hypothalamic factor neuropeptide Y (NPY) were examined. In the study, LH levels were measured in female rats treated with saline (control), NPY alone, GnRH alone, or GnRH in combination with NPY.

C. !!!!! This is stated in the purpose of the passage

Based on the passage, β-catenin most likely has: A.multiple subunits. B.very few disulfide bonds. C.a nuclear localization sequence. D.a high proportion of surface-exposed nonpolar residues. passage: and β-catenin, which activates expression of Wnt target genes. CK1 and GSK3 sequentially phosphorylate β-catenin, which targets it for ubiquitination. Figure one shows that it is activating transcription factors.

C. Activating transcription factors needs a nuclear localization sequence.

Which cells harvested from adult mice were most likely used as the highly proliferative benchmark in the experiment that generated the data shown in Figure 3? A.Adipocytes B.Cardiac muscle cells C.Gastrointestinal epithelial cells D.Neurons

C. Key word here is proliferative. Proliferative cells grow or multiply by rapidly producing new tissue, parts, cells, or offspring. This happens A LOT in the stomach. POE: A- no. mature adipocytes do not divide B- cardiac MUSCLE cells do not divide once matured. D- neurons NEVER divide. Once you lose a neuron, it's gone forever.

What are the primary myelin-forming cells in the peripheral nervous system? A.Microglia B.Astrocytes C.Schwann cells D.Oligodendrocytes

C. YUZ QUEEN

citric acid cycle substrates

Can I Keep Selling Sex For Money Office? Can- citrate I- isocitrate K- alpha ketoglutarate S- succinyl CoA S- succinate F- fumarate M-malate O- oxaloacetate *fadh2 conversion happens with the enzyme succinate dehydrogenase in the conversion of succinate to fumarate *GTP made between succinate and succinyl Co A

Which process moves chlorine ions into the cells of the green algae? A.Osmosis B.Diffusion C.Active transport D.Facilitated diffusion graph shows that chlorine is higher inside of the cell than outside of the cell.

Chlorine is a negatively charged ion. It is going to use active transport to need to get into the cell! ions must be moved into the cell against their concentration gradient, which requires energy. ------> active transport. - In the other processes, ions would move along their concentration gradient, either with or without the help of transport proteins.

According to the cross-bridge model of muscle contraction, the muscles stiffen after death because ATP is unavailable to bind and directly release: A. ADP from the actin head. B. ADP from the myosin head. C. the actin head from the myosin filament. D. the myosin head from the actin filament.

D because during normal muscle contraction, ATP is required to break the bonds between the actin filament and the myosin head. After death, no new ATP is generated, so the myosin head cannot be released from the actin filament, resulting in stiffening of muscles.

Secretory lysosomes are classified as lysosomes because secretory lysosomes have some functional components in common with conventional lysosomes. Given this, secretory lysosomes most likely contain: A. ribosomes. B. Krebs cycle enzymes. C. RNA and DNA polymerases. D. degradative enzymes that function at low pH.

D because lysosomes are defined as membrane-bound organelles that contain hydrolytic enzymes activated by a low pH. These enzymes are capable of degrading many kinds of biomolecules.

Which molecule is NOT formed during the citric acid cycle? A. Malate B. Succinate C. α-Ketoglutarate D. Phosphoenolpyruvate

D phosphoenolpyruvate is a product of glycolysis, not the citric acid cycle.

The information in the passage suggests that in mice CRY1 most likely affects XPA by: A.activating XPA protein activity. B.activating translation of XPA-encoding transcripts. C.repressing replication of the XPA-encoding gene. D.repressing transcription of the XPA-encoding gene. passage: the nucleotide excision repair system (NER) recognizes, removes, and replaces segments of DNA strands that contain CPDs and (6-4) PPs. Figure 1 shows how mouse skin levels of XPA (a damage-recognition and rate-limiting factor in NER) and clock regulatory protein CRY1

D. the graph shows that while CRY1 is up, XPA levels are down. so it is going to depress the activity. XPA is said to be a damage-recognition and rate limiting factor in NER, therefore you can deduce that it is going to be an enzyme=protein=transcription.

After a section of a DNA strand containing a UVR-induced lesion is removed and resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA strand by what type of bond? A.Disulfide B.Hydrogen C.Peptide D.Phosphodiester

Dna is joined together by phosphodiester bonds :) - phosphodiester bonds link the 3ʹ carbon atom of one deoxyribose and the 5ʹ carbon atom of another deoxyribose within the DNA molecules. - requires knowledge of DNA strand synthesis through the polymerization of nucleotides.

What is the best experimental method to analyze the effect of tdh2 gene deletion on the rate of histone acetylation? Comparing histone acetylation in wild-type and Δtdh2 cells by: A.Western blot B.Southern blot C.Northern blot D.RT-PCR

HISTONES ARE PROTEINS. Western blots look at proteins. - posttranslational modification of proteins such as histone acetylation is analyzed by Western blotting.

Based on the passage, which statement describes Wnt proteins? A.They are composed of multiple subunits. B.They have a positive charge. C.They are synthesized in the smooth endoplasmic reticulum. D.They fold into their tertiary structure in the cytoplasm. Passage: Wnt proteins are a family of palmitoylated secretory proteins with isoelectric points around 9 that bind and activate the G protein-coupled receptor Frizzled, whose structure includes seven transmembrane α-helical domains.

Isoelectric point is around 9. This means that the protein will be protonated at pH 7. protonated= positive charge! A is incorrect because there is no information in the passage to support this response. C is incorrect because secretory proteins are synthesized in the rough endoplasmic reticulum. D is incorrect because folding of secretory proteins occurs in the rough endoplasmic reticulum.

Assume that S. typhi immediately enters the bloodstream from the small intestine. Of the following, which would be the first major organ that bloodborne S. typhi would encounter? A.Stomach B.Pancreas C.Large intestine D.Liver

LIVER!!!! Liver is going to remove the toxins :) - blood from the small intestine is transported first to the liver, which regulates nutrient distribution and removes toxins from the blood.

Which of the following graphs likely illustrates the changes in blood estrogen levels that occur in female rats after treatment with NPY alone? table: treatment and blood LH levels: saline 0.32 NPY alone 0.28

LOOK AT YOUR DAMN TABLE. With NPY alone, there was barely ANY change of LH. Therefore, the graph should be a straight line.

When researchers determined the total cellular concentration of ATP in AlP-exposed rat liver cells, they found the concentration to be equal to the control value. Which conclusion about the metabolic state of the cell is best supported by these data? A.Glycolytic flux is increased after AlP treatment. B.Glycolytic flux is decreased after AlP treatment. C.Citric acid cycle flux is increased after AlP treatment. D.Citric acid cycle flux is decreased after AlP treatment. passage: It was found that AlP exposure resulted in a 65% decrease in ATP levels and a 48% decrease in the rate of ATP synthesis. In addition, the effect of AlP on the activity of three ETC complexes was determined (Table 1). The activities of these complexes were determined independently of each other.

Mitochondrial ATP synthesis has been decreased based on the passage and the data in the table. Therefore, most of the atp that will be made will have to be in the cytosol outside of the mitochondria. This indicates that there will be an increase in glycolytic flux after AIP treatment. REMEMBER: ATP is NOT directly produced by the citric acid cycle. Those answer choices should be immediately crossed out.

A large carbohydrate is tagged with a fluorescent marker and placed in the extracellular environment around a macrophage. The macrophage ingests the carbohydrate via phagocytosis. Which cellular structure is most likely to be fluorescently labeled upon viewing with a light microscope soon after phagocytosis? A.Nucleus B.Golgi apparatus C.Lysosome D.Endoplasmic reticulum

Okay steps: macrophage eats the shit macrophage need to breakdown the shit because it is dangerous. macrophage needs help from organelle that can break down the shit. This organelle is the lysosome. - because when a macrophage ingests foreign material, the material initially becomes trapped in a phagosome. The phagosome then fuses with a lysosome to form a phagolysosome. Inside the phagolysosome, enzymes digest the foreign object. Of the cell structures listed, the labeled carbohydrate is most likely to be microscopically visualized within a lysosome (phagolysosome).

Why was it necessary for the researchers to determine the activity of the complexes independent of one another? A.Complex stability is lost if the complexes are able to interact structurally. B.The complexes have different cellular locations, and it is not feasible to isolate them together. C.The complexes all use the same substrates, so their use must be monitored separately. D.The reactions catalyzed by the complexes are coupled to one another.

POE: A- no this makes absolutely no sense. B- no they are all located in the inner mitochondrial membrane where oxidative phosphorylation takes place. C- they DO NOT use the same substrates. Complex two starts using fadh2 whereas complex one does not use FADH2 at all. D- Absolutely. inhibition of complexes I and II affects the activity of Complex III, which affects the activity of Complex IV

Which interpretation(s) is(are) consistent with the observations in the passage? Surface amino acids of Protein X are mostly hydrophilic in aqueous solution. Surface amino acids of Protein X are mostly hydrophilic in presence of DPC. Surface amino acids of Protein X are mostly hydrophobic in presence of DPC. A.I only B.II only C.III only D.I and III only

POE: Looking at roman numeral 1 first. in aqueous solutions, what is going to be exposed? Hydrophilic ish. Hydrophilic=water loving=aqueous. (cross out b and C) 3- this would be right. In figure two, there was a increase in fluorescence when DPC was attached. Hydrophobic= more fluorescence!!!!!! D. Figure 2 indicates that Protein X has few exposed hydrophobic surfaces in aqueous solution. In the presence of DPC, hydrophobic amino acid residues are exposed on the surface, as evidenced by ANS binding and increased fluorescence level

A prion is best described as an infectious: A.prokaryote. B.transposon. C.protein. D.virus.

PROTEINNNN

enzymes of gluconeogenesis

Pyruvate carboxylase PEP carboxykinase Fructose-1,6-phosphatase Glucose -6-phosphatase

Subunits of Protein X are linked covalently by bonds between the: A.thiol groups of methionine residues. B.thiol groups of cysteine residues. C.hydroxyl groups of serine residues. D.hydroxyl groups of threonine residues. The purified protein was analyzed by gel electrophoresis under native, denaturing, and denaturing/reducing conditions as shown in Figure 1.

Reducing agents are going to kill the disulfide bonds. Disulfide bonds are only made by cysteine residues. This goes directly with answer choice B.

Based on the results in Figure 2, what effect does DPC have on the hydrophobic amino acids in Protein X? Figure 2 shows that in the presence of DPC, Protein X has increased fluorescence. A. DPC phosphorylates these amino acids. B. DPC hydrolyzes these amino acids. C. DPC exposes these amino acids. D. DPC suppresses these amino acids.

Remember, hydrophobic residues include aromatic residues which are going to have some pretty huge fluorescence. Since the graph shows an increase in fluorescence, DPC has adopted a conformation that exposes hydrophobic residues on its surface

Which image shows an example of a cyclobutane pyrimidine dimer?

Remember: pyrimidines are one ring. purines are two rings. Dimer=2 fused together.

Based on Figure 1, how do the deletions of the hst3 or the rtt genes affect the lifespan of yeast? A.The deletion of the hst3 gene compensates for the ∆rtt-induced decrease in lifespan. B.The deletion of the hst3 gene has no effect on the ∆rtt-induced increase in lifespan. C.The deletion of the rtt gene compensates for the ∆hst3-induced decrease in lifespan. D.The deletion of the rtt gene has no effect on the ∆hst3-induced decrease in lifespan.

Rtt gene literally follows the wild type. Therefore, nothing really occurs.

Scientists are hopeful that NPY can be used in combination with GnRH to treat certain cases of female infertility. Individuals with a deficiency in what receptor system would be most likely to benefit from such a treatment? A.GnRH B.LH C.NPY D.LH and NPY

So from the passage, we see that in the table that LH secretion is not altered with only NPY (cross out LH and NPY) NPY enhances the normal function of GnRH in stimulating blood LH levels as seen in the table. Therefore, NPY would be great for patients with GnRH deficiency.

Based on the data presented in Table 1, which amino acid residues of PRR are involved in binding to prorenin? Residue 109 Residue 140 Residue 201 Residue 269 I and II only I, II, and III only II and III only II, III, and IV only

So, the table shows that residue 109 barely changes the binding affinity, but the others chang binding drastically. therefore, all of the others would be involved in binding.

G542X is another CFTR allele. If a female heterozygous for G542X bears a child fathered by a male heterozygous for the ΔF508 allele, what is the probability that the child would be homozygous for the G542X allele, given that neither parent has CF? A.0.00 B.0.25 C.0.75 D.1.00

The answer to this question is A because both parents would need to carry the G542X allele in order for a child to be homozygous for the G542X allele. Father does not carry this allele.

Based on the passage, myosin Va most likely directly binds: A. tubulin. B. actin. C. melanin. D. Rab27a.

The answer to this question is B because based on the passage, myosin Va is a motor protein. Motor proteins such as myosin Va move along microfilaments through interaction with actin.

The middle region, as opposed to either end, of each of the seven α-helical domains of Frizzled is most likely to contain a high proportion of which type of amino acid residue? A.Nonpolar B.Polar uncharged C.Positively charged D.Negatively charged passage: Frizzled, whose structure includes seven transmembrane α-helical domains.

Transmembrane domains cross the phospholipid bilayer and will ALWAYS be nonpolar and have a high proportion of hydrophobic residues.

In the absence of Frizzled activation, β-catenin is covalently modified and: A.bound by a proteasome to initiate degradation into short peptides. B.translocated into the Golgi body for secretion through exocytosis. C.engulfed by a lysosome where it is hydrolyzed by proteases. D.stored in vesicles until the signaling pathway is activated. Passage: CK1 and GSK3 sequentially phosphorylate β-catenin, which targets it for ubiquitination. Ubiquitination= proteolysis via a proteasome.

Ubiquitination= proteolysis via a proteasome. This goes directly with answer choice A.

Under anaerobic conditions, how many net molecules of ATP are produced by the consumption of 5 moles of glucose? A. 3 × 10^24 B. 6 × 10^24 C. 9 × 10^24 D. 1.2 × 10^25

Under anaerobic conditions, 2 moles of ATP are produced from each mole of glucose. Thus, 10 moles of ATP would be generated from 5 moles of glucose. Since there are 6 × 1023 molecules per mole, 10 moles of ATP is equal to 6 × 1024 molecules. - you must know production of ATP under anaerobic conditions and Avogadro's number relating molecules to moles to scale the answer to the correct value and units.

Dendrotoxin from the mamba snake blocks voltage-gated potassium channels in somatic motor neurons that regulate skeletal muscle contraction. In what way would initial exposure to dendrotoxin affect the ability of a somatic motor neuron to propagate an electrical signal in response to a stimulus? A.It would inhibit the initiation of an action potential. B.It would shorten the refractory period. C.It would prolong the action potential. D.It would prevent depolarization.

VG potassium (k) channels are related to hyper polarization of action potential. hyperpolarizing is stopping the action potential of the cell and repolarizing it. POE: A- this would happen if the question said Na+ channels, not K+ B- actually it would prolong the refractory period. refractory period happens right after stimulation. C- YES! the action potential would continue on until hyperpolarization occurs. If the VG K+ channels are turned off, then hyper polarization won't happen. D- is essentially saying the same thing as A. because if potassium ion channels are blocked, the membrane would fail to repolarize, extending the length of the action potential and simulating excessive muscle contractions.

Vasopressin regulates the insertion of aquaporins into the apical membranes of the epithelial cells of which renal structure? A.Collecting duct B.Proximal tubule C.Bowman's capsule D.Ascending loop of Henle

Vasopressin= ADH Antidiuretic hormone modulates how much water is excreted in urine. the collecting duct is the LAST STOP before the peepee comes out, so it will control it here! ADH receptors are specifically located are on the distal convoluted tubule (DCT) and collecting duct (CD).

Ubiquitination

a postranslational modification that targets a protein for degradation (proteolysis) by a proteasome.

AlP exposed to an aqueous solution in which pH range will result in the largest amount of PHOSPHINE production? A.pH < 4 B.4 < pH < 7 C.7 < pH < 10 D.pH > 10 equation: AIP + 3H+ ---->AI+ + 3PH3

according to lechatlier's principle, when you increase the 3h+ (making PH hella acidic) the equation is going to shift to the right and make more phosphine.

Researchers have noted that chloramphenicol (a commonly used antibiotic) is becoming less effective in treating typhoid fever. The best explanation for this observation would be selection: A.against chloramphenicol in ΔF508 heterozygotes. B.against chloramphenicol in wild-type homozygotes. C.for chloramphenicol resistance in populations of S. typhi. D.for ΔF508 CFTR, which cannot bind chloramphenicol.

c. according to the passage, S. typhi is the causative agent of typhoid fever. As S. typhi becomes resistant to chloramphenicol, this antibiotic will become less effective in treating typhoid fever. ANTIBIOTIC RESISTANCE= LESS EFFECTIVE

Based on the passage, which amino acid will most likely react with phosphine? A.Met B.Cys C.Ser D.Thr passage: It is thought that phosphine reacts with important sulfhydryl groups

cysteine has sulfhydryl groups and can make disulfide bonds. Methionine is the other amino acid that is sulfur containing, but cannot make disulfide bonds. Therefore cysteine is the correct answer.

what do proteases do?

estion into smaller protein fragments

enzymes shared by glycogenolysis

glycogen phosphorylase phosphoglucomutase glucose 6 phosphatase

low Km

high affinity for substrate low [S] required for 50% enzyme saturation

Which of the following best describes the phenotype of an individual who is heterozygous with one ΔF508 and one wild-type CFTR allele? A.More susceptible to typhoid fever than wild-type homozygotes and has CF B.More susceptible to typhoid fever than ΔF508 homozygotes and does not have CF C.More resistant to typhoid fever than ΔF508 homozygotes and has CF D.More resistant to typhoid fever than wild-type homozygotes and does not have CF

individuals homozygous for the ΔF508 allele have CF. Therefore, an individual who is heterozygous for the ΔF508 allele does not have CF. However, this individual is more resistant to typhoid fever because the passage states that human epithelial cells expressing ΔF508 ingest significantly fewer S. typhi than do epithelial cells that express wild-type CFTR.

Vmax equation

kcat(enzyme concentration)

enzyme efficiency

kcat/km

NPY is not classified as a releasing factor because it has: A.no effect on LH secretion. B.no effect on LH secretion without GnRH. C.no effect on LH secretion in combination with GnRH. D.the same effect on LH secretion as GnRH.

looking at the table, we see that LH doesn't do something by itself, but does combined with GnRH So you are left with B!

high Km

low affinity for substrate high [S] is required for enzyme to be half-saturated

Vmax

maximum initial velocity or rate of an enzyme-catalysed reaction.

Based on the passage, CatB and CatL most likely act on EGP in which of the following cellular compartments to facilitate membrane fusion? A.Endoplasmic reticulum B.Golgi apparatus C.Endosomes D.Cytosol Passage: The infectivity of VSV-EGP, vesicular stomatitis virus (VSV) particles engineered to contain EGP instead of VSV glycoprotein in the viral envelope, was reduced more than 99-fold by inhibitors of the mammalian proteases CatB and CatL. ALSO the passage literally says it: Recognizing the similarities between how EGP and the glycoproteins of related viruses are processed during viral infection, as well as how inhibitors of ENDOSOMAL acidification block EboV infection, researchers hypothesized that acid-dependent proteases trigger the conformational changes in EGP that are necessary for fusion.

the entry of the virus into the host cell requires CatB and CatL proteases and involves endocytosis through the fusion of the viral membrane with the host cell membrane. Internalization of viral particles through endocytosis is mediated by endosomes. C.

The enzyme encoded by the tdh2 gene catalyzes the reversible conversion of: A.3-phosphoglycerate to 1,3-bisphosphoglycerate. B.glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate. C.fructose-1,6-bisphosphate to glyceraldehyde-3-phosphate. D.2-phosphoglycerate to 3-phosphoglycerate. "Cells with single deletion of the tdh2 gene (Δtdh2), which encodes glyceraldehyde-3-phosphate dehydrogenase (GAPDH),"

this would go with answer choice B. glyceraldehyde 3 phosphate is converted to 1,3 BPG enzymes of glycolysis: Harry Potter Plows A Trashy Ginger Prostitute Pumping Ejaculatory Penises.

How many molecules of reduced electron carrier are generated during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle? One Two Three Four

three can i keep selling sex for money officer k---> S :nadh s----> f :fadh2 (done by succinate dehydrogenase) m---> o :nadh

Based on the data presented in Figure 1, overexpression of PRR: increases blood pressure in part through an angiotensin II-dependent pathway. increases blood pressure in part through an angiotensin II-independent pathway. decreases blood pressure in part through an angiotensin II-dependent pathway. I only II only III only I and II only

when looking at the graph, there is an increase in losartan (angiotensin II antagonist) indicating that there is independent pathway ish going on. there is also an increase in prorenin, but the next bars is prorenin+ losartan. these bars are brought back down. ---> this would be the dependence shown.

Melanosomes most likely move along microtubules that originate in and radiate from the: A. centrosome. B. kinetochores. C. Golgi apparatus. D. microfilaments under the plasma membrane. passage: Melanocytes use secretory lysosomes called melanosomes to transport melanin pigments. Melanosomes are transported along microtubules to the cell periphery where the melanosomes transfer to microfilaments prior to the release of melanin from the cell

A because microtubules are cellular structures that originate from centrosomes. - recall the organization of microtubules and their role in cellular transport.

The precursor of EGP is translated from a transcript that has had one nontemplated nucleotide added to the open reading frame. This change does not create or eliminate a stop codon. Compared with the protein sGP, which is produced from the unedited transcript, EGP most likely has the same primary: A. amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence. B. carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence. C. sequence as sGP except that EGP has one additional amino acid. D. sequence as sGP except that EGP has one less amino acid.

A because the addition of one nucleotide to the open reading frame of EGP results in a frameshift mutation and a diff carboxy-terminal domain

In humans, the characteristic tissue of which of the following organs is NOT derived from mesoderm? A. Brain B. Heart C. Kidney D. Skeletal muscle

A because the brain is part of the central nervous system, which is derived from ectoderm. Heart, kidney, and skeletal muscle are derived from mesoderm.

Based on the passage, Foxp3 affects ErbB2 expression in noncancerous mammary epithelium most likely by directly: A. inhibiting synthesis of ErbB2 mRNA. B. stimulating synthesis of ErbB2 mRNA. C. inhibiting synthesis of ErbB2 protein. D. stimulating synthesis of ErbB2 protein.

A because the passage states that cancerous mammary epithelium in female Foxp3sf/+ mice has 8- to 12-fold more ErbB2 mRNA than noncancerous mammary epithelium. Because the passage states that Foxp3 binds the ErbB2 promoter, Foxp3 regulation of ErbB2 expression occurs at the transcriptional, not translational level. Noncancerous epithelium has reduced ErbB2 mRNA expression compared with cancerous epithelium, which suggests that in noncancerous epithelium, Foxp3 inhibits synthesis of ErbB2 mRNA.

Lytic granules are generally released from CTLs when the T-cell receptors on these cells bind specifically to: A. viral antigens presented on the surface of virus-infected cells. B. growth factors secreted by helper T lymphocytes. C. B-cell receptors on activated B lymphocytes. D. constant regions of secreted antibodies.

A. because cytotoxic T lymphocytes target virus-infected cells by recognizing the viral antigen presented on the cell surface.

Scientists have hypothesized that NPY is necessary for the generation of the preovulatory LH surge, a hormonal event that triggers ovulation. Which of the following findings best supports this hypothesis? A.When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur. B.NPY release from the hypothalamus increases just prior to the preovulatory LH surge. C.NPY can enhance GnRH-stimulated LH secretion in female rats during the preovulatory period. D.NPY has no effect on GnRH-stimulated LH secretion in male rats.

A. this is a standalone question from the passage. Don't get confused!!!!

Which experimental approach(es) can be used to analyze the effect of ROS on the lifespan of yeast? Comparing the lifespans of: I. wild-type yeast versus yeast lacking antioxidant enzymes II. wild-type yeast versus yeast overexpressing antioxidant enzymes III. yeasts growing in the presence or absence of hydrogen peroxide A.I only B.II only C.II and III only D.I, II, and III

ALWAYS ALWAYS ALWAYS use the wild type as the control. all listed options influence ROS levels in yeast and can be used to analyze the role of ROS in regulating the lifespan of yeast.

Inhibition of phosphofructokinase-1 by ATP is an example of: allosteric regulation. feedback inhibition. competitive inhibition. A.I only B.III only C.I and II only D.II and III only

Absolutely feedback inhibition. (cross out A and B) This is absolutely not competitive inhibition. ATP does NOT bind to the active site. It binds to a spot other than the active site when is known as the allosteric site. because ATP, the end product of glycolysis, downregulates through feedback inhibition the activity of phosphofructokinase-1 by binding to a regulatory site other than the active site of the enzyme (allosteric regulation). In contrast, competitive inhibition involves competition for binding to the active site. - how the product of glycolysis (ATP) inhibits a regulatory enzyme of glycolysis (phosphofructokinase-1).

A homodimeric protein was found to migrate through SDS polyacrylamide gel electrophoresis (SDS-PAGE) with a mobility that matched that of a 45-kDa standard. What change in the experiment would increase the chances of observing the mobility expected for the 22.5-kDa monomer? A. Increasing the gel running time B. Adding a reducing agent C. Using a higher voltage D. Removing the SDS

B because adding a reducing agent would eliminate any disulfide bridges and allow the monomers to run separately-thus leading to a migration expected for the 22.5-kDa protein.

In humans, eggs and sperm are most similar with respect to: A. cell size. B. genome size. C. the time required for development. D. the numbers produced by a single individual.

B because both eggs and sperm contain a haploid number of chromosomes and therefore they are most similar with respect to their genome size.

Human breast cancer patients whose tumors overexpress HER2, the human homolog of ErbB2, may be treated with trastuzumab, an antibody that was developed to be highly specific for the extracellular domain of HER2. Given this, which of the regions of trastuzumab shown in the figure is(are) most likely highly specific for the extracellular domain of HER2? A. Regions 1 and 2 only B. Regions 1 and 3 only C. Regions 3 and 4 only D. Region 4 only

B because regions 1 and 3 correspond to the variable portion of the light and heavy chains, respectively, of an antibody. The variable region of an antibody will enable recognition of a particular antigen, such as HER2, which typically has elevated expression in breast cancer tumors. - the tips of the Y