9/26 flashcards week 1

Why was Ca2+ added to the oxygen-infused, electrolyte rich bath? (Where the experiment took place with isolated muscle cells) To allow Ca2+ to: A) be sequestered inside the sarcoplasmic reticulum of the muscle fibers B) enter the mitochondria to promote sufficient ATP production by muscle fibers C) increase the affinity of muscle fiber myoglobin for oxygen D) facilitate the binding of Ach to its receptor on the muscle fiber

A Why A? Calcium is a mineral that plays an important role in skeletal muscle contraction. The sarcoplasmic reticulum is in muscle fiber thats responsible for tightly regulating intracellular calcium concentration by sequestering calcium ions during periods of relaxation and releasing them into the cytosol during periods of contraction Not B? Muscle fibers primarily generate ATP through glucose metabolism, not really involving calcium Not C? Iron facilitates oxygen binding by the proteins myoglobin and hemoglobin. Not calcium. Not D? At the neuromuscular junction, the entry of calcium ions into the presynaptic neurons leads to the release of Ach. BUT because the muscle fibers in the electrolyte bath were completely isolated, no presynaptic neurons were present, so calcium had no job to fulfill.

Given that 1 unit of activity equals 1nmol of substrate converted per minute, what was the specific activity of purified TMPK toward dTMP in exp 1? (3 micrograms of TMPK was used. reaction proceeded for 30 minutes, and it was determines that 0.3nmol of dTMP and 0.15 nmol of UMP became phosphorylated) A) 1.7x10^-3 units/microgram B) 3.3x10^-3 units/migrogram C) 5.0x10^-3 units/microgram D) 1.0x10^-2 units/microgram

B Why B? 0.3nmol/30min = 0.01 units of activity. 0.01units/3micrograms protein = 3.3x10^-3 units/microgram A, C, and D wrong math

If the titation of H2PO4- in a urine sample was continued until all of the acid in the solution was neutralized, how many equivalents of NaOH would be needed to fully neutralize the solution? (1M NaOH) HPO2- + H+ <-> H2PO4- A) 1 B) 2 C) 3 D) 4

B Why B? H2PO4- has 2 protons to give up to be neutralized Not A C D? too little or too many protons

NDPK and TMPK (they use ATP to phosphorylate things) are members of a class of enzymes called: A) phosphatases B) kinases C) phosphorylases D) ligases

B Why B? Kinases transfer the y-phosphate from ATP to other molecules. Not A? Phosphatases remove phosphate groups from molecules by hydrolysis Not C? Phosphorylases break bonds by adding inorganic P across them. Since NDPK and TMPK use ATP, not inorganic P, this is wrong Not D? Ligases link 2 molecules using ATP hydrolysis, releasing Pi

VEGF signaling also leads to the dilation of existing blood vessels. One function of vasodilation is to A) increase body temperature in cold environments B) increase blood flow to the intestines following a meal C) decrease blood flow to the kidneys during dehydration D) maintain blood pressure following an episode of fluid loss

B Why B? increasing blood flow promotes delivery of oxygen and nutrients to tissues. This promotes digestion. Not A? Cold environments make you vasoconstrict Not C? Vasodilation would increase blood flow rate to the kidney Not D? vasodilation would further decrease BP

DAB most likely inhibits which reaction? (graph on back) A) phosphorylation of glucose B) Phosphorolysis of glycogen C) export of lactate from cells D) hydrolysis of glycogen

B Why B? phosphorolysis is degradation of glycogen bonds. 2A shows that after exercise to exhaustion, glycogen stores in the hippocampus are partially depleted. This suggests the DAB prevents glycogen degradation. So, inhibiting phosphorolysis is the only option that lead to decreased glycogen degradation Not A? inhibiting phosphorylated glucose would inhibit g6p available, so glucose degradation would increase to replenish the g6p needed for glycolysis Not C? Inhibition of lactate export would cause an increase in lactate concentration. 2B shows that lactate is present at DECREASED levels in rats with DAB Not D? Figure 2 indicates that DAB inhibits glycogenolysis. However, the first step in glycoegen onlysis breaks a bond in glycogen by adding phosphate across it (phosphorolysis) rather than adding water to it, which would be hydrolysis.

To further study anti-nAChR antibody function, the researchers wanted to replicate the experiments described in the passage in other mammalian animal models, but their research proposal was rejected by other scientists. Why was the proposal rejected? (Exp1: isolated skeletal muscled in a Ca2+ and glucose bath, action potentials used to evaluate contractile function) (Exp 2: 3 muscle sample groups, anti-nAChR antibodies, ACh, or both) A) Because resting muscle tension cannot be used as a control for post infusion tension B) Because the electrolyte rich baths were infused with inconsistent amounts of Ach C) Because skeletal muscles in the body would not be exposed to autoimmune antibodies D) Because animal models cannot be used to study human diseases

B Why B? use concentrations x Ach used, get double the result Not A? This control has to be used Not C? autoimmune antibodies are synthesized by B cells and transported to all body tissues, including skeletal muscle Not D? mice and other rodents are frequently used as animals models

Surgical removal of the ovaries is one treatment for endometriosis. Complete removal of the ovaries would most likely result in which of the following? A) Significant increase in bone mass B) Degeneration of breast tissue C) Severe immune system impairment D) Increased frequency of menstruation

B Why B? withdrawl of estrogen and progesterone leads to irritability, fatigue, anxiety, breast atrophy, infertility, and decreased bone density Not A? Withdrawl of female sex hormones leads to greater osteoclast activity - diminishing bone mass instead of increasing it Not C? Ovarian function is not related to immune response, would not have an effect Not D? Removal of ovaries would lead to NO menstrual cycles

Acetylation of lysine residues in histones increases gene expression because: A) DNA is tightly bound to negatively charged amino acids on histones B) the carboxyl oxygen atoms in acetyl groups form h-bonds with nitrogenous bases C) the salt bridges between charged amino acids and phosphate groups are disrupted D) lysine residues in histones associate with positively charged phosphate groups in DNA

C Why C? DNA winds around histones to create nucleosomes. Genes that are actively transcribed are unwound stretches (euchromatin). DNA is predominantly negatively charged due to the phosphate groups on the backbone. Histones associate with DNA by forming salt bridges between positive amino acids and negative phosphates. Acetylation of these disrupts this. Not A? Negatively charged amino acids would be REPELLED by the negatively charged phosphate groups on DNA Not B? The carboxyl oxygen atoms in the acetyl groups form hydrogen bonds with WATER not DNA Not D? Lysine residues are positively charged, and associate with NEGATIVE phosphate groups

Activity assays were performed to determine the best purification method for a cell-free extract of Protein X. The results of the assays are shown on the backside. Based on the data, what was the highest purification yield of Protein X? A) 0.5% B) 25% C) 50% D) 75%

C Why C? Total activity of unpurified extract (800mg)(0.5u/mg) = 400u. Total activity of alcohol prep (200mg)(1u/mg)=200u. 400/200 = 50% A B and D wrong math

Strontium is converted to yttrium by which of the following processes? ( strontium-90 goes to Yttrium-90) A) electron capture B) positron emission C) gamma ray emission D) beta emission

D Why D? beta emission is the ejection of an electron or positron from the nucleus. Most common is emission of a nuclear electron, which results from a neutron decaying into a proton. This results in the atomic number to increase by one Not A and B? emission of a positron and electron capture both result in a decrease of atomic numbers, which would convert strontium to rubidium Not C? gamma ray emittion does not effect atomic number or mass of the nucleus

If the proposed mechanism in the passage is correct (glycogenolysis), which of the following metabolic reactions must occur for astrocytes to convert glycogen into a fuel for neurons? I. Glucose-1-phosphate to glucose-6-phosphate II. Glucose to glucose-6-phosphate III 3-phosphoglycerate to 2-phosphoglycerate IV pyrucvate to acetyl-CoA V Pyruvate to lactate

I III and V Why I? first step of glycogenolysis produces g-1-p which is then converted to g-6-p Why III? g-6-p then enters glycolysis which involves the conversion of 3-pg to 2-pg Why V? final step of glycolysis produces pyruvate, which is then fermented into lactate Not II? in glycogenolysis, glycogen is converted to g-1-p and then to g-6-p Not IV? pyruvate is converted into acetyl-CoA prior to the TCA cycle, once acetyl-CoA is formed, it can't be converted into lactate

Fission of 235U produces multiple elements. Based on Figure 1, which of the following is the atomic number of the most common fission product? (on back) a. 38 b. 55 c. 92 d. 133

b why? 133ish on graph - look at periodic table okay Cs atomic weight is 132.9 and the ATOMIC NUMBER IS 55. not a or c? not the greatest peak not d? that's the atomic weight not number

This type of uranium required for Reaction 1 can be formed by the emission of an alpha particle from which of the following isotopes? (reaction 1: 235U -> 90Sr + Z) a. 239 U b 239 Pu c 235 U d 235 Pa

b why? alpha decay is emission of 2 neutrons and 2 protons. the resulting nucleus has a mass number that is 4 less than the mass number of the original. the daughter has an atomic number that is 2 less than the original not a? alpha decay of a would = 235Th not c? alpha decay would lead to 231Th not d? alpha decay would result in 231Ac

Which of the following changes could be made to this experiment to increase the relative signal intensity of peaks that correspond to smaller m/z values? a. increase the pH of the sample solvent b. decrease the pH of the sample solvent c. decrease the concentration of the sample d. increase the concentration of the sample

b why? in MS signal intensity corresponds to the relative quantity of ions. Lowering pH increases the concentration of protons, and results in a greater number of multiply charged molecules not a? increase in pH would cause less protons, would do opposite not c? decreasing the concentration of the sample would decrease the amount of sample present and cause a decrease in the overall signal intensity not d? increasing the concentration of the sample would increase the amount of the sample present, but would not affect the relative heights of the peaks because ALL the peaks would increase

Thin-film interference is observed when polychromatic light is incident on an interface formed by two semitransparent media. Which of the following best explains why thin-film interference generates a multicolored array when 2 semitransparent fluids are used? a. the density of the top fluid varies along the fluid interface b. the thickness of the top fluid varies along the fluid interface c. the osmolarity of the top fluid varies along the fluid interface d. the presence of 2 semitransparent fluids decreases the frequency of reflection

b why? thin-film interference = multicolored arrays generated by reflections of 2 layers of semitransparent media. thin film interference is influenced by the thickness of the film because light waves of a specific wavelength that interfere constructively or deconstructively at one film thickness may no longer interfere in the same fashion at a different thickness. not A or C? variations in fluid density and osmolarity cant explain changes to the interference pattern of light that are necessary to produce multicolored arrays. not D? layering 2 semitransparent fluids INCREASES the frequency of reflection events

A concave mirror has a 4m radius of curvature. This mirror will focus distant objects at a location that is approximately: a. 4m in front of the mirror b. 4 m behind the mirror c. 2 m in front of the mirror d. 2 m behind the mirror

c why? concave mirrors will form an inverted, real image when outside of the focal point. inside the focal point, it will do an upright virtual image. convex always makes Upright virtual. focal length = r/2 = 2m. for concave, the point of focus is in FRONT of the mirror. the opposite for convex.

If researchers needed to synthesize Compound 1 before beginning the synthesis of Compound 5, which of the following sugards would requore the fewest steps if used as a starting material? (Compound 1 on other side) a. fructose b. ribose c. mannose d. glucose

c why? mannose hayworth projection is D U U D not A? aldose, anomeric carbon at c2 Not B? 5 carbons Not d? glucose is D D U D

Prior to adduct formation, DNA was isolated from a cellular mixture. Membrane lipids were first extracted using an organic solvent, which was then evaporated, and DNA was precipitated from the aqueous layer using cold ethanol and sodium acetate. Which of the following mechanisms allowed for DNA precipitation? a. sodium acetate lowers the pH of the solution, changing the conformation of the DNA structure b. ethanol has a greater polarity than water, pulling DNA out of the aqueous solution c. cold ethanol lowers the temperature of the solution, precipitating frozen DNA d. sodium acetate forms ionic bonds with DNA, neutralizing its charge

d why? DNA is polar, so will stay in the aqueous phase because like dissolves like. To fix this, the charge must be neutralized. Gentle mixing with ethanol disrupts the hydration shell around DNA molecules. Then sodium cations neutralize DNA's charge via ionic bonding with phosphate groups. not A? sodium acetate is a base, would increase the solution's pH not b? ethanol has a lower polarity than water. this isn't what happens. not c? cold ethanol is used because its less likely than warm to disrupt h-bonds in DNA, but it's not cold enough to freeze DNA.

Suppose that a simple circuit comprising one voltage source and one metallic resistor yields current I. if the resistor were replaced with another resistor that is identical except that it has 75% lower conductivity, then I would a. increase by a factor of 2 b. increase by a factor of 4 c. decrease by a factor of 2 d. decrease by a factor of 4

d why? metal conductibity = 1/resistivity. I = V/R, 1/4R - 1/4 original value