MCAT Bio: Ch. 3 Drill Q&A Flashcards
Grading the drills
Why is ATP known as a “high energy structure” at neutral pH?
A) It exhibits a large decrease in free energy when it undergoes hydrolytic reactions.
B) The phosphate ion released from ATP hydrolysis is very reactive.
C) It causes cellular processes to proceed at faster rates.
D) Adenine is the best energy storage molecule of all the nitrogenous bases.
A) It exhibits a large decrease in free energy when it undergoes hydrolytic reactions.
Choice A is the best answer because it directly addresses the energetics of ATP hydrolysis.
Choice B discusses the reactivity of the released phosphate ion and not the structure of ATP (eliminate).
Choice C describes rate of cellular processes and not energy (eliminate)
Choice D talks about adenine structure and is not related to how ATP is a good energy storage molecule (eliminate).
Which of the following best describes the secondary structure of a protein?
A) Various folded polypeptide chains join together to form a larger unit.
B) The amino acid sequence of the chain.
C) The polypeptide chain folding upon itself due to hydrophobic/hydrophilic interactions.
D) Peptide bonds hydrogen-bonding to one another to create a sheet-like structure.
D) Peptide bonds hydrogen-bonding to one another to create a sheet-like structure.
The secondary structure of proteins is the initial folding of the polypeptide chain into alpha helices or beta-pleated sheets.
Eliminate Choice A b/c it describes quaternary protein formation.
Eliminate Choice B b/c it describes the primary protein structure.
Eliminate Choice C b/c it describes tertiary protein structure
Phenylketonuria (PKU) is an autosomal recessive disorder that results from the deficiency of the enzyme phenylalanine hydroxylase. This enzyme normally converts phenylalanine into tyrosine. PKU results in intellectual disability, growth retardation, fair skin, eczema, and a distinct musty body odor. Which of the following is most likely true?
A) Treatment should include a decrease in tyrosine in the diet.
B) The musty body odor is likely caused by a disorder in aromatic amino acid metabolism.
C) Patients with PKU should increase the amount of phenylalanine in their diet.
D) PKU can be acquired by consuming too much aspartame (artificial sweetener with high levels of phenylalanine).
B) The musty body odor is likely caused by a disorder in aromatic amino acid metabolism.
Phenylalanine and its derivatives are aromatic amino acids and high levels of this compound lead to musty body odor.
Use POE to get to this answer
A defect in phenylalanine hydroxylase (THB factor) would result in the buildup of phenylalanine. This means an excess of byproducts such as phenylacetate, phenylactate, and phenylpyruvate. The important part is the decrease in tyrosine.
A) Eliminate because patients with PKU should increase the amount of tyrosine in their diet.
C) Eliminate because patients with PKU should decrease the amount of phenylalanine in their diet.
D) Eliminate because PKU is a genetically acquired disorder (autosomal recessive) so it cannot be acquired through too much consumption of aspartame.
A genetic regulator is found to contain a lysine residue that is important for its binding to DNA. If a mutation were to occur such that a different amino acid replaces the lysine at that location, which of the following resulting amino acids would likely be the least harmful to its ability to bind DNA?
A) Glycine
B) Glutamate
C) Aspartate
D) Arginine
D) because it is basic
Lysine is basic and best at binding negatively charged DNA. Thus a mutation leading to another basic amino acid would cause the least change in ability to bind to DNA and cause less harm.
———————–
Basic: Histidine, arginine, and lysine
Acidic: Glutamate and aspartate
Neutral: Glycine
B & C) Eliminate bc they are acidic and will cause the most harm to its ability to bind DNA
A) Eliminate bc it is a neutral amino acid
Increasing the amount of cholesterol in a plasma membrane would lead to an increase in:
A) membrane permeability
B) atherosclerotic plaques
C) membrane fluidity at low temperatures but a decrease in membrane fluidity at high temperatures.
D) membrane fluidity at high and low temperatures
C) because cholesterol at low temperatures increases fluidity and cholesterol at high temperatures decreases fluidity
ring structure at low temperatures prevents packing and increases fluidity
A) Eliminate bc cholesterol fills the “holes” between the fatty acid tails and decreases membrane permeability, not increase
B) Eliminate b/c Atherosclerotic plaques relate to high levels of blood cholesterol, not membrane cholesterol
D) Eliminate bc temperature affects fluidity in the presence of cholesterol
A human space explorer crash lands on a plane where the naive inhabitants are entirely unable to A human space explorer crash-lands on a planet where the native inhabitants are entirely unable to digest glycogen, but are able to digest cellulose. Consequently, they make their clothing out of glycogen-based material. The starving space explorer eats one of the native inhabitants’ shirts and the natives are amazed. Based on this information, which of the following is/are true?
I. The explorer can digest alpha-glycosidic linkages
II. The native inhabitants can digest alpha-glycosidic linkages
III. The native inhabitants can digest starch.
A) I only
B) I and III only
C) II and III only
D) I, II, and III
A) because glycogen (and starch) are alpha-glycosidic linkages
Item I is true because humans can diest alpha glycosidic linkages like those found in glycogen. (Choice C eliminated)
Item II is false: cellulose contains beta glycosidic linkages. If the natives digest cellulose but not glycogen they cannot digest alpha glycosidic linkages. (Choice D eliminated)
Item III is false: starch also contains alpha glycosidic linkages. If the natives cannot digest glycogen then they likely cannot digest starch either. (Choice B eliminated).
Chapter 3 Passage Based Question (Pg 59)
In concluding that the hexachloroplatinate ions were bound to Photosygem 1 due to the attraction of opposite charges, the researchers apparently assumed that the structure of the membrane was:
A) determined solely by hydrophobic bonding
B) positively charged
C) covalently bound to the platinate
D) negatively charged
B) positively charged
The passage states that the ion is attracted to Photosystem I by the attraction of opposite charges (positively charged photosystem and negatively charged hexachloroplatinate ion).
Figure 1 indicates that:
A) photoactivation of the chloroplast membrane results in the reduction of the anhydride-containing molecule NADP+
B) electrons are lost from Photosystem 1 through conversion of NADPH to NADP+ and are replaced by electrons from Photosystem 2.
C) there is a net gain of electrons by the system
D) electrons are lost from Photosystem 1 through the conversion of NADP+ to NADPH but are not replaced by electrons from Photosystem 2.
A) photoactivation of the chloroplast membrane results in the reduction of the anhydride-containing molecule NADP+
B) electrons are lost from Photosystem 1 through conversion of NADPH to NADP+ and are replaced by electrons from Photosystem 2.
The main result of the light phase as depicted in Fig 1, is the reduction of NADP+ to make NADPH.
B) Eliminate because NADP+ is converted into NADPH not vice versa
C) Eliminate because any system, mass, and charge are conserved. Electrons move from one molecule to another but they are not created or destroyed in a chemical reaction.
D) Eliminate since Fig 1 depicts electrons moving from Photosystem 2 to Photosystem 1.
In addition to NADPH, the photosynthetic light phase must supply the dark phase with another molecule that stores energy for biosynthesis. Among the following, the molecule would most likely be:
A) ADP
B) CO2
C) Inorganic phosphate
D) ATP
D) ATP
the passage states that the light reactions supply the dark reactions with a high-energy substrate. The most likely candidate among the choices is ATP.
If NADP+ is fully hydrolyzed to its component bases, phosphates, and sugars, what type of monosaccharide would result?
A) A three carbon triose
B) A hexose
C) A pentose
D) An alpha D glucose
C) a pentose
NADPH contains ribose, a pentose
If in a given cell the photosynthetic dark phase were artificially arrested while the light phase proceeded, the cell would most likely experience:
A) decreased levels of NADPH
B) increased levels of NADPH
C) increased levels of carbohydrate
D) increased activation of the chloroplast
B) increased levels of NADPH
The light phase makes NADPH and the dark phase consumes it. In the absence of the dark phase, NADPH will continue to be produced but none will be consumed making NADPH levels rise.
C) eliminate because the dark phase is responsible for biosynthesis such as carbohydrate production so this will decrease not increase.
D) eliminate b/c amount of light and photoactivation should remain the same.
To determine the primary structure of the protein portion of Photosystem 1, a series of cleavage reactions was undertaken. To break apart the protein, the most logical action to take would be to
A) decarboxylate free carboxyl groups
B) hydrolyze peptide bonds
C) repolymerize peptide bonds
D) hydrolyze amide branch points.
B) hydrolyze peptide bonds
Proteins are composed of amino acids residues which are joined together by peptide bonds during the translation process. To split the protein into smaller pieces, proteases and chemical reagents act to hydrolyze the peptide bond, reversing the biosynthetic process.
A researcher examined a sample of the principal substance produced by the photosynthetic dark phase and concluded that he was working with a racemic mixture of glucose isomers. Which of the following experimental findings would be inconsistent with such a conclusion?
A) The sample is composed of carbon, hydrogen, and oxgen only.
B) The sample consists of an aldohexose
C) The sample rotates the plane of polarized light to the left.
D) The sample is optically inactive
C) The sample rotates the plane of polarized light to the left.
A racemic mixture contains equal quantities of two stereoisomers that rotate plane-polarized light in opposite directions. Since there are equal quantities of both, racemic mixtures are optically inactive.
C) is inconsistent with this conclusion that the sample is racemic.
All other choices are consistent with the conclusion that the sample is a racemic mixture of glucose.
A) Eliminate because carbs (glucose included) are made of only carbon, hydrogen, and oxygen.
B) Eliminate because glucose with six carbons and a carbonyl group on the 6th carbon is an aldohexose.
D) eliminate because racemic mixtures do not rotate light.
