BIEN135 Midterm 1 Discussion Questions

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1.1) When two atoms approach each other closely, the energy goes up because the nuclei of the atoms repel each other (True/False)

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1.1) When two atoms approach each other closely, the energy goes up because the nuclei of the atoms repel each other (True/False)

False

<p>False</p>

1.3) An N-H•••O=C hydrogen bond has optimal energy when it: A. is bent B. is linear C. has a donor-acceptor distance of 4Å D. has a donor-acceptor distance of 2Å

B. is linear Hydrogen bonds have optimal energy when the dipoles are colinear

1.11) When two atoms approach closer than _________ the interaction energy goes up very sharply.

their van der Waals radii

<p>their van der Waals radii</p>

1.13) At room temperature, the value of _________ is about 2.5 kj*mol-1.

thermal energy

1.17) Order the following elements from lowest electronegativity to highest: P, C, N, H, O, S. Use your ordering of the atoms to rank the following hydrogen bonds from weakest to strongest: A. SH----O B. NH----O C. OH----O

A < B < C H = P < C = S < N < O

<p>A &lt; B &lt; C H = P &lt; C = S &lt; N &lt; O</p>

1.25) What chemical properties have led to DNA being selected through evolution as the information molecule for complex life forms instead of RNA?

  • The 2'-OH in an RNA can react to break the backbone by forming a cyclic 2'-3' phosphodiester bond. DNA lacks a 2'-OH group.

  • In the event of DNA on one strand being lost or damaged it can be replaced or repaired using the other strand. RNA is single-stranded.

<ul><li><p>The 2&apos;-OH in an RNA can react to break the backbone by forming a cyclic 2&apos;-3&apos; phosphodiester bond. DNA lacks a 2&apos;-OH group.</p></li><li><p>In the event of DNA on one strand being lost or damaged it can be replaced or repaired using the other strand. RNA is single-stranded.</p></li></ul>

2.3) H-bond acceptor, H-bond donor, H-bond acceptor, methyl group is the pattern of potential interactions at the edge of which Watson-Crick base pair? A. A-T B. G-C C. A-A D. C-G E. T-A

A. A-T

<p>A. A-T</p>

2.16) In water, why is base stacking relatively more important than hydrogen bonding to forming a DNA double helix?

  • Water can make H bonds with the DNA base. The energetic gain in interactions when those H bonds are satisfied by other nucleotides is not great.

  • The many small polar interactions along the nucleotide rings that are formed upon base stacking are energetically more favorable.

2.19) Why do most transcription factors interact with the major groove of B-form DNA rather than the minor groove?

  • The major groove has more distinguishing interactions than the minor groove.

  • The depth of the major groove is compatible with the interaction with protein structural elements such as α helix

<ul><li><p>The major groove has more distinguishing interactions than the minor groove.</p></li><li><p>The depth of the major groove is compatible with the interaction with protein structural elements such as α helix</p></li></ul>

3.3) Many sugars are structural isomers of other sugars (True/False)

True

3.5) Which of the following fatty acids are unsaturated? (see image) A. B. C. D. E. F.

A, B, E

  • saturated: all C-C single bonds

  • unsaturated: with C=C double bonds

<p>A, B, E</p><ul><li><p>saturated: all C-C single bonds</p></li><li><p>unsaturated: with C=C double bonds</p></li></ul>

3.10) Inositol is an unusual six-carbon sugar because it does not contain an _______ atom in its ring.

oxygen

<p>oxygen</p>

3.11) A lipid _____ is a section of membrane with both composition and physical properties that are distinct from the surrounding membrane.

raft

3.12) In eukaryotic cells, intracellular compartments bounded by membranes are called _____________.

Organelles - subcellular structure or chamber often bounded by a membrane that is specialized for particular functions

3.14) What information can be gained more readily from the Fischer representation than from the Haworth representation?

  • Fischer projection is convenient for comparing stereochemistry of different sugar isomers

  • Haworth representation is convenient for three-dimensional structures.

<ul><li><p>Fischer projection is convenient for comparing stereochemistry of different sugar isomers</p></li><li><p>Haworth representation is convenient for three-dimensional structures.</p></li></ul>

3.17) The O blood type group is known as the universal donor. What property of the glycosylation pattern of O-type individual allows for the blood to be transfused into individuals of any blood type without eliciting an immune response?

  • A, B, and AB contain the same tetrasaccharide that defines the O blood type

  • No immune reaction.

3.18) What molecular forces drive the formation of bilayers when phospholipids are immersed in water?

  • The charged phospho-head groups point out to form a favorably charged interface with water

  • The hydrophobic alkyl chains

4.2) The secondary structure of a protein refers to the extent and order of its α helices and β sheets. (True/False)

True

4.3) Which of the following statements regarding protein domains is NOT true? A. Secondary structural elements of a domain generally pack so that a hydrophobic core is formed. B. A protein domain normally contains 50-200 residues. C. Protein domains are units of tertiary protein structure. D. Proteins are only comprised of one protein domain.

D

4.5) Most protein conformational changes involve breaking and reforming several covalent bonds along the polypeptide chain. (True/False)

False

4.7) The least restricted φ and ψ angles are found in polypeptides in which class of secondary structure? A. right-handed α helix B. β sheet C. left-handed α helix D. loop

D

4.9) Soluble proteins have mostly _______ chains on the inside and mostly ______ side chains on the outside

Hydrophobic; hydrophilic

4.11) SH2 domains bind to ________ and SH3 domains bind to ________.

phosphotyrosine peptide, polyproline peptide

4.14) A group of scientists isolate a novel strain of bacteria. ...In this strain, the lipid bilayer is twice as thick (70 Å) as that of a normal strain of bacteria. ...If you isolated a single transmembrane helix from a protein in this strain, how long would you expect it to be?

  • length of strain = 70 Å

  • length of single residue = 1.5 Å

  • number of residues = 70 / 1.5 = 47

4.16) Using the hydrophobocity scale in Figure 4.74, calculate the hydrophobicity index for the 19 contiguous residue window defined by the following sequence: Pro-Gly-Ala-Val-Val-Ile-Trp-Phe-Val-Val-Met-Ser-Ala-Ile-Ile-Phe-Tyr-Ala-Thr Could this segment be part of a transmembrane helix?

0.4 + 4.5 + 2.1 - 2.1 - 2.1 - 4.5 - 8.8 - 7.1 - 2.1 - 2.1 - 2.9 + 2.1 + 2.1 - 4.5 - 4.5 - 7.1 - 2.9 + 2.1 + 0.8 = -36.6 -36.6/19 = -1.9 This segment could be part of a transmembrane helix as the hydrophobicity index is negative and there are no charged residues.

<p>0.4 + 4.5 + 2.1 - 2.1 - 2.1 - 4.5 - 8.8 - 7.1 - 2.1 - 2.1 - 2.9 + 2.1 + 2.1 - 4.5 - 4.5 - 7.1 - 2.9 + 2.1 + 0.8 = -36.6 -36.6/19 = -1.9 This segment could be part of a transmembrane helix as the hydrophobicity index is negative and there are no charged residues.</p>

4.17) Why are isolated secondary structural elements not usually stable in isolation, even though all backbone hydrogen bonds are satisfied?

  • the hydrophobic effect is the major force driving

  • isolated secondary structural elements cannot bury hydrophobic residues away from water

4.19) The simplest form of the Ramachandran diagram (shown in the figure) is calculated by ignoring hydrogen bonding, interactions with water, and the hydrophobic effect. Why is this simple form of the diagram still an effective predictor of protein backbone conformation?

  • The simplest Ramachandran diagram shows which conformations are prevented due to interatomic collisions

  • The repulsive energy of such collisions is so high that it dominates the forbidden regions of the diagram

  • Taking the other forces into account modulates the favorability of some regions of the Ramachandran diagram, but the overall forbidden and allowed zones are not changed.

<ul><li><p>The simplest Ramachandran diagram shows which conformations are prevented due to interatomic collisions</p></li><li><p>The repulsive energy of such collisions is so high that it dominates the forbidden regions of the diagram</p></li><li><p>Taking the other forces into account modulates the favorability of some regions of the Ramachandran diagram, but the overall forbidden and allowed zones are not changed.</p></li></ul>

4.25) A scientist isolates a membrane protein that transports sodium ions from inside the bacterium (where it is at low concentration) into solution (where the sodium ion concentration is high). Does this protein use active or passive transport? What are two energy sources for accomplishing this transport?

  • Active transport

  • ATP hydrolysis and light energy

5.2) An environment profile in the 3D-1D profile method compares: I. the stability of the amino acid in varying solvents II. the burial of each amino acid in the structure III. the hydrophobicity surrounding each amino acid IV. the type of secondary structure element containing each amino acid

A. I, II, III, and IV B. I and IV C. II, III, and IV D. II and IV

C.

<p>C.</p>

5.3) Protein domains can be assembled together in many different ways, because surface sidechains can be mutated easily without losing protein stability. (True/False)

True

  • Protein molecules are dynamic systems in which the positions of the atoms fluctuate considerably

  • The flexibility of the protein structure allows the substitution of sidechains with quite different shapes

5.4) In contrast to ribonuclease, some proteins cannot fold without the assistance of proteins known as molecular chaperons. This means the thermodynamic hypothesis of protein folding does not apply to these proteins. (True/False)

False

  • Chaperones prevent aggregation by repeatedly binding to and releasing protein chains until they are folded properly

  • The acquisition of specific structure by a particular protein is still a function of the sequence of the protein itself

5.7) The core of a protein generally contains residues from the __________ class of amino acids.

hydrophobic/nonpolar

5.8) Globin proteins bind the iron-containing ________ cofactor.

heme

5.10) According to the BLOSUM substitution matrix, the most conservative mutation from tryptophan (W) other than to itself, is to __________, which has a score of __________.

Tyrosine (Y), 2

<p>Tyrosine (Y), 2</p>

5.15) A tryptophan is substitute by a tyrosine than a tryptophan is substituted by a cysteine?

Likelihood = 2^(Score/2) W-Y: 2^(2/2) = 2 W-C: 2^(-2/2) = 0.5 W-C is 2/0.5 = 4 times more likely W-Y substitution is more conservative than W-C

<p>Likelihood = 2^(Score/2) W-Y: 2^(2/2) = 2 W-C: 2^(-2/2) = 0.5 W-C is 2/0.5 = 4 times more likely W-Y substitution is more conservative than W-C</p>

5.17) Why is the sequence similarity generally higher than comparing two globins from mammals than when comparing a globin from a mammal and a globin from a plant?

  • both the mammalian and plant globin could have been derived from a single ancestral globin and expanded through gene duplication

  • the mammalian globin share a common ancestor with each other that is more recent

5.19) What characteristics define a protein domain?

  • contain 50-200 residues

  • have a distinct topology

  • a self-contained hydrophobic core

    • the components forming a quaternary structure can be domains (myoglobin is a domain itself does not have domains)

5.22) How many folds are represented below? Describe what CATH class each fold belongs to and how the secondary structure is arranged in each fold.

  • there are four folds

  • A & E have the same Rossmann like mixed α/β fold

  • B has a unique mixed α/β fold

  • C has an all β sheet structure arranged in a barrel

  • D & F have a four helix bundle

<ul><li><p>there are four folds</p></li><li><p>A &amp; E have the same Rossmann like mixed α/β fold</p></li><li><p>B has a unique mixed α/β fold</p></li><li><p>C has an all β sheet structure arranged in a barrel</p></li><li><p>D &amp; F have a four helix bundle</p></li></ul>