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MCAT > Biochem: Ch 6, 7 > Flashcards

Biochem: Ch 6, 7 Flashcards

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1
Q

DNA

A

macromolecule that stores genetic info in all living organisms

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2
Q

nucleoside structure

A

five carbon sugar bonded to nitrogenous base

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3
Q

nucleotide structure

A

nucleosides with 1-3 phosphate groups added

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4
Q

nucleotides in DNA contain

A

deoxyribose

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5
Q

nucleotides in RNA contain

A

ribose

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6
Q

chargaff’s rules

A

purines and pyrimidines are equal in number in a DNA molecule

because of base pairing, A = T and C = G

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7
Q

what can cause denaturation of DNA?

A

things that can disrupt the hydrogen bonding and base pairing of DNA

alkaline pH, chemicals like formaldehyde and urea

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8
Q

backbone of DNA

A
  • alternating sugar and phosphate groups
  • formed as nucleotides are joined by 3’-5’ phosphodiester bonds
    • phosphate group links the 3’ carbon of one sugar to the 5’ phosphate group of the next incoming sugar in the chain
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9
Q

DNA and RNA strands have an overall ___ charge

A

negative

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10
Q

purine structure

A

contains two rings

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11
Q

pyrimidine structure

A

one ring

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12
Q

purine names

A

adenine, guanine

(PURe As Gold)

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13
Q

pyrimidine names

A

cytosine, uracil, thymine

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14
Q

aromatic rules

A
  1. cyclic
  2. planar
  3. conjugated
  4. Huckel’s rule: 4n+2 pi electrons
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15
Q

Huckel’s rule

A

4n+2 pi electrons

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16
Q

aromatic compounds

A

stabilized due to delocalized pi electrons

unreactive

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17
Q

nucleic acid stability

A

purines and pyrimidines contain nitrogen in their aromatic rings –> even more stable

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18
Q

watson-crick model

A
  1. the two strands of DNA are antiparallel
  2. sugar-phosphate backbone is on the outside of the helix with the nitrogenous bases on the inside
  3. complementary base pairing
  4. chargaff’s rules
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19
Q

double helix of DNA

A

two linear polynucleotide chains of DNA are wound together in spiral orientation along common axis

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20
Q

complementary base pairing

A

A - T via 2 hydrogen bonds

G - C via 3 hydrogen bonds

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21
Q

complementary base pairing advantages

A

G-C 3 hydrogen bonds –> strong

hydrogen bonds and hydrophobic interactions between bases provide stability to double helix

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22
Q

B-DNA

A

right handed helix

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23
Q

Z-DNA

A

left handed helix

unstable

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24
Q

denatured, single stranded DNA can be reannealed if

A

the denaturing condition is slowly removed

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25
Q

probe DNA

A

DNA with known sequence

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26
Q

in eukaryotes, DNA is wound around ____(_) to form _____, which may be stabilized by ___(_)

A

histones (H2A, H2B, H3, H4)

nucleosomes

another histone protein (H1)

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27
Q

DNA and its associated histones make up

A

chromaatin

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28
Q

heterochromatin

A

dense, transcriptionally silent DNA that appears dark under light microscopy

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29
Q

euchromatin

A

less dense, transcriptionally active DNA that appears light under light microscopy

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30
Q

telomeres

A

ends of chromosomes

contain high GC content to prevent unraveling of the DNA

slightly shortened during replication

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31
Q

can shortened telomeres be reversed?

A

slightly by enzyme telomerase

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32
Q

centromeres

A

located in middle of chromosome

hold sister chromatids together until they are separated during anaphase in mitosis

contain high GC content to maintain a strong bond between chromatids

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33
Q

nucleoproteins

A

proteins that associate with DNA

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34
Q

replisome

A

replication complex

set of specialized proteins that assist the DNA polymerases

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35
Q

DNA is first unwound at _____ by _____, which produces ____

A

origin of replication

helicases

replication forks on either side of the origin

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36
Q

prokaryotic chromosome

A

circular

contains only one origin of replication

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37
Q

eukaryotic chromosome

A

linear

contain many origins of replication

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38
Q

how are unwound strands of DNA kept from reannealing or being degraded?

A

single-stranded DNA binding proteins

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39
Q

single-stranded DNA binding proteins

A

keep unwound strands of DNA from reannealing or being degraded

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40
Q

supercoiling

A

causes torsional strain on DNA molceule

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41
Q

DNA topoisomerases

A

create nicks in DNA molecule, allowing relaxation of torsional pressure

release supercoiling

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42
Q

how is supercoiling released?

A

DNA topoisomerases

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43
Q

DNA replication is ____

A

semiconservative

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44
Q

semiconservative

A

one old parent strand and one new daughter strand is incorporated into teach of the two new DNA molecules

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45
Q

DNA cannot be synthesized without ___, so ___

A

an adjacent nucleotide to hook onto

small RNA primer is put down by primase

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46
Q

DNA polymerase

eukaryotes vs prokaryotes

A

eukaryotes: alpha, delta, e
prokaryotes: III

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47
Q

DNA polymerase

A

read the template DNA 3’ to 5’ and synthesizes new strand 5’ to 3’

proofread its work and excises incorrectly matched bases

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48
Q

DNA strands during replication

A

leading strand: requires only one primer and can then be synthesized continuously in its entirety

lagging strand: requires many primers and is synthesized in discrete sections called Okazaki fragments

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49
Q

DNA polymerase I

A

prokaryotes

remove RNA primers

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50
Q

RNase H

A

eukaryotes

remove RNA primers

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51
Q

DNA ligase

A

fuses the DNA strands together to create one complete molecule

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52
Q

DNA replication in eukaryotes steps

A
  1. helicase unwinds DNA
  2. primase puts down small RNA primer
  3. DNA polymerase reads and synthesizes new strand 5’ to 3’
    1. leading strand made continuously
    2. lagging strand made in okazaki fragments
  4. RNase H removed RNA primers
  5. DNA polymerase fills in DNA
  6. DNA ligase fuses DNA strands together
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53
Q

helicase

A

unwinds DNA into two single strands

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54
Q

leading strand

A

strand that is copied in continuous fashion, in the same direction as the advancing replication fork

read 3’ to 5’ and complement synthesized in 5’ to 3’

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55
Q

lagging strand

A

copied in a direction opposite the direction of the replication fork

since DNA polymerase can only synthesize in 5’ to 3’ direction from a 3’ to 5’ template, okazaki fragments ar eproduced

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56
Q

primase

A

syntthesizes short primary in 5’ to 3’ direction to start replciatino on each strand

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57
Q

sliding clamp

A

strengthens interaction between DNA polymerases and template strand

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58
Q

oncogenes develop from

A

mutations of proto-oncogenes

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59
Q

oncogenes

A

promote cell cycling

may lead to cancer

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60
Q

cancer

A

unchecked cell proliferation with the ability to spread by local invasion or metastasize

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61
Q

metastasize

A

migrate to distant sites via the bloodstream or lymphatic system

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62
Q

tumor supressor genes

A

aka antioncogenes

code for proteins that reduce cell cycling or promote DNA repair

mutations can lead to cancer

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63
Q

mismatch repair

A

occurs during G2 phase

genes MSH2 and MLH1 detect and remove errors that were missed during S phase

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64
Q

nucleotide excision repair

A

occurs during G1 and G2

fixes helix deforming lesions of DNA (such as thymine dimers) via cut and patch process that requires an excision endonuclease

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65
Q

base excision repair

A

occurs during G1 and G2

fixes nondeforming lesions of DNA helix (such as cytosine deamination) by removing the base, leaving an apurinic/apyrimidinic (AP) site

AP endonuclease then removes the damaged sequence, which can be filled in with the correct bases

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66
Q

proofreading

A

DNA polymerase can detect the lack of stability of hydrogen bonds from incorrectly paired bases

incorrect base is excised and replaced

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67
Q

how can the enzyme discriminate between the template strand and the new one?

A

daughter strand is identified by its lack of methylation

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68
Q

cytosine deanimation

A

loss of an amino group from cytosine and results in conversion of cytosine to uracil

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69
Q

what is the key structural difference in the types of lesions corrected by nucleotide excision repair vs those corrected by base excision repair?

A

nucleotide excision repair: corrects lesions that are large enough to distort the double helix

base excision repair: corrects lesions that are small enough not to distort the double helix

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70
Q

recombinant DNA

A

DNA composed of nucleotides from two different sources

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71
Q

DNA cloning steps

A
  1. introduces a fragment of DNA into vector plasmid
  2. restriction enzyme cuts both the plasmid and the fragment, which are left with sticky ends
  3. once the fragment binds to the plasmid, it can be introduced into a bacterial cell and permitted to replicate, generating many copies of the fragment of interest
  4. once replicated, the bacterial cells can be used to create a protein of interest, or can be lysed to allow for isolation of the fragment of interest from the vector
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72
Q

DNA cloning

vectors contain

A

origin of replication, fragment of interest, and at least one gene for antibiotic resistance (to permit for selection of that colony after replication)

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73
Q

DNA libraries

A

large collections of known DNA sequences

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74
Q

genomic libraries

A

contain large fragments of DNA, including both coding and noncoding regions of the genome

cannot be used to make recombinant proteins or for gene therapy

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75
Q

cDNA libraries

A

aka expression libraries

contain smaller fragments of DNA

only include the exons of genes expressed by the sample tissue

can be used to make recombinant proteins or for gene therapy

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76
Q

hybridization

A

joining of complementary base pair sequences

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77
Q

types of hybridization

A

PCR, agarose gel electrophoresis, southern blotting

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78
Q

polymerase chain reaction (PCR)

A

millions of copies of a DNA sequence created from very small sample by hybridization

  • requires: primers that are complementary to the DNA that flanks the ROI, nucleotides, DNA polymerase, heat
  • DNA of interest is denatured, replicated, then cooled to allow reannealing
    • repeated several times until enough copies are available for further testing
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79
Q

agarose gel electrophoresis

A

separates DNA molecules by size

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80
Q

southern blotting

A

can be used to detect the presence and quantity of various DNA strands in a sample

after electrophoresis, the sample is transferred to a membrane that can be probed with a single stranded DNA molecules to look for a sequence of interest

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81
Q

DNA sequencing

A

uses dideoxyribonucleotides, which terminate the DNA chain because they lack a 3’ -OH group

resulting fragments can be separated by gel electrophoresis and the sequence can be read directly from the gel

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82
Q

dideoxyribonucleotides

A

terminate the DNA chain because they lack a 3’ -OH group

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83
Q

gene therapy

A

method of curing genetic deficiencies by introducing a functional gene with a viral vector

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84
Q

transgenic mice are created by

A

integrating a gene of interest into germ line or embryonic stem cells of a developing mouse

can be mated to select for the transgene

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85
Q

chimeras

A

organisms that contain cells from two different lineages (such as mice formed by integration of transgenic embryonic stem cell into a normal mouse blastocyst)

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86
Q

knockout mice are created by

A

deleting a gene of interest

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87
Q

safety and ethical issues in biotechnology

A

pathogen resistance and ethics of choosing individuals for specific traits

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88
Q

DNA cloning

A

technique that can produce large amounts of a desired sequence

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89
Q

restriction enzymes/endonucleases

A

enzymes that recognize specific double stranded DNA sequences

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90
Q

exons

A

coding regions of DNA

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91
Q

introns

A

noncoding regions of DNA

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92
Q

cDNA

A

complementary DNA

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93
Q

DNA electrophoresis

the longer the DNA strand, the ___ it will migrate in the gel

A

slower

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94
Q

transgene

A

cloned gene that is introduced into mice

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95
Q

what does pcr accomplish for a researcher?

A

increases the number of copies of a given DNA sequence

can be used for a simple containing very few copies of the DNA sequence

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96
Q

what does southern blotting accomplish for a researcher?

A

useful when searching for a particular DNA sequence because it separates DNA fragments by length and the probes for a sequence of interest

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97
Q

Which of the following is the proper name for the bond between phosphate groups (e.g. between the gamma and beta phosphate groups in ATP)?

(A) Anhydride
(B) Hydride
(C) Ester
(D) Phosphodiester

A

(A) Anhydride

Between multiple covalently-bound phosphate groups are an anhydride linkage.

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98
Q

Which of the following statements about Base Pairing are true?

I. The Adenine-Thymine and Cytosine-Guanine pairing is referred to as Watson-Crick base pairing
II. There is also “Wobble Base Pairing”, where two nucleotides that are not part of the Watson-Crick Base Pairing can pair.
III. There are instances where DNA can undergo Wobble Base Pairing.

(A) I only
(B) I and II only
(C) I and III only
(D) I, II and III

A

(B) I and II only

Each of the following statements are true:

I. The Adenine-Thymine and Cytosine-Guanine pairing is referred to as Watson-Crick base pairing
II. There is also “Wobble Base Pairing”, where two nucleotides that are not part of the Watson-Crick Base Pairing can pair.
III. There are instances where RNA can undergo Wobble Base Pairing.

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99
Q

Which carbons of ribose are involved in the ribose ring formation?

(A) C-1 and C-2
(B) C-1 and C-4
(C) C-2 and C-3
(D) C-2 and C-5

A

(B) C-1 and C-4

The lone pair on the oxygen attached 4th carbon (C-4) forms a bond with C-1 which then leads to the C-1 oxygen forming a bond with a hydrogen (forming a hydroxyl group).

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100
Q

What is the structural difference in ribose vs deoxyribose?

A

The structural difference between ribose and deoxyribose is at the second carbon (C-2). The ribose has an -OH group at C-2, while deoxyribose has a -H at C-2.

101
Q

The structural backbone of DNA has been compromised by a rare pathogen. This pathogen must have disrupted what type of bond?

(A) Covalent Bonds
(B) Hydrogen Bonds
(C) Dipole-dipole Interactions
(D) Ionic Bonds

A

(A) Covalent Bonds

The pathogen of this rare disease most likely targets and destroys phosphodiester linkages (covalent bonds), which hold the DNA structural backbone together.

102
Q

Where do the nitrogenous bases bind to the backbone of DNA?

(A) C-1 of deoxyribose
(B) C-2 of deoxyribose
(C) C-5 of deoxyribose
(D) O of Phosphate

A

(A) C-1 of deoxyribose

The nitrogenous bases bind to the first carbon (C-1) on the sugar group of the backbone.

103
Q

CRB Compare Heterochromatin and Euchromatin, focusing on the role of Histones.

A

Heterochromatin is tightly packed around histones, preventing transcription.
Euchromatin is transcriptionally active and not tightly packed around Histones, and is often seen during Interphase.

104
Q

Which of the following are functions of telomeres?

I. Protect the chromosome from deterioration.
II. Prevent the chromosomes from sticking together.
III. Prevent the loss of genetic information with each round of replication.

(A) I Only
(B) I and II Only
(C) II and III Only
(D) I, II, and III

A

(D) I, II, and III

Telomeres are the cap ends of chromosomes and they serve as a protection for the chromosomes from deterioration. Telomeres also prevent the chromosomes from sticking to each other. They also help prevent the loss of genetic information in each round of replication, because telomeric sequences indicate to the enzyme that it has reached the end of the chromosome.

105
Q

True or False? Telomeres are usually found in both eukaryotes and prokaryotes.

A

False. Telomeres are found in the chromosomes of eukaryotic cells and NOT prokaryotic cells because prokaryotic cells have circular chromosomes that do not have ends.

106
Q

What does the enzyme telomerase do?

A

The enzyme telomerase lengthens the telomeres and brings them back to their original length.

107
Q

What would happen to a cell that loses all of its telomeres?

A

When a cell loses all of its telomeres, the cell does not divide anymore and will die.

108
Q

In which type of DNA would you find an organism’s most important genetic information?

(A) Single copy
(B) Slightly repetitive (10+ repeats)
(C) Highly repetitive (100+ repeats)
(D) Extremely repetitive (1,000+ repeats)

A

(A) Single copy

An organism’s most important genetic information is mainly found on the chromosome where there are high amounts of single copy DNA.

109
Q

Telomeres are extremely repetitive. They consist of which sequence repeated over and over again?

(A) 5’GGGTTA3’
(B) 5’GTTTAA3’
(C) 5’GGGTTG3’
(D) 5’GCTAAG3’

A

(A) 5’GGGTTA3’

Mnemonic: “Go Go Great Telomeres! Telomeres abound!”

110
Q

Which of the following statements about Transposons are FALSE?

(A) Transposons all code for Transposase proteins that can catalyze the movement of Transposon sequences in the genome.
(B) Transposases come in three different structures.
(C) Transposases are found only in Eukaryotes.
(D) Transposases in Eukaryotes are thought to be degenerate retroviruses.

A

(C) Transposases are found only in Eukaryotes.

Transposases are found in both Prokaryotes and Eukaryotes.

111
Q

What enzyme is responsible for adding nucleotides to the DNA template strands?

(A) DNA Primase
(B) DNA Polymerase
(C) DNA Ligase
(D) DNA Helicase

A

(B) DNA Polymerase

DNA polymerase is the enzyme responsible for adding nucleotides to the DNA template strands.

112
Q

How does Topoisomerase make Helicase’s job easier?

A

Topoisomerase is responsible for removing DNA supercoils, unwinding the tightly wound DNA helix, making it easier for the enzyme helicase to cut through the middle of DNA.

113
Q

What enzyme is responsible for adding RNA primers?

(a) DNA Primase
(b) RNA Polymerase
(c) RNA Ligase
(d) DNA Helicase

A

DNA primase is the enzyme is responsible for adding RNA primers, which will allow DNA Polymerase to bind
to the DNA and start adding nucleotides in the 5’-3’ direction.

114
Q

What is conservative DNA replication?

A

The theory of conservative DNA replication describes when the two original template DNA strands stay together in a double helix and produce a copy composed of two new strands.

115
Q

What is dispersive DNA replication?

A

The theory of dispersive DNA replication describes DNA replication that results in two pairs of double-helix DNA and each of the 4 DNA strands contains some old DNA and new DNA.

116
Q

On which side of the gel electrophoresis (negative electrode or positive electrode) would you place samples of DNA fragment?

A

You would place samples of DNA fragments on the same side of the gel as the negative electrode. This way, DNA (with its negatively-charged phosphate backbone) will migrate across the gel towards the positive electrode.

117
Q

What is a DNA ladder and why do we need one when running a DNA gel electrophoresis?

A

DNA ladder is a set of standardized DNA molecules of known size that are used to determine the size of an unknown DNA sample molecule run on a gel during electrophoresis.

118
Q

In what direction do the anions and cations move towards in an electrical field?

A

The anions always move toward the anode while the cations always move toward the cathode in an electrical field.

119
Q

Place the steps of a polymerase chain reaction (PCR) in the correct order:

I. Annealing
II. Extension
III. Denaturation

(A) III > II > I
(B) II > III > I
(C) III > I > II
(D) I > II > III

A

(C) III > I > II

The polymerase chain reaction (PCR) happens subsequently in the following order:

(1) Denaturation
(2) Annealing
(3) Extension

120
Q

Describe what happens during each step of PCR and why the temperature is set to what it is during each step:

(1) Denaturation (96° C)
(2) Annealing (55° C)
(3) Extension (72° C)

A

(1) Denaturation - The double stranded DNA that you are interested in studying is denatured as the temperature is increased to 96° C.
(2) Annealing - As the temperature is decreased to 55° C, the primer anneals to the end of the region that you want to copy on each strand of DNA.
(3) Extension - DNA Polymerase (which is most active at a temperature of 72° C) begins extending the DNA from the primer.

121
Q

Which of the following statements about cDNA libraries are true?

I. cDNA Libraries will lack all introns typically removed by alternative splicing.
II. cDNA Libraries can vary based on the type of cell the RNA was taken from.
III. cDNA Libraries isolated from one cell/cell type can be called Expression Libraries.

(A) I only
(B) I and II only
(C) II and III only
(D) I, II and III

A

(D) I, II and III

Each of the following statements are true:

I. cDNA Libraries will lack all introns typically removed by alternative splicing.
II. cDNA Libraries can vary based on the type of cell the RNA was taken from.
III. cDNA Libraries isolated from one cell/cell type can be called Expression Libraries.

122
Q

Once DNA or cDNA is inserted into a cloning vector, how is that genetic material “amplified”?

A

Amplification of the DNA/cDNA is accomplished through inserting the cloning vector into a bacteria that can replicate that genetic material over and over again. From there, we can sequence the gene of interest.

123
Q

Put the following steps of DNA Cloning in order from first to last:

I. Uptake
II. Cut
III. Amplify
IV. Paste

(A) I > III > II > IV
(B) III > I > II > IV
(C) II > IV > I > III
(D) II > IV > III > I

A

(C) II > IV > I > III

The steps of DNA Cloning are as follows:

(1) Cut
(2) Paste
(3) Uptake
(4) Amplify

124
Q

Which of the following would happen if you restriction digested a vector at exactly one Restriction Site?

(A) 2 fragments form
(B) The Vector is linearized and 2 fragments form
(C) The vector is linearized as 1 fragment.
(D) None of the above.

A

(C) The vector is linearized as 1 fragment.

Think of this like making one cut in the band of a wedding ring. It would no longer be an intact circle shape, but all of the metal would still be connected!

125
Q

Why do the plasmids used in DNA Cloning contain an antibiotic resistance gene?

A

Many bacteria in the petri dish will not take in the plasmid. We want these bacteria to die so that we know we only have bacteria on our plate that contain the gene of interest. To kill off all bacteria without a plasmid, we add an antibiotic. Only bacteria that contain the plasmid will survive since the plasmid contains an antibiotic resistance gene that makes them immune to that specific antibiotic.

126
Q

There are four directional types of Blots in general. Match the type of blot with the information it provides.

I. Southern
II. Northern
III. Eastern
IV. Western

(A) Post-Translational Protein modifications
(B) DNA strands present
(C) Protein levels
(D) mRNA levels

A

I. Southern - (B) DNA strands present

II. Northern - (D) mRNA levels

III. Eastern - (A) Post-Translational Protein modifications

IV. Western - (C) Protein levels

I like to think of these as a bizarro-compass rose. Instead of “North South East West”, I think of these as “South North West East”.

127
Q

Describe the four steps of DNA Cloning:

(1) Cut
(2) Paste
(3) Transform
(4) Amplify

A

(1) Cut - Restriction Enzymes are used to cut the gene of interest out of a greater body of DNA.
(2) Paste - That gene is then inserted into a plasmid.
(3) Transform - The plasmid is placed in a solution with bacteria. Heat is applied, which causes the bacteria to take in the plasmid. This uptake process is termed “transformation.”
(4) Amplify - The bacteria are grown on a plate, increasing the amount of the gene of interest.

128
Q

Put the following steps of a Southern Blot in the correct order from first to last:

I. Gel Electrophoresis
II. Cut
III. Probe
IV. Filter Paper
V. X-ray

(A) II, I, IV, III, V
(B) II, I, IV, V, III
(C) I, III, V, IV, II
(D) II, IV, I, III, V

A

(A) II, I, IV, III, V

The steps of a Southern Blot are as follows:

(1) Cut
(2) Gel Electrophoresis
(3) Filter Paper
(4) Probe
(5) X-ray

129
Q

Describe each of the 5 steps of a Southern Blot:

(1) Cut
(2) Gel Electrophoresis
(3) Filter Paper
(4) Probe
(5) X-ray

A

(1) Cut - Restriction Enzymes are used to cut the DNA into smaller pieces.
(2) Gel Electrophoresis - DNA fragments are placed in wells and will travel from one end of the gel to the other, getting separated based on their size.
(3) Filter Paper - Filter paper is place on the gel, and the fragments will transfer over to the filter paper.
(4) Probe - A radio-labelled single-stranded DNA fragment is added to the filter paper. It will hybridize to the gene of interest.
(5) X-ray - In order to visualize the radio label, the filter paper is exposed to an x-ray, which will only show a band for the gene of interest.

130
Q

Put the steps of DNA Sequencing in the correct order:

I. Gel Electrophoresis
II. PCR with ddNTPs
III. Analysis
IV. PCR

(A) II > IV > I > III
(B) IV > II > I > III
(C) I > III > I > IV
(D) III > I > IV > I

A

(B) IV > II > I > III

The steps of DNA Sequencing are as follows:

(1) PCR
(2) PCR with ddNTPs
(3) Gel Electrophoresis
(4) Analysis

131
Q

Describe the steps of DNA Sequencing:

(1) PCR
(2) PCR with ddNTPs
(3) Gel Electrophoresis
(4) Analysis

A

1) PCR - PCR is used to amplify the gene of interest.
(2) PCR with ddNTPs - The sample is separated into four different containers, one containing regular nucleotides plus radio-labelled ddGTP, another container with ddCTP, another with ddTTP, and another with ddATP. PCR is then continued in each of the four containers.
(3) Gel Electrophoresis - The sample is place in a well and the DNA strands travel across the gel, becoming separated by size.
(4) Analysis - The DNA strands are analyzed in order from shortest to longest fragment.

132
Q

central dogma

A

DNA is transcribed to RNA, which is translated to protein

133
Q

degenerate code

A

allows multiple codons to encode for the same amino acid

134
Q

initiation codon

A

AUG

135
Q

termination codons

A

UAA, UGA, UAG

136
Q

wobble

A

third base in codon

137
Q

what allows mutations to occur without effects in the protein?

A

redundancy and wobble

138
Q

point mutations can cause

A
  • silent mutations
  • nonsense (truncation) mutations
  • missense mutations
139
Q

silent mutations

A

have no effect on protein synthesis

140
Q

nonsense mutations (truncation)

A

produce a premature stop codon

141
Q

missense mutations

A

produce a codon hat codes for.a different amino acid

142
Q

frameshift mutations result from

A

nucleotide addition or deletion

143
Q

RNA is structurally similar to DNA except:

A
  • substitution of a ribose sugar for deoxyribose
  • substitution of uracil for thymine
  • single stranded instead of double stranded
144
Q

messenger RNA (mRNA)

A

travels into cytoplasm to be translated

only type of RNA that contains info that is translated

145
Q

transfer RNA (tRNA)

A

translates the codon into the correct amino acid

brings in amino acids and recognizes the codon on the mRNA using its anticodon

146
Q

ribsomal RNA (rRNA)

A
  • makes up the ribosome and is enzymatically active - can function as ribozymes
  • help catalyze the formation of peptide bonds
  • important in splicing out its own introns
  • synthesized in nucleus
147
Q

gene

A

unit of DNA that encodes a specific protein or RNA molecule

148
Q

mRNA is aynthesized in __ direction

A

5’ to 3’

149
Q

codons

A

three nucleotide segments

represents one amino acid

150
Q

monocistrionic

A

mRNA

each molecule translates into only one protein product

151
Q

aminoacyl tRNA synthase

A

activates amino acids in tRNA

152
Q

ribozyme

A

enzymes made of RNA molecules instead of peptides

153
Q

why did the wobble develop?

A

protect against mutations in coding regions of DNA

mutations in wobble tend to be silent or degenerate

154
Q

reading frame

A

3 nucleotides of a codon

155
Q

cystic fibrosis

A

caused by frameshift mutation

156
Q

transcription steps

A
  1. helicase and topoisomerase unwind the DNA double heliz
  2. RNA polymerase II binds to TATA box in promoter region of gene
  3. hnRNA synthesized from DNA template (antisense) strand
157
Q

posttranscriptional modifications

A
  • 5’ cap added (7-methylguanylate triphosphate cap)
  • 3’ poly-A tail added
  • intron/exon splicing
    • introns removed and exons ligated together
  • inc variability of gene products
    • prokaryotes: thru polycistrionic genes
    • eukaryotes: thru alternative splicing
158
Q

splicing

A

done by snRNA and snRNPs in spliceosome

snRNPs couple with snRNA

introns removed in lariat structure

exons ligated together

159
Q

polycistrionic genes

A

starting transcription in different sites within the gene leads to different gene products

160
Q

alternative splicing

A

combining different exons in modular fashion to acquire different gene products

allows organism to make more different proteins from limited number of genes

161
Q

transcription

A

creation of mRNA from DNA template

produces only one copy of the two strands of DNA

162
Q

transcription factors

A

help RNA polymerase locate and bind to promoter region of DNA during transcription

search for promotor and enhancer regions in DA

163
Q

hnRNA

A

heterogeneous nuclear RNA

pre processed mRNA

mRNA is derived from hnRNA via posttranscriptional modifications

164
Q

snRNA

A

small nuclear RNA

165
Q

RNA polymerase I

A

located in nucleolus

synthesizes rRNA

166
Q

RNA polymerase II

A

transcribes mRNA - synthesizes hnRNA and snRNA

binds to promoter region (TATA box)

located in nucleus

does not require primer to start

167
Q

RNA polymerase III

A

located in nucleus

synthesizes tRNA and some rRNA

168
Q

RNA polymerase travels along template strand in __ direction

A

3’ to 5’

169
Q

RNA vs DNA polymerase

A

RNA polymerase does not proofread its work

170
Q

snRNPs

A

small nuclear ribonucleoproteins

171
Q

5’ cap

A

added during transcription

recognized by ribosome as the binding site

protects mRNA from degradation in cytoplasm

172
Q

poly A tail

A
  • protects mRNA against rapid degradation
  • composed of adenine bases
  • the longer the poly A tail, the more time the mRNA will be able to survive before being digested in cytoplasm
  • assists with export of mRNA from nucleus
173
Q

where does translation occur?

A

ribosomes

174
Q

stages of translation

A
  1. initiation
  2. elongation
  3. termination
175
Q

initiation

A

step 1 of translation

  • prokaryotes:
    • 30S ribsome attaches to shine-dalgarno sequence and scans for start codon
    • lays down N-formylmethionine in P site of ribosome
  • eukaryotes:
    • 40S ribsome attaches to 5’ cap and scans for start codon
    • lays down methionine in P site of the ribosome
176
Q

elongation

A

2nd stage of translation

  • A site of ribosome holds the incoming aminoacyl-tRNA complex
  • P site holds tRNA that carries the growing polypeptide chain
    • where first amino acid binds
    • peptide bond is formed as polypeptide is passed from tRNA in P site to tRNA in A site (GTP used for energy)
  • unchanged tRNA pauses in E site before exiting the ribosome
177
Q

termination

A

stage 3 of translation

when codon in A site is stop codon, release factor places a water molecule on polypeptide chain –> allows peptidyl transferase and termination factors to hydrolyze the completed polypeptide chain from the final tNRA –> polypeptide chain released

178
Q

posttranslational modifications

A
  • folding by chaperones
  • formation of quaternary structure
  • cleavage of proteins or signal sequences
  • covalent addition of other biomolecules (phosphorylation, carboxylation, glycosylation, prenylation)
179
Q

DNA –> DNA

+ direction

A

replication

synthesized in 5’ - 3’ direction

180
Q

DNA –> RNA

+ direction

A

transcription

synthesized in 5’-3’ direction

181
Q

RNA –> protein

+ direction

A

translation

read in 5’ to 3’ direction

182
Q

binding sites in ribosome for tRNA

A
  • A site: aminoacyl
  • P site: peptidyl
  • E site: exit
183
Q

initiation factors (IF)

A

assists initiation

184
Q

elongation factors (EF)

A

locate and recruit aminoacyl-tRNA along with GDP during elongation

help to remove GDP when energy has been used

185
Q

peptidyl transferase

A

forms the peptide bond during elongation

186
Q

chaperones

A

assist in protein folding process

187
Q

posttranslational processing

phosphrylation

A

addition of phosphate group by protein kinases to activate or deactivate proteins

most commonly seen with serine, threonine, and tyrosine

188
Q

posttranslational processing

carboxylation

A

addition of carboxylic acid groups, usually to serve as calcium binding sites

189
Q

posttranslational processing

glycosylation

A

addition of oligosaccharides as proteins pass through ER and golgi apparatus to detrmine cellular destination

190
Q

posttranslational processing

prenylation

A

addition of lipid groups to certain membrane bound enzymes

191
Q

jacob monod model

A

explains how operons work

operon contains structural genes, an operator site, promoter site, and regulator gene

192
Q

operons

A

inducible or repressible clusters of genes transcribed as a single mRNA

193
Q

inducible systems

A

bonded to a represser under normal conditions

can be turned on by an inducer pulling the repressor from operator site

194
Q

lac operon is a _____ system

A

inducible

195
Q

repressible systems

A

transcribed under normal conditions

can be turned off by corepressor coupling with the repressor and the binding of this complex to the operator site

196
Q

trp operon is a ___ system

A

repressible

197
Q

operator site

A

binding site for repressor protein

upstream of structural gene

198
Q

promotor site

A

provides place for RNA polymerase to bind

upstream of operator

199
Q

regulator gene

A

upstream of promoter site

codes for repressor protein

200
Q

why is lac operon an inducible system?

A

bacteria can digest lactose, but it is more energetically expensive than digesting glucose

bacteria only want to do this if lactose is high and glucose is low

201
Q

lac operon is induced by presence of

A

lactose

202
Q

catabolite activator protein (CAP)

A

transcriptional activator used by e coli when glucose levels are low to signal that alternative carbon sources should be used

203
Q

how lac operon is turned on

A

falling levels of glucose –> inc in cAMP –> binds to CAP –> CAP conformational change –> binds to promoter region of operon –> transcription of lactase gene

204
Q

how trp operon works

A

trp high in local environment –> trp acts as corepresssor –> binding of repressor and corepressor to repressor –> repressor binds to operator site –> cell turns off

205
Q

corepressor

A

activates repressor

206
Q

structural gene

A

gene of interest

its transcription depends on repressor being absent from operator site

207
Q

promotors are located

A

within 25 base pairs of transcription start site

208
Q

enhancers are located

A

more than 25 base pairs away from transcription start site

209
Q

modulation of chromatin structure

A

affects the ability of trascriptional enzymes to access the DNA through histone acetylation or DNA methylation

210
Q

histone acetylation

A

increases accessibility of DNA

decreases the positive charge on lysine residues and weakens the interaction of the histone with DNA, resulting in open chromatin conformation

211
Q

DNA methylation

A

decreases accessibility of DNA

often linked with the silencing of gene expression

212
Q

transcription factor structure

A

DNA binding domain: binds to specific nucleotide sequence in protein region or to DNA response element to help in recruitment of transcriptional machinery

activation domain: allows for binding of several transcription factors and other important regulatory proteins

213
Q

DNA response element

A

sequence of DNA that binds only to specific transcription factors

214
Q

gene amplification

A

accomplished through enhancers and gene duplication

215
Q

enhancers

A

increase the likelihood of genes to be amplified

216
Q

gene duplication

A

increases the expression of a gene product

  • different ways:
    • duplicated in series on same chromosome, yielding many copies in a row of the same genetic info
    • duplicated in parallel by opening the gene with helicases and permitting DNA replication only of that gene
217
Q

histone acetylases

A

acetylate lysine residues found in amino terminal tail regions of histone proteins

218
Q

histone deacetylases

A

remove acetyl groups from histones, which results in closed chromatin conformation and overall decrease in gene expression levels in the cell

219
Q

DNA methylases

A

add methyl groups to cytosine and adenine nucletides

220
Q

Describe the process of Alternative Splicing. Does it happen in Prokaryotes, Eukaryotes, or Both?

A

Alternative Splicing occurs in Eukaryotes, and is when exons (as well as introns) could be removed from the RNA during processing to form mRNA. This allows multiple proteins to be formed from the same sequence of DNA.

221
Q

What is the difference between the template strand and coding strand in the process of transcription?

A

The template strand is the strand of DNA that the RNA Polymerase reads as it generates mRNA.

The coding strand is not read by RNA Polymerase. Its sequence resembles the newly synthesized strand of mRNA except that the coding strand contains Thymine instead of Uracil.

222
Q

Which of the following nitrogenous bases is NOT found in mRNA?

(A) A
(B) T
(C) C
(D) G

A

(B) T

You will not find the base Thymine in RNA. Instead, Adenine will pair with Uracil (A-U). The base pair Cytosine-Guanine (C-G) remains the same.

223
Q

Which of the following is not a modification made to pre-mRNA before it becomes mature mRNA?

(A) The addition of 5’ cap
(B) The addition of the poly-A tail
(C) The splicing of introns
(D) The removal of telomeric mRNA

A

(D) The removal of telomeric mRNA

After RNA polymerase makes pre-mRNA, the pre-mRNA goes through some modifications such as:

  • the addition of a 5’ cap
  • the addition of the poly-A tail
  • the splicing of introns
224
Q

What is the function of an micro RNA (miRNA)?

A

The function of miRNA is to silence mRNA via complementary base pairing.

225
Q

What is the function of the spliceosome?

A

The spliceosome is responsible for splicing introns out and ligating exons together in making mature mRNA.

226
Q

What are the primary components of a ribosome?

I. tRNA
II. rRNA
III. Proteins

(A) I and II Only
(B) I and III Only
(C) II and III Only
(D) I, II, and III

A

(C) II and III Only

A ribosome is composed of proteins and ribosomal RNA (rRNA).

227
Q

What is the difference between a codon vs. an anti-codon?

A

A codon is composed of three nucleotides found on the mRNA that code for an amino acid.

An anticodon is part of transfer RNA (tRNA) that complements the mRNA, positioning the tRNA and its attached amino acid for addition to the polypeptide. tRNA are specific for each amino acid.

228
Q

The t-RNA with the anticodon of UAC will initially bind at which site within the ribosome?

(A) A-site
(B) E-site
(C) P-site
(D) S-site

A

(C) P-site

UAC is the anticodon for AUG, which is the start codon. This tRNA molecule carries methionine into the P site to start the translation process.

229
Q

At which site on the ribosome will an incoming tRNA (other than the tRNA with the anticodon of UAC) bind?

(A) A-site
(B) E-site
(C) P-site
(D) S-site

A

(A) A-site

The A-site on the ribosome is where the tRNA initially enters carrying an amino acid. The name “A-site” comes from the word “amino acid.”

230
Q

As the tRNA in the P-site shifts to the E-site, which of the following is true?

I. The ribosome shifts towards the 5’ end of the mRNA.
II. The tRNA in the A-site shifts to the P-site.
III. The amino acid from the A-site moves to bind to the polypeptide chain in the P-site.

(A) I Only
(B) II Only
(C) I and II Only
(D) I, II, and III

A

(B) II Only

As the tRNA in the P-site shifts to the E-site…

  • The ribosome shifts towards the 3’ (NOT 5’) end of the mRNA.
  • The tRNA in the A-site shifts to the P-site.
  • The polypeptide chain in the P-site moves to bind to the amino acid in the A-site.
231
Q

CRB True or false? There are multiple Hairpin Loops typically seen in tRNA.

A

There are multiple Hairpin Loops typically seen in tRNA.

232
Q

What site does the ribosome recognize and initially bind to in eukaryotic vs. prokaryotic cells?

A

The ribosome recognizes and binds to the 5’ cap of eukaryotic mRNA.

The ribosome recognizes and binds to the Shine-Delgarno Sequence of prokaryotic mRNA.

233
Q

True or False? Prokaryotic mRNA does not contain any non-coding regions.

A

False. From the 5’ to 3’ end, prokaryotic mRNA contains non-coding mRNA, the Shine-Delgarno Sequence, non-coding mRNA, the start codon, coding mRNA, the stop codon, and then more non-coding mRNA.

234
Q

Why doesn’t prokaryotic mRNA need the 5’ cap and poly-A tail?

A

The prokaryotic mRNA does not have the 5’ cap and poly-A tail because transcription and translation both happen simultaneously in the cytosol; thus, the mRNA does not need to “travel.”

For eukaryotic mRNA, on the other hand, it needs to exit the nucleus and travel to the ribosome, and the 5’ cap and poly-A tail are added on as extra protection.

235
Q

What is the difference between exonuclease activity and endonuclease activity?

A

Enzymes engage in exonuclease activity when they remove an incorrect nucleotide from the end of a DNA strand.

Enzymes engage in endonuclease activity when they remove an incorrect nucleotide from the middle of a DNA strand.

236
Q

After replication is completed, there is a thymine base-paired with a guanine. What will fix this issue?

(A) The DNA Polymerase III Proofreading Mechanism
(B) The Base Excision Repair Mechanism
(C) The Mismatch Repair Mechanism
(D) The Nucleotide Excision Repair Mechanism

A

(C) The Mismatch Repair Mechanism

The mismatch repair mechanism happens after DNA replication. Through this mechanism, mismatches in the DNA sequence missed by the DNA polymerases during replication will be fixed.

237
Q

In which of the following stages of the Cell Cycle would you expect Mismatch Repair to be active?

(A) G1
(B) G2
(C) S
(D) M

A

(B) G2

In the G2 phase (right after S phase, where transcription occurs) is where the Mismatch Repair Mechanism is active.

238
Q

Put the following steps of the Mismatch Repair Mechanism in order:

I. DNA Ligase repairs the DNA backbone.
II. Exonuclease removes the misplaced nucleotide.
III. DNA Polymerase adds the correct nucleotide.
IV. The mismatched DNA is marked with a cut.

(A) IV, I, II, III
(B) I, IV, II, III
(C) IV, II, III, I
(D) I, II, III, IV

A

(C) IV, II, III, I

Mismatch Repair Mechanism entails the following steps in order:

  1. The mismatched DNA is marked with a cut.
  2. Exonuclease removes the misplaced nucleotide (and perhaps some of its neighbors) from the newly synthesized strand.
  3. DNA Polymerase adds the correct nucleotide.
  4. DNA Ligase repairs the DNA backbone by binding the newly placed nucleotide to the neighboring nucleotides.
239
Q

What is the difference between DNA mutations and DNA damage? Think of an example of each.

A

DNA mutations change the order of the DNA sequence. An example would be a mutation from 5’-ATCG-3’ to 5’-AACG-3’.

DNA damage is the structural damage to DNA that leaves nucleotides in the correct order. An example of this is a pyrimidine dimer.

240
Q

There are only three steps to Nucleotide Excision Repair. What are they?

A

(1) Endonuclease removes pyramidine dimer and surrounding nucleotides.
(2) DNA Polymerase will fill in the gap with the proper nucleotides.
(3) DNA Ligase will repair the backbone by connecting the new nucleotides to the old nucleotides.

241
Q

Compare senescence versus apoptosis.

A

To say a cell is senescent means that the cell no longer divides.

Apoptosis is programmed cell death in which a cell digests/destroys itself.

242
Q

Gene expression is regulated at the level of:

(A) Replication
(B) Transcription
(C) Translation
(D) Post-translational Modification

A

(B) Transcription

Gene expression is regulated at the level of transcription, meaning that at certain times only certain DNA will be expressed and transcribed into RNA. This way, we only produce the proteins we need at a given time.

243
Q

What is the similarity and difference between deacetylation and methylation?

A

Both deacetylation and methylation inactivate the expression of DNA.

The difference is that methylation is a permanent method of down-regulating genes while deacetylation is a reversible mechanism.

244
Q

Methylation typically occurs in the DNA regions rich in which nucleotide?

(A) A
(B) U
(C) T
(D) C

A

(D) C

Methylation is most common in cytosine and guanine rich sequences called CpG Islands.

245
Q

As cells divide and differentiate from stem cells into specific tissues which gene regulatory mechanism is commonly involved? Why?

(A) Acetylation
(B) Deacetylation
(C) Methylation
(D) Both A and B

A

(C) Methylation

Methylation is involved as cells divide and differentiate from stem cells into specific tissues because methylation alters gene expression in a stable (more permanent) manner.

246
Q

How does methylation affect DNA transcription directly? Indirectly?

A

Methylation affects DNA transcription directly by impeding the binding of transcriptional proteins on the genes.

Methylation can affect DNA transcription indirectly by attracting proteins called methyl cpg-binding domain proteins (MBDs), which recruit additional chromatin remodeling proteins such as histone deacetylases (HDACs) to the gene site, resulting in the condensation of the chromatin into heterochromatin.

247
Q

Which of the following is NOT a component of the basic transcription apparatus?

(A) General transcription factors
(B) Enhancers
(C) RNA polymerase
(D) mediator multiple protein complex

A

(B) Enhancers

General transcription factors (GTFs), RNA polymerase, and mediator multiple protein complex all constitute the basic transcription apparatus found in prokaryotic cells.

Enhancers are DNA sequences, not proteins!

248
Q

CAP is an example of an activator protein. Describe the role and mechanism of action of Catabolite Activator Protein (CAP) in e. Coli.

A

In E. Coli, cyclic adenosine monophosphate (cAMP) is produced during glucose starvation. Consequently, cAMP binds to a protein called CAP which causes a conformational change that allows the CAP protein to bind to a DNA site adjacent to the promoter. Then the CAP protein recruits RNA polymerase to the promoter..
It is an example of an activator protein because it interacts with the RNA polymerase to help it attach to the promoter sequence thus encouraging the expression of a gene.

249
Q

Compare enhancers and promoters.

A

Enhancers sequences of DNA that are distant from the gene(s) they regulate. They are bound by activators which loop the DNA in a certain way, bringing other transcription factors and mediator proteins to the promotor, enhancing the expression of that DNA.

Promotors, on the other hand, are sequences of DNA located immediately upstream from the gene(s) that they regulate. RNA Polymerase and activator proteins may bind here.

MCAT (26 decks)

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