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Biology Module 3 > Module 3 > Flashcards

Module 3 Flashcards

1
Q

DNA

A

Deoxyribonucleic Acid

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2
Q

DNA ____ and ____ genetic information

A

Stores, transmits

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3
Q

What happened to a mouse when injected with virulent / non virulent bacteria. What would happen if heat was added to each bacteria? What does this conclude?

A

Virulent - mouse dies of pneumonia
Nonvirulent - mouse remains healthy

Heads added (darkens and kills bacteria)

Killed virulent - mouse remains healthy
Killed virulent and live nonvirulent - mouse dies of pneumonia

  • concludes that a molecule was transmitting virulence (Frederick Griffith 1928)
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4
Q

How do we know DNA transmits virulence?

A
  1. Made a virulent bacteria extract with DNA, RNA, and protein that was extracted from heat-killed virulent cells, and purified to make a solution
  2. Got 4 solutions of nonvirulent bacteria and added the following to each container:
  3. Solution only
  4. Solution + RNase
  5. Solution + protease
  6. Solution + DNase
  7. Containers 1, 2, and 3 became a solution of virulent and nonvirulent bacteria, container 4 remained a nonvirulent bacteria solution
  8. This indicated that DNA was removed by enzyme DNase, and therefore the cause of virus transmission
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5
Q

Where is genetic info stored?

A

In genes

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6
Q

What are nucleic acids made of?

A

Nucleotides

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7
Q

What is a nucleotide?

A
  • building blocks of nucleic acids (DNA)
  • made up of a base (A, C, T, or G), a pentose sugar, and at least 1 phosphate group
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8
Q

What makes each nucleotide distinct?

A

The bases

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9
Q

Characteristics of bases

A
  1. Purines (double-ring)
    - Adenine or Guanine
  2. Pyrimidines (single-ring)
    - Thymine or Cytosine
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10
Q

Nucleoside vs. Nucleotide

A

Nucleoside: base + sugar
Nucleotide: base + sugar + phosphate (gives (-) charge)

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11
Q

how are nucleotides linked together?

A

through the phosphate group on the 5’ carbon of nucleotide 2 attacking the 3’ hydroxyl (OH) of nucleotide 1 = phosphodiester bond

phosphodiester bond: C-O-P-O-C

  • result is nucleotide 1 has 2 (-) charges on the phosphate, nucleotide 2 has 1 on the left oxygen of phosphate group
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12
Q

DNA has _____. The phosphodiester bonding results in a free ___ phosphate group at the top and a free ___ hydroxyl (OH) group at the bottom.

A

polarity, 5’, 3’

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13
Q

What is the structure of DNA?

A

double helix

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14
Q

what 3 conclusions allowed the double helix of DNA to be proved?

A
  1. complementary base pairing probably between A + T and C + G (not proved until 2nd finding)
    - OH and NH (hydrogen bonds link bases)
    - 1 OH and 1 NH between A and T (2 bonds)
    - 2 OH and 1 inner NH between G and C (3 bonds)
  2. Chargaff’s Rule - regardless of cell type:
    %A = %T and %G = %C
  3. X-Ray Crystallography of DNA - DNA is a helical, repetitive structure
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15
Q

Watson-Crick model

A

10 base pairs per complete turn (3.4 nm)
diameter of DNA: 2 nm
1. sugar phosphate backbone in rightward direction
2. major groove - large turn
3. minor groove - small turn

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16
Q

base stacking

A

non-polar, flat sides of bases face inwards and stack tightly
- sugar phosphate backbone is polar
- DNA strands are antiparallel

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17
Q

steps to go from DNA helix -> chromosomes

A
  1. DNA helix 2nm diameter
  2. DNA wrapped twice around histone protein (+) charge, makes nucleosome beads
  3. Nucleosome beads linked create chromatin fiber (30nm diameter)
  4. Chromatin fiber is coiled (300nm diameter)
  5. Coils even more (700nm diameter)
  6. Condensed chromatid (1400nm diameter)
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18
Q

2 theories of DNA replication

A
  1. Conservative - first replication makes 2 parental strands, 2 new ones
    - 2nd replication makes 1 parental, 3 new ones
  2. Semi-conservative - first replication makes 2 half parental half new strands
    - 2nd replication makes 2 new strands, 2 half parental half new ones
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19
Q

Is DNA conservative or semi-conservative? How was this tested?

A

DNA is semi-conservative
- radioactive N isotopes used to label DNA in bacteria
1. Parental strands made with 15N label = heavy DNA
2. Daughter strands made with 14N label = light DNA
3. Measured density of DNA after each replication

Finding was also confirmed in eukaryotes using fluorescent nucleotides after 2 rounds of replication
- daughter strand had fluorescence, parental had none (daughter appeared black on screen, parent was lighter colour)

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20
Q
  1. density of DNA
  2. density of DNA after 1 replication
  3. density of DNA after 2 replication
A

1) 1.722 gm/cm^3
2) 1.715
3) 1.708 daughter, 1.715 parent

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21
Q

Does A-T bond have 2 NH bonds?

A

No, it has 1 OH and 1 NH because NH is a + bond, do not want them to repel

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22
Q

Steps for DNA synthesis / replication

A
  1. Helicase unwinds DNA 5’ -> 3’ (breaks H-bonds)
    - forms replication fork
    - replication always occurs from 5’ -> 3’
  2. RNA Primase binds to make RNA primer’s on both strands
    - primer allows DNA polymerase to bind and replicate
  3. DNA polmyerase makes complimentary DNA from RNA primer
    - adds nucleotides/reads in 5’ -> 3’
  4. A different DNA Polymerase removes RNA primers to connect Okazaki Fragments on lagging strand
  5. DNA Ligase attaches all fragments in lagging strand (primers removed and replaced w/ DNA)
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23
Q

What are the stabilizers in DNA replication?

A
  1. Topoisomerase II - stabilizes unwound DNA during the process (behind helicase)
    - relieves stress of unwounding
  2. Single-strand binding protein (ssBP) - stabilizies DNA strands being replicated
    - keep DNA apart
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24
Q

Trombone Loop

A

lagging strand forms this so both strands are elongated together
- allows the 2 RNA polymerases to stay in contact without strand interference

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25
Q

What enzyme is used when proofreading for misplaced/incorrectly added nucleotides?

A

DNA Polymerase - turns on exonucleic function by removing incorrect nucleotide, then adds in the correct nucleotide

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26
Q

Origins of replication in:
a) prokaryotes
b) eukaryotes

A

a) prokaryotes: one origin of replication
- replication starts at origin and moves around circular chromosome in both directions

b) eukaryotes: multiple origins of replication (ORI)
- 2 replication forks in one replication bubble move in opposite directions
- 2 replication bubbles meet to make one large bubble

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27
Q

telomere

A
  • caps at the end of each chromosome
  • chromosomes contain 1000s of these repeating sequences (5’ -> 3’)
  • function is to prevent chromosome shortening during cell division
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28
Q

direction of leading and lagging strand

A
  • leading strand points TOWARDS replication fork (where Helicase is)
  • lagging strand points AWAY from replication fork
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29
Q

True or False: The parental strand and newly synthesized DNA strand are the same length on the leading and lagging strand

A

False: due to lack of priming, DNA polymerase cannot fill the gap at the very end of the chromosome for the lagging strand
True: for the leading strand

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30
Q

expected vs. observed result of chromosomes after cell division

A

expected: each time a cell divides, the chromosomes shorten until it bursts and becomes damaged

observed: the chromosomes are protected, length stays the same due to telomeres

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31
Q

how are telomeres incorporated into DNA strands within chromosomes?

A
  1. Telomerase protein has an RNA template built into it to synthesize the telomere repeat sequence
  2. DNA polymerase can fill in the shortened part of the linear DNA by complementary base pairing the new strand with the telomere repeat strand
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32
Q

Telomerase protein activity in:
1. Adult cells
2. Germ and Stem cells

A
  1. Adult cells: almost no Telomerase activity
  2. Germ and Stem cells: has Telomerase activity
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33
Q

consequences of low vs. high Telomerase activity in Germ/Stem cells

A

Low activity: linked to aging and rare inherited diseases
High activity: associated with cancer

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34
Q

Up to _____ telmorere repeats can be replicated to divide

A

100

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35
Q

A lack of telomerase activity limits what?

A

the number of times a cell can divide

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36
Q

PCR and its requirements

A

Polymerase Chain Reaction: amplification of 1 copy of DNA
Requirements:
- template DNA (to amplify this)
- DNA polymerase
- nucleotides
- 2 Oligonucleotide primers (fwd + reverse), complimentary to sequences of gene of interest (these replace RNA primers) -> the two primers point inwards to each other

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37
Q

Steps for a PCR

A
  1. Denaturation of double strand DNA
    - using high heat (95 degrees) to break H-bonds -> under b.p.
  2. Annealing of primers to complimentary sequences (60-70 degrees)
    - using Hydrogen bonds
    - must have excess primers
  3. Extension of the DNA from the primer w/ Taq polymerase
    - results in final amplified DNA

-> process repeated 20-30 cycles
Every cycle = 2^n copies
(n = number of cycles)

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38
Q

why does PCR need excess primers created?

A

to ensure that template DNA doesn’t reanneal

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39
Q

Taq Polymerase

A

a type of DNA polymerase from Thermus aquaticus (lives in hot springs)

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40
Q

what plays the role of Helicase in PCR?

A

heat

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41
Q

how does electrophoresis help visualized PCR products?

A
  • agarose gel has pores to place them in
  • small fragments move faster
  • large fragments move slower
    -> DNA is therefore separated by size
  • electrophoretic buffer well contains ions
  • pores block large fragments
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42
Q

which enzyme synthesizes the RNA primer needed to initiate DNA synthesis?

A

Primase

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43
Q

deoxynucleotide vs. dideoxynucleotide

A
  • deoxynucleotide has a hydroxyl (OH) group on the 3’ carbon, allowing this end to be elongated
  • dideoxynucleotide lacks the 3’ hydroxyl group (just H at this end), thus cannot be elongated b/c no OH to attack incoming nucleotide triphosphate
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44
Q

use of dideoxynucleotides

A

the different fluoreescent labels on the end of each dideoxynucleotide (daughter strand) helps find unknown sequence of parent

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45
Q

microarray

A

a chip with different DNA sequences with known locations (single strands)
- helps identify disease mutations

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46
Q

cDNA probe

A

one denatured strand of DNA cut into smaller fragments
- use 2 samples (normal and tumor)
- label w/ fluorescent dyes and combine them
- hybridized probe to microarray and scan for divergences in gene sequences

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47
Q

GWAS

A

Genome wide association study: helps identify specific disease mutations
- CHIPS can be used to genotype 500,000 - 5 million SNPs

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48
Q

SNP

A

single nucleotide polymorphism
- a point mutation

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49
Q

DNA damage examples

A
  1. ssDNA breaks
  2. cross-link T bases (DNA adducts)
  3. missing base
  4. bulky side group attached to a base
  5. dsDNA breaks
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50
Q

list all DNA repair mechanisms

A
  1. Mismatch repair
  2. Base-excision repair
  3. Nucleotide-excision repair
  4. dsDNA break repair

first 3 are transcription-coupled/global genome repair, 4th is homologous recombination/non-homogolous end-joining

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51
Q

what are the DNA damaging agents?

A
  1. Base mismatch: DNA replication stress
  2. ssDNA breaks: oxygen radicals, ionizing radiation, chemotherapeutics
  3. DNA adducts: polyaromatic hydrocarbons, UV light
  4. dsDNA breaks: ionizing radiation, chemotherapeutics
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52
Q

mismatch repair

A
  1. MutS protein recognizes mismatched bases, initiates repair process
  2. MutL and MutH proteins are recruited, MutH breaks backbone some distance away (cleaves DNA)
  3. exonuclease enzyme removes nucleotides between the 2 proteins
  4. DNA polymerase fills in missing nucleotides, and DNA ligase joins the backbones
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53
Q

Base excision repair

A

repairs DNA when a base is damaged (e.g. cytosine loses a nitrogen group)
1. cytosine easily loses amino group, forming a base called uracil (after DNA replication)
2. uracil cannot base pair w/ guanine
3. DNA Uracil glycosylase cleaves uracil base from deoxyribose sugar
4. AP endonuclease cleaves DNA backbone to remove sugar
5. DNA polymerase fills gap and DNA ligase seals it

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54
Q

AP endonuclease

A

cleaves DNA backbone for base excision repair to remove sugar

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55
Q

DNA Uracil glycosylase

A

enzyme that cleaves uracil base from sugar for base excision repair

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56
Q

MutL, MutH, MutS

A

MutL - associates with MutS

MutH - cleaves DNA

MutS - recognizes mistmatched bases in DNA, initiates repair process

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57
Q

Nucleotide excision repair

A

(similar to mismatch repair)

  1. UV radiation can make 2 T’s bind to each other incorrectly
  2. enzymes cleave DNA at sites of damage
  3. DNA polymerase fills gap, DNA ligase seals DNA
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58
Q

CRISPR

A

clustered regularly interspaced short palindromic repeats
-> CRISPR-Cas uses short guide RNA (20 nucleotides) to bind to specific DNA sequences
-> Cas9 protein cuts/edits anything that resembles guide RNA (works like a pair of molecular scissors)

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59
Q

2 ways DNA editing used for disease

A
  1. allow cell itself to repair cut in DNA, leads to gene’s f’n being turned off
  2. researchers can insert, repair, or edit a gene by designing small DNA template, changing genome code
    -> this maintains gene f’n and replaces a mutation
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60
Q

Benefits of DNA/Genome editing?

A

CRISPR-Cas9 gene editing helps fight sickle-cell disease in 2 ways:

  1. template-style editing of hemoglobin
    - using guide RNA, Cas9 enzyme repairs faulty Beta-globin gene (subunit of hemoglobin)
  2. remove a protein to prevent hemoglobin production
    - Cas9 promotes production of fetal haemoglobin by breaking gene that encodes a repressor
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61
Q

Cons of DNA/Genome editing

A
  1. Off-target effects: there can be many sites in genome that look like guide RNA
  2. On-target effects: after Cas9 cuts DNA, repair mechanisms of cell are unpredictable (can be repaired perfectly -> religation, or some letters inserted/deleted)
  3. Mosaicism:
    (1) if a developing embryo contains a few cells with risky mutations, a biopsy (to see if DNA editing needed) picks up that mutated cell and may lead to unnecessary manipulations

(2) CRISPR-Cas9 treatment may leave too many cells uncorrected to treat the disease (some mutated cells remain)

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62
Q

___ gene = ___ protein

A

1,1

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63
Q

Central dogma

A

DNA (in nucleus) -> RNA (in nucleus) -> Protein (in cytoplasm)

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64
Q

DNA vs. RNA

A

DNA RNA
Sugar: deoxyribose (H on 2’) ribose (OH on 2’)
Bases: A,T,C,G A,U,C,G
5’ end: monophosphate triphosphate
Size: very large smaller (1 gene)
Strands: double single (less stable)

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65
Q

Thymine vs. Uracil

A

T: base is CH3
U: base is H

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66
Q

RNA

A

ribonucleic acid

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67
Q

3 stages of DNA transcription

A
  1. Initiation: promotor sequence spans few 100bps for general transcription factors to bind
    -> transcription starts 25bp downstream of promotor
    -> transcription activator proteins bind enhancor sequence
    -> general transcription factors recruit RNA Polymerase II, and transcription activator proteins recruit mediator complex to bind to RNA polymerase II + general transcr. factors
  2. Elongation: RNA made from 3’ -> 5’ strand of DNA (RNA transcript grows 5’ -> 3’)
    -> no replication fork b/c transcript only being made in 1 direction
    -> done on transcription bubble on template strand
  3. Termination: terminator sequence allows Polymerase to fall off DNA template (thus RNA transcription is complete)
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68
Q

coding vs. noncoding template strand

A

coding - nontemplate strand
noncoding - template strand

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69
Q

Regulation

A

occurs before initiation of DNA transcription:

  1. transcription starts at promotor sequence (TATATA sequence b/c easy to open apart) -> dictates what proteins are expressed in cell
  2. transcription ends at terminator sequence
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70
Q

housekeeping genes

A

needed all the time and are transcribed consistently

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71
Q

transcription factor in prokaryotes vs. eukaryotes

A

prokaryotes: sigma factor
eukaryotes: general transcription factor

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72
Q

enhancor sequences

A

require activation proteins to bind, it is upstream of gene of interest

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73
Q

general transcription factors

A

6 proteins that initiate transcription when binding to promotor region, recruit RNA Polymerase II

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74
Q

upstream + examples

A

anything BEFORE specific gene to be transcribed (promotors, enhancers)

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75
Q

downstream + examples

A

anything AFTER specific gene to be transcribed (terminators)

76
Q

transcriptional activator proteins

A

bound to enhancer sequences to transcribe then fall off, recruits mediator complex

77
Q

Mediator complex

A

binds to RNA Polymerase and general transcription factors, DNA folds back

78
Q

when can transcription begin?

A

can begin once full complex assemples (signals to RNA polymerase II to start transcribing)

79
Q

What would happen if an enhancer sequence were mutated so that its binding partner was always bound and recruiting the RNA polymerase complex?

A

transcription would occur continuously

80
Q

DNA Polymerase vs. RNA Polymerase

A

DNA Polymerase RNA Polymerase
Template: DNA DNA
Primer?: Yes No
Opening of No - needs helicase Yes
DNA temp?:
Okazaki fragm? Yes No - only 1 strand
Proofreading?: Yes No

81
Q

why doesn’t RNA Polymerase require a primer?

A

complex only binds to where a promotor is located

82
Q

how does RNA Polymerase open the DNA template?

A

breaks H-bonds between DNA strands to open transcription bubbles

83
Q

Why is it okay for RNA Polymerase not to proofread?

A

RNA is short-lived and unstable (does not pass to offspring), lasts a few hours
-> thus can make diff. sequences proteins with mutations present
-> also, RNA Polymerase makes a few errors (1 ever 10,000 nucleotides)

84
Q

DNA Transcription in Prokaryotes

A

Transcription and Translation are coupled, translation begins before transcription is completed (ribosome makes protein)

  1. Template DNA allows for making primary transcript = mRNA
85
Q

RNA Processing

A

[RNA (aka primary RNA transcript/ pre-mRNA) must be processed into Messenger RNA (mRNA) to leave nucleus]

1) 5’ cap: addition of modified nucleotide to 5’
-> does not compl. base pair w/ RNA strand, but protects 5’ end of this RNA transcript
-> triphosphate bridge between 5’ cap and 1st nucleotide of RNA transcript (replaces phosphodiester bond)

2) Polyadenylation: addition of poly (A) tail to 3’
-> adenine nucleotides

3) Removal of introns (non-protein coding regions) and splicing of exons (protein coding regions)
-> 1. middle of intron attacks 5’ splice site
-> 2. cleaved 5’ splice site attacks 3’ splice site
-> 3. lariat breaks down quickly into individual nucleotides
-> 4. spliced exons are adjacent in the processed RNA

86
Q

5’ cap

A

a modified G nucleotide ( has extra methyl group on base)

87
Q

Roles of 5’ cap and 3’ poly (A) tail

A

5’ cap: for protein translation (ribosome attachment)
3’ poly(A) tail: for export of RNA transcript to cytoplasm
Both: needed for stabilization of mRNA

88
Q

spliceosome

A

complex of RNA and protein, catalyzes the removal of introns from pre-mRNA
-> binds to introns of RNA transcript to loop

89
Q

lariat

A

cleaved 5’ splice site attacks 3’ splice site to fully remove intron (makes lariat) -> the loop of intron

90
Q

___% of human genes have an intron. ___% of human guenes are alternatively spliced

A

90%, >80%

91
Q

Alternative RNA processing, what is the majority of RNA and their functions?

A

not all RNA is protein-coding mRNA, majority of RNA (>90%) becomes:

  1. ribosome RNA (rRNA): protein translation
  2. transfer RNA (tRNA): protein translation
  3. small nuclear RNA (snRNA): part of spliceosome
  4. microRNA (miRNA) / small interfering RNA (siRNA): regulation of transcription/translation (regulatory)
92
Q

amino acid

A

comprise an alpha carbon covalently bonded to 4 groups:
1. amino group (NH2)
2. carboxyl group (COOH)
3. H atom
4. R group (varies with each amino acid) -> aka “side chain”

-> building blocks of proteins

93
Q

what is the pH of inside cells? What happens to amino acid in this pH?

A

pH = 7, amino acids modified to have charges:
-> H atom gains 1 (+) charge
-> COOH carboxyl group loses H+ (becomes COO-)

94
Q

Categories of R groups on amino acids and their properties

A
  1. Hydrophobic (nonpolar): go inside protein, aggregate through vaan der waals forces to avoid H2O
  2. Hydrophilic (polar): no charge
  3. Hydrophilic (positive/basic)
  4. Hydrophilic (negative/acidic)
  5. Unique:
    (1) Cysteine (C): sulfide bridges R group, very strong bonds for protein structure
    (2) Glycine (G): H as R group, flexible b/c H can be turned/move easily -> smallest R group
    (3) Proline (P): bulky ring on R group -> proteins fold around it
95
Q

how are amino acids linked?

A

at pH 7, (-) carboxyl on one amino acid binds with amino group (NH3+) on another, releasing H2O
-> results in amino end (NH2) and carboxyl end (COOH)

COO- + NH3+ = CO-NH and H2O

96
Q

determining number of water molecules produced with amino acids

A

(n-1) = number for bonds, n is the number of amino acids, therefore H2O molecules produced are always 1 less than amino acids present

97
Q

polypeptide

A

PRIMARY STRUCTURE: peptide bonding between 2 or more amino acids (string of amino acids)
-> has N-terminal (amino end -> NH2), and C-terminal (carboxyl end -> COOH)
-> aka free amino group, free carboxyl group

98
Q

alpha helix + beta sheet

A

SECONDARY STRUCTURE: interactions between amino acids in primary structure

(1) alpha helix: H-bonding between carbonyl (C=O) of amino acid, and amide group that is 4 amino acids away (N-H)
-> O and H bond
-> coiled, protein structure (right-hand), each turn = 3.6 amino acids, R-groups exposed from helix

(2) beta sheet: H-bonding between amino acids on parallel and anti-parallel strands of polypeptide (folds back on itself)
-> forms pleated sheet of protein with 4-10 pleats (strands)

99
Q

Tertiary structure of proteins and ways to represent them

A

interactions between R-groups in polypeptide with secondary structure, forms folded 3D structure

Ball and stick model: every atom in each amino acid (primary)

Ribbon model: secondary structures (how alpha + beta sheets interact)

Space filling model: 3D shape with folds of proteins and where pocket is for enzymes to attach

100
Q

interactions in a tertiary protein structure

A
  1. Hydrophobic interactions: weak vaan der waals forces (WITHIN protein)
  2. H-bonds and ionic bonds: polar R-groups, ionic bonds are oppositely charged R-groups (NH3+ binds with O-),
    H-bonds are (O-H)
  3. Disulfide bond/bridge/linkage (STRONGEST BOND): sulfied R-groups on cysteines
    -> disulfide is COVALENT bonding
101
Q

Quaternary structure of proteins and functions

A

proteins with multiple polypeptides (multi-subunit proteins)
e.g. Hemoglobin

-> comprises 4 polypeptides (globins), each has iron-containing heme group that binds oxygen: 2 alpha-globin, 2 beta-globin
-> makes bigger pores for ions to enter

102
Q

over ___ genes required for translation

A

100

103
Q

how do nucleotides get translated to amino acid?

A

with the 3 nucleotide sequence codons that correspond to amino acids

104
Q

start codon

A

AUG (in mRNA) -> makes first amino acid in every protein (MET)
-> AUG codon is NONPOLAR

105
Q

stop codon

A

UAA, UAG, UGA -> do not code for an amino acid and end protein synthesis

106
Q

Every amino acid has multiple codons except ______

A

the start codon: Methionine = Met

107
Q

reading frame

A

3 types, used to break mRNA sequence in separate codons (like sentences with words)

108
Q

True or False: because there are 3 different possible reading frames in a mRNA, most mRNAs can be translated in a cell into 3 different proteins

A

False, only 1 reading frame contains the start codon

109
Q

tRNA in protein translation

A

transfer RNA, bind to a particular amino acid
-> have an anti-codon loop (3 nucleotides), that is complimentary sequence to codon for amino acid
-> anti-codon binds to mRNA to bring correct amino acid

110
Q

codon sequence and anti-codon directions

A

codon: 3’ <- 5’
anticodon: 5’ <- 3’

111
Q

how to know what anticodon looks like from an mRNA sequence?

A

translate codon on mRNA to complimentary codon (using U instead of T), and translate again into the anticodon

112
Q

aminoacyl tRNA synthetases

A

enzymes that promote covalent bonding between tRNA and amino acid (based on codon)

113
Q

ribosomes

A

comprise ribosomal RNA and proteins to make complex for translation (initiation site for protein synthesis)
-> E (exit site), P (peptidyl site), A (aminoacyl site) reading frames

114
Q

Eukaryotic vs. Prokaryotic Translation Initiation Step and mRNA name

A

Eukaryotic: ribosome recognizes and binds mRNA at 5’ cap, starts protein synthesis at start codon (AUG)

Eukaryotic mRNA = monocistronic (1 gene sequence per mRNA)

Prokaryotic: ribosome recognizes and binds at Shine-Dalgarno sequences (no cap)

Prokaryotic mRNA = polycistronic (1 mRNA can have multiple protein-coding gene sequences)

115
Q

Protein Translation steps

A
  1. Initiation: initiation factor proteins recruit small subunit of ribosome and tRNA^Met to scan mRNA for start codon
    -> ribosome binds at 5’ cap
    -> Large subunit of ribosome recruited, and tRNA^Met binds P site, intiaition factor proteins are released when start codon found
  2. Peptide Bond Formation: next complimentary tRNA binds to next codon in A site, Met is transferred from P site to A site, peptide bond between first 2 amino acids occurs
  3. Ribosome Movement and Elongation: ribosome moves over 1 codon, uncharged tRNA moves to E site and leaves ribosome
    -> process repeats to elongate polypeptide
  4. Termination: stop codon encountered, Protein Release Factor binds A site, breaking covalent bond between last charged tRNA and amino acid
    -> ribosome subunits break apart
116
Q

true or false: gene number and genome size are predictors of genome complexity

A

false: number of protein coding genes do not correlate with complexity of an organism

117
Q

c-value and c-value paradox

A

c-value: amount of DNA in a cell, does nothing to complexity of organism

c-value paradox: contradiction between genome size and complexity

118
Q

why do we need regulation of gene expression?

A

allows for control on which proteins every cell synthesizes/expresses
-> can also give rise to multiple proteins/proteins that perform different functions (in stomach, muscles, etc.)

119
Q

the human body contains ~200 major cell types. They look and function differently because each:

A

expresses a different set of genes

120
Q

why are some genomes larger?

A

due to polypolidy or chromosome duplication (more than 2 sets of chromosomes)
e.g. flowering plants are polyploid (bananas, coffee)
-> e.g. Triploid (3N), Tetraploid (4N)

121
Q

what is the cause of the c-value paradox?

A

due to large non-protein coding regions
1) alpha satellite DNA: tandem DNA copies near centromeres of chromosome
2) transposable elements: DNA transposons and retrotransposons
-> no function, cause harm to proteins

122
Q

actual protein coding exons is ____% of genome

A

2.5%

123
Q

Retrotransposons vs. DNA transposons

A

retrotransposons: copy and paste mechanism
-> multiply DNA by reverse transcription (RNA -> DNA)

DNA transposons: cut and paste mechanism (when DNA is damaged)
-> excision, then gene jumps to a different spot on DNA (“jumping genes”)

124
Q

Gene expression: DNA and Chromatin remodeling

A

Epigenetic modifications (pass on to daughter cells)
-> DNA wraps around (+) histone proteins (nucleosome) to start chrosome condensation
-> chromatin remodelled to expose DNA sections for transcription of genes to occur

125
Q

define histone protein tails, how are they modified?

A

histones are chemically modified on protein tails
protein tail: string of amino acids exposed from histone protein

modification: adding or removing methyl (CH3) or acetyl (COCH3) groups on (+) charge amino acid Lysine

-> lysine becomes either: monomethyl lysine, trimethyl lysine, or acetyl lysine

126
Q

what occurs when there is heavy DNA methylation?

A

transcriptional repression

127
Q

CpG island

A

clusters of C and G nucleotides near promotor sequence of gene

128
Q

how is X-chromosome gene expression done?

A
  1. X-inactivation: 1 X chromosome is randomly inactivated in cells during embryogenesis (leads to patches/mosaic b/c maternal/paternal x-chromosome varies)
  2. Inactivation begins with transcription in XIC
    -> Xist RNA binds to X-chromosome DNA on XIC and spreads outwards
    -> Xist is spliced and coats entire chromosome, repressing gene transcription (e.g. heavy methylation)
129
Q

Xist gene

A

X-inactivation specific transcript (non-protein coding RNA)

130
Q

XIC

A

X-chromosome inactivation center

131
Q

When does Xist expression increase?

A

before X-inactivation

132
Q

Alternative splicing may be considered a mechanism of gene regulation because it:

A

results in different protein products

133
Q

Gene expression: DNA Methylation

A

allows/represses transcription
-> bases in nucleotides can be chemically modified:
-> addition of methyl (CH3) group on Cytosine

134
Q

Gene expression: X-chromosome

A

uses imprinting (sex-specific silencing of gene expression)
-> regulation of X-chromosome gene expression for people with 2 X-chromosome (double)

135
Q

Gene expression: transcription

A

promotor and enhancers on mRNA regulate gene expression through Regulatory Transcription Factors (TFs) that bind on multiple enhancer sequences on a gene
-> TFs can also bind silencer sequences to prevent transcription

136
Q

Gene expression: RNA Processing

A

insulin receptor mRNA spliced based on cell type

137
Q

Gene expression: RNA editing

A

allows for temporary changes in gene/protein sequence (unlike DNA editing)
-> DNA Base modification: activating DNA repair mechanisms
(1) Deamination of A to I (losing NH2, becomes O)
(2) Deamination of C to U (losing NH2, becomes O)

138
Q

Gene expression: Translation

A

UTR = untranslated region aka poly(A) tail -> regulates translation
(5’ UTR behind ribosome) and (3’ UTR after ORF) -> open reading frame
-> they contain regions for protein binding and small regulatory RNA binding sites (ssRNA)

139
Q

translation initation requires ___ proteins

A

25

140
Q

Adding or removing 5’ cap on mRNA regulates gene activity in translation because the 5’ cap is necessary for:

A

translational initiation

141
Q

Gene expression: mRNA stability

A

enzymes in cytoplasm make mRNA unstable and degrade it (harsh environment)
-> RISC (RNA induced silencing complex) degrades mRNA by chromatin remodeling, RNA degradation or translational inhibition

142
Q

Gene expression: post-translational modification

A

regulate/alter protein structure or function

143
Q

how can mutations be inherted?

A

through cell divison (daughter cells) or passed on from parents to child

144
Q

What 2 factors affect rate of mutation?

A

depends on size of genome, and # of cell divisons per generation (humans = highest)

145
Q

average mutation rates for a single nucleotide are low in organisms with ___

A

proofreading (DNA Polymerase)

146
Q

human genome can tolerate high mutation rates because

A

most mutations are in non-protein coding regions

147
Q

hotspots

A

certain regions or nucleotides in a genome are prone to mutation (10x higher mutation rate)

148
Q

germ cells vs. somatic cells and relation to mutations, and how many chromosomes in each?

A

germ: reproductive cell (mutation will be in every cell in body) -> haploid (23 chromosomes)
somatic: non-reproductive (DNA passed on to daughter cells in same individual) -> diploid (45 chromosomes)
-> mutation only occurs in certain body regions for somatic

149
Q

____ mutations contribute to cancer

A

somatic

150
Q

mutations are ____, not based on ____

A

random, needs

151
Q

A population of mosquitoes is exposed to the pesticide DDT for several
generations. At the end of that time, most individuals in the population are
resistant to DDT. The most likely reason is that:

A

some indviduals in original population had mutations that led to resistance

152
Q

small-scale mutations

A
  1. point: one base changes/wrong base added
    - e.g. SNPs
  2. nonsense: codon becomes stop codon
  3. deletion: loss of 1 or a few nucleotides in a row (loss of 1 codon)
  4. frameshift: insertion or deletion of nucleotides that are not in multiples of 3
153
Q

most common mutation

A

point

154
Q

2 types of point mutations

A
  1. silent (synonymous): mutations has no protein change
  2. missense (nonsynonymous): protein changes (potentially making protein nonfunctional)
155
Q

When the DNA sequence of the gene that codes for the peptide hormone insulin is compared in two mammals (e.g., humans and rats), most of the sequence differences
are synonymous mutations. These far outnumber sequence differences that result in amino acid substitutions. Why might this be?

A

Amino acid substitutions often result in proteins that have lost or compromised function, thus are selected against

156
Q

Which mutations in an animal somatic cell are inherited by the next generation?

A

none, mutations in somatic cells are not inherited by offspring

157
Q

almost ___% of human genome made of Transposable Elements (TEs)

A

50%

158
Q

transposable element mutation, and how is it moved?

A

disrupt genes (changing pigment in corn b/c no more purple pigment gene)
-> only moved by active Transposase enzyme

159
Q

Chromosomal Mutations (result in large changes in nucleotides)

A
  1. Duplication or Deletion: most common
    -> duplication less harmful than deletions (loss of genes)
  2. Duplication and Divergence: divergence is accumulation of mutations on extra gene copy -> results in new protein
  3. Copy Number Variations: difference in number of gene copies across individuals (both protein and non-coding protein regions)
  4. Tandem Repeats: a type of CNV -> small nucleotide sequences (20-50 bp) repeated next to each other)
    - child can expand tandem repeats
  5. Inversion (not as harmful)
  6. translocation: 2 chromosomes switch parts
    -> BAD for genes in germ cells, may not go in same gamete, thus losing genetic info
    -> amt of transcription is different
160
Q

If deletion is in only chromsome, will it survive? Why or why not?

A

Yes, if the deletion is in one chormosome, somatic cell can be saved), germ cells only have 1 chromosome = detrimental

161
Q

blending inheritance

A

progeny is average of both parents (wrong)

162
Q

principle of segregation

A

2 alleles of gene pair separate equally into gametes so
1/2 = allele 1
1/2 = allele 2
-> heterozygotes

163
Q

probability of 0 and 1 in genotype occurences

A

0 = no genotype passed on
1 = genotype will be passed on

164
Q

principle of independent assortment

A

segregaation of 1 set of alleles is seperated by another (yellow wrinkly and green round)

165
Q

With independent assortment, the ratio of phenotypes in the F2 generation of a cross between true-breeding strains (AA bb × aa BB) can be described as 9:3:3:1
when A and B are dominant over a and b. To what phenotype does the “9” in the ratio refer?

A

dominant for both traits

166
Q

epistasis

A

some genes interact affecting phenotype ratios

167
Q

pedigree

A

diagram of family history

168
Q

cause of recessive traits emerging and affecting individuals

A

mating between relatives

169
Q

dominant vs. recessive traits on pedigrees

A

parents: dominant usually 1 affected, recessive 1 or none affected
special circumstances: dominant 1 parent affected, results in half offspring affected
recessive occurs more with mating between 1st cousins

170
Q

incomplete penetrance and causes

A

some individuals with genotype do not show phenotype
- causes can be due to environment or gene interactions, and disease-risk mutations (most common)
- typically dominant, does not skip generations

171
Q

variable expressivity

A

phenotype shown at different severities

172
Q

autosomes

A

non-sex chromosomes

173
Q

In mammals, males inherit their X chromosome from their _____ and transmit it to their _____.

A

mothers, daughters

174
Q

nondisjunction

A

failure of chromosomes to separate during meiosis

175
Q

genetic linkage

A

genes really close together on chromosome (do not independently assort)

176
Q

nonrecombinant

A

progeny with same genotypes as parents

177
Q

closer genes = ___ likely chance of recombination

A

less

178
Q

The frequency of recombination during meiosis is a function of:

A

the distance between genes: the farther apart the genes are, the more frequent the recombination between them.

179
Q

when is frequency of recombination inaccurate? Why?

A

distances larger than 15 map units (increased chance of double crossovers)

180
Q

Y-chromosome haplotype

A

every new combo of nucleotides in Y chromosome

181
Q

From which parent do most animals inheirt mitochondrial genes?

A

Maternal

182
Q

MERRF

A

mutation that results in abnormal mitochondria

183
Q

most complex traits follow ____ distribution, even w/ environment

A

normal

184
Q

complex trait offspring

A

regression towards mean

185
Q

genotype by environment interaction

A

cannot infer phenotype from genotype unless environment conditions are known

186
Q

genotype by environment interaction

A

cannot infer phenotype from genotype unless environment conditions are known

Biology Module 3 (1 decks)

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