MT2 Material Flashcards

Question 1

Q

What is the sugar in ribonucleosides?

Question 2

Q

What is the sugar in deoxyribonucleosides?

Answer

A

A derivative of D-ribose: 2-deoxy-D-ribose

Question 3

Q

What is the basic structure of a nucleotide?

Answer

A

Composed of 3 main parts:

Five membered sugar (aldopentose ß-furanose)
Nitrogenous base - attached to C1’ of the sugar via N glycosidic linkage
Phosphate group - negatively charged - attached to the 5’ carbon of the sugar.

Question 4

Q

What is the sugar in nucleotides?

Answer

A

aldopentose ß-furanose

Question 5

Q

How are ribose nucleotides and deoxyribose nucleotides differentiated?

Answer

A

Ribose has -OH at 2’ carbon

(component of ribonucleotides and RNA)

Deoxyribose has -H at 2’ carbon

(component of deoxyribonucleotides and DNA)

Question 6

Q

Why is RNA less stable than DNA?

Answer

A

The oxygen of the 2’ OH group of the ribose in RNA can serve as a nucleophile attacking phosphodiester bond in RNA in base catalyzed reaction. Thus, RNA is much less stable than DNA which has a 2’ H group, and as such RNA must be stored at -80°C.

Question 7

Q

How is the nitrogenous base attached to the nucleotide?

Answer

A

Via N to the C1’ of the sugar.

Question 8

Q

6 membered ring with two nitrogens

Answer

A

Pyrimidines

Question 9

Q

Fused 5 and 6 membered rings

Question 10

Q

What are the pyrimidines?

Answer

A

Cytosine (C) - DNA and RNA

Thymine (T) - DNA

Uracil (U) - RNA

“CUT the PY”

Question 11

Q

What are the purines?

Answer

A

Adenine (A) - DNA and RNA

Guanine (G) - DNA and RNA

“PUre AGony”

Question 12

Q

Differentiate between nucleoside and nucleotide.

Answer

A

Base + sugar = nucleoside

Base + sugar + phosphate = nucleotide

Question 13

Q

Answer

A

Cytosine (C)

* is link to sugar

Question 14

Q

Answer

A

Thymine (T)

(DNA only)

* is link to sugar

Question 15

Q

Answer

A

Uracil (U)

(RNA only)

* is link to sugar

Question 16

Q

Answer

A

Adenine (A)

* is link to sugar

Question 17

Q

Answer

A

Guanine (G)

* is link to sugar

Question 18

Q

Answer

A

Adenosine

Ribonucleoside

Question 19

Q

Answer

A

Adenylate (AMP)

Ribonucleotide

Question 20

Q

Question 21

Q

Question 22

Q

Answer

A

Guanosine

Ribonucleoside

Question 23

Q

Answer

A

Guanylate (GMP)

Ribonucleotide

Question 24

Q

Question 25

Q

Answer

A

Cytidine

Ribonucleoside

Question 26

Q

Answer

A

Cytidylate (CMP)

Ribonucleotide

Question 27

Q

Question 28

Q

Answer

A

Uridine

Ribonucleoside

Question 29

Q

Answer

A

Uridylate (UMP)

Ribonucleotide

Question 30

Q

Answer

A

Deoxyadenosine

Deoxyribonucleoside

Question 31

Q

Answer

A

Deoxyadenylate (dAMP)

Deoxyribonucleotide

Question 32

Q

Answer

A

Deoxyguanosine

Deoxyribonucleoside

Question 33

Q

Answer

A

Deoxyguanylate (dGMP)

Deoxyribonucleotide

Question 34

Q

Answer

A

Deoxycytidine

Deoxyribonucleoside

Question 35

Q

Answer

A

Deoxycytidylate (dCMP)

Deoxyribonucleotide

Question 36

Q

Answer

A

Deoxythymidine

Deoxyribonucleoside

Question 37

Q

Answer

A

Deoxythymidylate (dTMP)

Deoxyribonucleotide

Question 38

Q

How is the phosphoric acid attached to nucleotides?

Answer

A

The 5’-OH of the (deoxy)ribose is esterified to phosphoric acid.

Question 39

Q

1P

Answer

A

monophosphate

Question 40

Q

2P

Answer

A

diphosphate

Question 41

Q

3P

Answer

A

triphosphate

Question 42

Q

Compare nomenclature for talking about individual molecules versus sequences of DNA/RNA.

Answer

A

We use AMP, ADP when talking about individual molecules, and single letter codes when talking about sequences of DNA/RNA.

e.g., adenosine 5’-monophosphate (AMP) a.k.a. adenylate

adenosine triphosphate (ATP)

Note: For DNA it is the same nomenclature, but we add ‘deoxy’ at the beginning (or ‘d’ in abbreviations)

Question 43

Q

The atoms in nitrogenous bases in purine synthesis come from a single source.

True or false?

Answer

A

False.

Atoms in nitrogenous bases come from different sources.

Question 44

Q

Purine synthesis requires aerobic reactions that do not cost ATP.

True or false?

Answer

A

False.

Purine synthesis requires anabolic reactions that cost lots of ATP.

Question 45

Q

What is the common precursor of purine synthesis?

Answer

A

Inosine monophosphate (IMP)

Question 46

Q

Purines are synthesized in the form of nucleosides.

True or false?

Answer

A

False.

Purines are synthesized in the form of nucleotides.

Question 47

Q

Ribonucleotides are synthesized first, then if needed they can be converted to deoxy.

True or false?

Question 48

Q

What is the committed step of purine synthesis?

Answer

A

Step 2) The pyrophosphate is replaced by amino group of Glutamine, forming 5-phospho-ß-D-ribosylamine, PPi, and Glutamate.

Question 49

Q

What is the first step of purine synthesis?

Answer

A

Ribose-5-phosphate (R5P) from the pentose phosphate pathway is phosphorylated using 1 ATP (but 2 ATP equivalents) to form 5-phosphoribosyl 1-pyrophosphate (PRPP) by the enzyme phosphate pyrophosphokinase (aka PRPP synthetase).

Question 50

Q

What enzyme phosphorylates ribose-5-phosphate (R5P)?

Answer

A

Ribose phosphate pyrophosphokinase (aka PRPP synthetase)

Question 51

Q

What serves as a scaffold for building bases?

Answer

A

5-phosphoribosyl 1-pyrophosphate (PRPP)

Question 52

Q

Answer

A

α-D-Ribose-5-Phosphate (R5P)

Question 53

Q

Answer

A

5-Phosphoribosyl-α-pyrophosphate (PRPP)

Question 54

Q

Answer

A

α-D-Ribose-5-phosphate (R5P)

Question 55

Q

Answer

A

5-Phosphoribosyl-α-pyrophosphate (PRPP)

Question 56

Q

What is the 2nd step of purine synthesis?

Answer

A

The committed step.

The pyrophosphate is replaced by the amino group of Gln, forming 5-phospho-ß-D-ribosylamine, PPi, and Glu.

Question 57

Q

What happens in the third step of purine synthesis?

Answer

A

In a series of 10 reactions, 5-phospho-ß-D-ribosylamine is converted to the nucleotide inosinate (IMP) - the common intermediate for both AMP and GMP.

Question 58

Q

What happens in the fourth step of purine synthesis?

Answer

A

IMP is converted to AMP or GMP (in 2 steps for either option)

Question 59

Q

How is IMP converted to AMP?

Answer

A

Aspartate is added forming adenylosuccinate (occurs with GTP hydrolysis to GDP and Pi)
Then fumarate is removed forming AMP

Question 60

Q

How is IMP converted to GMP?

Answer

A

IMP is hydrated by water and oxidized by NAD+ to xanthylate (XMP)
XMP is converted to GMP by changing carbonyl to amino group using Glutamine as a nitrogen source (occurs with ATP hydrolysis to AMP and PPi (i.e., 2 ATP equivalents))

Question 61

Q

Pyrimidine synthesis costs more energy than purine synthesis.

True or false?

Answer

A

False.

Pyrimidine synthesis is more simple than purine synthesis and does not cost as much energy.

Question 62

Q

In purine synthesis, how many steps involve ATP?

Question 63

Q

In pyrimidine synthesis, the base is made first and then attached to PRPP.

True or false?

Answer

A

True.

UTP is synthesized first, then CTP and TTP are synthesized from UTP.

Question 64

Q

Describe the first step of pyrimidine synthesis.

Answer

A

Carbamoyl Phosphate Synthetase 2 (CPS-2) forms carbamoyl phosphate from bicarbonate, Gln (converted to Glu in the process), and 2 ATP (hydrolyzed to ADP)

Answer 57

A

Carbamoyl phosphate reacts with Asp forming carbamoyl aspartate.

Answer 58

A

Carbamoyl aspartate (in a series of reactions) is converted to the base called orotate.

Answer 59

A

Orotate is transferred onto PRPP forming orotidylate (OMP) and PPi by action of the enzyme pyrimidine phosphoribosyl-transferase.

(UMP synthetase in humans)

Answer 60

A

OMP is decarboxylated to form uridylate (UMP) by orotidine-5’-decarboxylase.

(UMP synthetase in humans)

Answer 61

A

Nucleoside monophosphate kinase (UMP kinase) converts UMP to UDP using ATP.

Answer 62

A

Step 7) Broad specificity nucleoside diphosphate kinase converts UDP to UTP.

Note: Nucleoside diphosphate kinase catalyzes the reaction:

XDP + YTP = XTP + YDP

Answer 63

A

XDP + YTP = XTP + YDP

(i.e., In step 7 of pyrimidine synthesis it converts UDP to UTP)

Answer 64

A

UTP is converted to CTP by replacing carbonyl with amino group from Gln with 1 ATP consumed.

Answer 65

A

nucleoside monophosphate kinases: BMP + ATP ⇌ BDP + ADP

nucleoside diphosphate kinases: BDP + ATP ⇌ BTP + ADP

where B – anything but A

Answer 66

A

Interconversion of nucleotides by monophosphate and diphosphate kinases.

Answer 67

A

In the last step of pyrimidine synthesis UTP is converted to CTP by replacing the carbonyl with an amino group from Gln with 1 ATP consumed.

Answer 68

A

2 ATPs are needed - both are used in the first step.

One transfers phosphate, the other is hydrolyzed to ADP and Pi.

Also when UMP is converted to UTP and also when UTP is converted to CTP.

Answer 69

A

There are 2: one to form carbamoyl aspartate and the other to form dihydroorotate

Answer 70

A

It is catalyzed by OPRT; PRPP provides ribose-5-P

PPi splits off of PRPP (irreversible)

Answer 71

A

Pyrimidine synthesis is decreased and excess orotic acid is excreted in the urine.

Treatment - UMP supplements

Answer 72

A

Catalyzed by ribonucleotide reductase
Complex mechanism (don’t need to know details for BIOC 302) using free radicals
Converts rNDPs to dNDPs.
dNDPs are converted to dNTPs by nucleoside disphosphate kinase

Answer 73

A

dUDP > dUTP by nucleoside diphosphate kinase
dUTP > dUMP by UTPase (aka dUTP diphosphatase)
dUMP > dTMP by thymidylate synthase using N5,N10-methylene tetrahydrofolate as methyl donor producing dihydrofolate.
Dihydrofolate is converted back to N5,N10-methylene tetrahydrofolate using NADPH (dihydrofolate reductase) and serine (serine-hydroxymethyl transferase).

Note: Serine is a source of methyl group on dTMP.

Answer 74

A

N⁵,N¹⁰-Methylene-tetrahydrofolate

Answer 75

A

7,8-Dihydrofolate

Answer 76

A

Tetrahydrofolate

Answer 77

A

Tetrahydrofolate

Answer 78

A

Tetrahydrofolic acid

Answer 79

A

Nucleoside diphosphate kinase

Answer 80

A

UTPase (aka dUTP diphosphatase)

Answer 81

A

Thymidylate synthase

Answer 82

A

N⁵,N¹⁰-methylene-tetrahydrofolate

Answer 83

A

Thymidylate (dTMP) and Dihydrofolate

Answer 84

A

Dihydrofolate is converted back to N⁵,N¹⁰- methylene tetrahydrofolate using NADPH (dihydrofolate reductase) and Serine (serine-hydroxymethyl transferase).

Answer 85

A

AZT (azidothymidine)

This is an antiretroviral medication used to treat and prevent HIV/AIDS.

Answer 86

A

Elevated uric acid levels in the blood, accumulation of uric acid causes gout (uric acid has low solubility)
Uric acid crystals will form in the extremeties with a surrounding area of inflammation
Can be caused by a defect in an enzyme of purine metabolism or by reduced secretion of uric acid into the urinary tract
Gout can be treated by allopurinol that acts as competitive inhibitor of xanthine oxidase leading to less uric acid production

Answer 87

A

Gout can be treated by allopurinol that acts as competitive inhibitor of xanthine oxidase leading to less uric acid production.

Answer 88

A

To maintain the balanced pool of nucleotides.

Answer 89

A

AMP, GMP, ADP, and IMP all inhibit PRPP synthesis allosterically, lowering the rate of conversion of R-5-P to PRPP.
AMP, GMP, and IMP inhibit glutamine-PRPP amidotransferase allosterically lowering the rate of conversion of PRPP to 5-P-ribosylamine
AMP inhibits allosterically adenylsuccinate synthetase, lowering conversion of IMP to adenylosuccinate
GMP inhibits allosterically IMP dehydrogenase, lowering conversion of IMP to XMP

Answer 90

A

True.

As does GMP, ADP, and IMP.

This lowers the rate of conversion of R-5-P to PRPP.

Answer 91

A

False.

AMP (and GMP and IMP) allosterically inhibit glutamine-PRPP amidotransferase, lowering the rate of conversion of PRPP to 5-P-ribosylamine.

Answer 92

A

False.

AMP allosterically inhibits adenylosuccinate synthetase, lowering conversion of IMP to adenylosuccinate.

Answer 93

A

A lower rate of conversion of R-5-P to PRPP.

Answer 94

A

Lowered rate of conversion of PRPP to 5-P-ribosylamine.

Answer 95

A

Lowered rate of conversion of IMP to adenylosuccinate.

Answer 96

A

Lowered rate of conversion of IMP to XMP

Answer 97

A

UMP allosterically inhibits CPS-2, lowering production of carbamoyl phosphate
CTP allosterically inhibits aspartate transcarbamoylase, lowering production of carbamoylaspartate
ATP allosterically activates aspartate transcarbamoylase, increasing production of carbamoylaspartate

Answer 98

A

False.

UMP allosterically inhibits CPS-2, lowering production of carbamoyl phosphate.

Answer 99

A

True.

This lowers production of carbamoylaspartate

Answer 100

A

True.

This increases production of carbamoylaspartate

Answer 101

A

Lowered production of carbamoyl phosphate

Answer 102

A

Lowered production of carbamoylaspartate

Answer 103

A

Primary allosteric site:
- Activated by rATP
- Inhibited by dADP
- Inhibited by dATP
Substrate specificity site - determines substrate specificity

Answer 104

A

CUGA order of preference:

High ATP (dATP) will stimulate CDP conversion to dCDP
High ATP (dATP) will stimulate UDP conversion to dUDP
dTTP will inhibit production of dUDP and dCDP at the same time stimulating switch in specificity to dGPD production
dGTP will stimulate switch in specificity to dADP production

Raja’s short answer: Production fo dCDP → dCTP activates production of dUDP → dTTP which activates production of dGDP →dGTP which activates production of dADP → dATP and the cycle continues

Answer 105

A

GMP and AMP in a series of reactions are degraded to xanthine
- Guanine to xanthine by deamination
- Adenosine > inosine > hypoxanthine > xanthine
Xanthine is converted to uric acid
In humans, uric acid is a final product of purine degradation
- Invertebrates can convert uric acid all the way to ammonia
- Sharks can convert uric acid to urea
- Most mammals can convert uric acid to allantoin by enzyme uricase

Answer 106

A

Most mammals can convert uric acid to allantoin by action of the enzyme uricase.

Answer 107

A

Intermediates of carbon metabolism (succinyl-CoA) and ammonia (NH₄⁺).

NH₄⁺is packaged as urea through the urea cycle and excreted.

Answer 108

A

Hypoxanthine Guanine Phosphoribosyltransferase (HGPRT) deficiency
X-linked genetic condition
Up to 20 times the uric acid in the urine than in normal individuals. Uric acid crystals form in the urine.
Treated with allopurinol

Answer 109

A

Similar to purine, nucleotidase removes P, and then nucleosidase removes sugar.
C > U or Me-C > T with NH₄⁺ released.
Reduction of U or T using NADPH
Opening of the ring
CO₂ and NH₄⁺ are released
Beta-alanine (from U degradation) and beta-aminoisobutyrate (from T degradation) are further metabolised and nitrogen is secreted in the form of urea

Answer 110

A

False.

CO₂ and NH₄⁺ are both released after the opening of the ring.

Answer 111

A

False.

In pyrimidine degradation, ß-alanine (from U degradation) and ß-aminoisobutyrate (from T degradation) are further metabolized and nitrogen is secreted in the form of urea.

Answer 112

A

They catalyze the addition of ribose-5-phosphate to the base from PRPP to yield a nucleotide.

Base + PRPP > Base-ribose-phosphate (BMP) + PPi

The reaction:

Adenine + PRPP > AMP + PP_i
- is catalyzed by adenosine phosphoribosyltransferase

The reactions:

Guanine + PRPP > GMP + PP_i
Hypoxanthine + PRPP > IMP + PP_i
- are catalyzed by hypoxanthine-guanine phosphoribosyltransferase

Answer 113

A

AZT is analog of thymidine, lacking 3’-OH
Prevents growth of DNA strand
HIVs reverse transcriptase has higher affinity to AZT than to dTTP, so AZT will prevent HIV replication cycle
Human DNApol has higher affinity for dTTP rather than AZT

Answer 114

A

AZT is analog of thymidine, lacking 3’-OH.

It prevents growth of DNA strand.

HIVs’ reverse transcriptase has higher affinity to AZT than to dTTP, so AZT will prevent replication cycle.

Human DNApol has higher affinity for dTTP rather than AZT.

Answer 115

A

AZT is analog of thymidine lacking 3’-OH and prevents growth of DNA strand.

HIVs’ reverse transcriptase has a higher affinity to AZT than to dTTP, so AZT prevents HIV replication cycle.

Human DNApol has higher affinity for dTTP rather than AZT.

Answer 116

A

Hypoxanthine Guanine Phosphoribosyltransferase (HGPRT) deficiency is an X-linked genetic condition that results in up to 20 times the uric acid in the urine compared to normal individuals. Uric acid crystals form in urine.

HGPRT is treated with allopurinol.

Answer 117

A

To make a dinucleotide, the 5’-phosphate is esterified to the 3’-OH of another nucleotide.

Answer 118

A

DNA and RNA are long polymers that consist of individual nucleotides linked by phosphodiester bonds.
Sugar-phosphate bonds form a ‘backbone’ and mostly planar bases ‘above’ the sugar and pointing away.
Chain of nucleotides has polarity: one end having 5’-phosphate and the other having a free 3’-OH.
Convention dictates that the sequence is written 5’ end on the left and 3’ end on the right (5’ > 3’)
RNA has 2’-OH that makes it less stable and prone to alkaline hydrolysis.
The sugar-phosphate backbone is negatively charged

Answer 119

A

DNA is a right-handed double helix double helix: two polynucleotide strands that are bound by H-bonding between bases.
The 2 polynucleotides (DNA strands) are complementary and antiparallel; the sequence of one determines the other
The sugar phosphate backbone is on the outside and the bases are on the inside of the helix.
The bases are almost perpendicular to the helix axis and are separated by 3.4Å.
The helix repeats itself every 34Å (there are ~10 bases per turn of the helix and a rotation of 360)
The diameter of the helix is 20Å.

Answer 120

A

The ratio of A:T and G:C in DNA is always around 1 - so A must bind T (2-H bonds) and G must bind C (3-H bonds)

Answer 121

A

Every 34Å (~10 bases per turn)

Answer 122

A

Electrostatic (repulsion)
Hydrophobic ‘forces’
H-bonds
Pi-stacking interactions (van der Waals interactions) between bases

Answer 123

A

A, B, and Z.

B form: 10.5 residues per turn of the helix, 1 turn is 36Å in length, major and minor groove due to glycosidic bonds not being perfectly opposite of each other.

Answer 124

A

10.5 residues per turn, each turn is 36Å in length.

Major (wider and deeper) and minor (narrower and shallower) grooves are formed because glycosidic bonds are not perfectly opposite to each other.

Answer 125

A

Usually single stranded (ss)
Some viruses have double stranded (ds) RNA
RNA can also fold on itself and form regions of ds structure (e.g., tRNA)

Answer 126

A

Disruption of the hydrogen bonding and hydrophobic interactions between the bases resultin the separation of the strands (dsDNA > 2x ssDNA)
This may be done by increased temperatures or modified pH.
The melting temperature is defined as the temperature at the midpoint in the denaturation curve.

Answer 127

A

dsDNA absorbs less UV light at 260nm than ssDNA, so we use absorbance as a measure of concentration of DNA.

A-T denatures faster, G-C denatures slower.

Answer 128

A

False.

A-T denatures faster, G-C denatures slower.

Answer 129

A

Certain sequences of DNA, RNA, and DNA/RNA hybrids can adopt this structure.
Also right handed by the helix is wider and shorter, 11bp per turn
Major groove is narrow and deep
Minor groove is very broad and shallow

Answer 130

A

Essential to biological processes.
Transcriptional control, DNA replication, and DNA repair are all events that depend on accurate interpretation of the DNA sequence by a protein

Answer 131

A

The major groove is larger and contains more information:

H-bond donors/acceptors
Methyl groups (thymine)

Answer 132

A

GC content: higher GC > higher T_m
Salt: higher cations concentration increase T_m, but if salt concentration is too high T_m will decrease
Low (<2.3) or high (>11.5) pH decreases T_m
Organic compounds

Answer 133

A

DNA can be renatured (i.e., have two strands come back together at a lower temperature - known as annealing).

Nucleation: the 2 strands find a region of complementarity and form a short double helix
Zippering: in either direction from the paired region of complementarity, the double helix is elongated

Answer 134

A

We can anneal ssDNA with another ssDNA or ssRNA to form a hybrid molecule (hybridization).

For example:

Transcription DNA-RNA hybrid double helix
In situ hybridization - using fluorescently labelled DNA oligonucleotides to visualize RNA molecules using hybridization of RNA/DNA

Answer 135

A

DNA is a large molecule that needs to be compactly packed but accessible when it is needed. B-DNA with 10bp per turn is ‘relaxed’. If we were to unwind DNA in one spot it becomes overwound in others.

Answer 136

A

Packaging of DNA
DNA replication
DNA transcription

Answer 137

A

Changes the linking number by +/- 1 by making ss break using Tyr in catalysis.

Answer 138

A

Change linking number by +/- 2 by making ds break (using 2 ATP molecules)

For example, bacterial gyrase: underwinds bacterial DNA - decreasing the linking number by 2

Answer 139

A

Bacterial gyrase is a special type II topoisomerase that underwinds bacterial DNA.

Answer 140

A

Very compact structures made of chromatin (DNA-protein complex)
The basic unit is a nucleosome
The nucleosomes can be packed together into higher order structures.
- 10 nm ‘zig-zag filament’
- 30 nm superhelix

Answer 141

A

DNA protein complexes consisting of 2 tetramers of histones (each containing H2A, H2B, H3, and H4) forming a barrel with DNA wrapped twice around it.
H1 serves as a ‘clamp’ holding incoming strands at the base of the nucleosome together.
Histones have a high percentage of basic amino acids.
By changing the charge of histones we can modify the accessibility of DNA for transcription factors
- Acetylation Lys makes DNA packing less tight - studied by epigenetics
The final form of DNA packaging in eukaryotes are chromosomes.

Answer 142

A

DNA protein complexes consisting of 2 tetramers of histones (each containing H2A, H2B, H3, and H4) form a barrel with DNA wrapped twice around it.
H1 serves as a ‘clamp’ holding incoming strands at the base of the nucleosome together.
Histones have a high percentage of basic amino acids.

Answer 143

A

The ends of linear chromosomes that consist of tandem repeats of a short (usually 6-8 nt) DNA sequence of TxGy in one strand (x,y =1-4).
‘T-loop’; tandem repeat binding factors (TRF) stabilize telomeres

Answer 144

A

Enzymes that can chance the coiling of dsDNA (underwound overwound) by changing linking number (topology)

Type I Topoisomerases
- Change linking number by +/- 1 by making ss break using Tyr in catalysis
Type II Topoisomerases
- Change linking number by +/- 2 by making ds break (uses 2 ATP molecules)

Answer 145

A

4.5 Mbp of DNA in a single circle.
The chromosome is highly compacted
~500 supercoiled loops
Histone-like proteins (HU proteins)
Nucleoid - region of cytoplasm with DNA in prokaryotes
- Can have additional smaller pieces of circular DNA - plasmids (often antibiotic resistance genes)

Answer 146

A

We can anneal ssDNA with another ssDNA or ssRNA to form a hybrid molecule (hybridization).

For example:

Transcription DNA-RNA hybrid double helix
In situ hybridization - using fluorescently labelled DNA oligonucleotides to visualize RNA molecules using hybridization of RNA/DNA

Answer 147

A

DNA is a large molecule that needs to be compactly packed but accessible when it is needed. B-DNA with 10bp per turn is ‘relaxed’. If we were to unwind DNA in one spot it becomes overwound in others.

Answer 148

A

Packaging of DNA
DNA replication
DNA transcription

Answer 149

A

Changes the linking number by +/- 1 by making ss break using Tyr in catalysis.

Answer 150

A

Change linking number by +/- 2 by making ds break (using 2 ATP molecules)

For example, bacterial gyrase: underwinds bacterial DNA - decreasing the linking number by 2

Answer 151

A

Bacterial gyrase is a special type II topoisomerase that underwinds bacterial DNA.

Answer 152

A

Very compact structures made of chromatin (DNA-protein complex)
The basic unit is a nucleosome
The nucleosomes can be packed together into higher order structures.
- 10 nm ‘zig-zag filament’
- 30 nm superhelix

Answer 153

A

DNA protein complexes consisting of 2 tetramers of histones (each containing H2A, H2B, H3, and H4) forming a barrel with DNA wrapped twice around it.
H1 serves as a ‘clamp’ holding incoming strands at the base of the nucleosome together.
Histones have a high percentage of basic amino acids.
By changing the charge of histones we can modify the accessibility of DNA for transcription factors
- Acetylation Lys makes DNA packing less tight - studied by epigenetics
The final form of DNA packaging in eukaryotes are chromosomes.

Answer 154

A

DNA protein complexes consisting of 2 tetramers of histones (each containing H2A, H2B, H3, and H4) form a barrel with DNA wrapped twice around it.
H1 serves as a ‘clamp’ holding incoming strands at the base of the nucleosome together.
Histones have a high percentage of basic amino acids.

Answer 155

A

The ends of linear chromosomes that consist of tandem repeats of a short (usually 6-8 nt) DNA sequence of TxGy in one strand (x,y =1-4).
‘T-loop’; tandem repeat binding factors (TRF) stabilize telomeres

Answer 156

A

Enzymes that can chance the coiling of dsDNA (underwound overwound) by changing linking number (topology)

Type I Topoisomerases
- Change linking number by +/- 1 by making ss break using Tyr in catalysis
Type II Topoisomerases
- Change linking number by +/- 2 by making ds break (uses 2 ATP molecules)

Answer 157

A

Bacterial supercoiling levels are determined by the balance between the activity of the topoisomerases.
Eukaryotic supercoiling is maintained by binding to histones and relaxation of DNA between nucleosomes.

Answer 158

A

“The sequential transfer of genetic information flows to the proteins and once transferred cannot be transferred back”

“DNA makes RNA, RNA makes protein” (narrow definition, not entirely true)

Answer 159

A

Replication must overcome high affinity fo DNA strands to each other.
The DNA must be unwound without being damaged.
Must rearrange chromosomal packaging of DNA to allow access of molecular machinery of replication.
Must be accurate.
Must be fast.
DNA polymerases that replication DNA can only do it in one direction 5’ → 3’, but strands run in opposite directions.
Replication errors need to be repaired.
DNA replication enzymes cannot replicate ends of linear chromosomes.

Answer 160

A

DNA is replicated semi-conservatively, which means each parental strand serves as a template for a daughter strand. (i.e., daughter molecule contains one original parental strand and one newly synthesized daughter strand.

Answer 161

A

dNTP base pairs with its complemental base on the parental template strand.
The 3’-OH of the growing primer strand acts as a nucleophile, attacking the innermost 5’-phosphate of the incoming dNTP.
A phosphodiester bond is formed with the release of pyrophosphate (PPi), which is then further hydrolyzed to 2 inorganic phosphates (Pi).

Note: The strand grows in the 5’ → 3’ direction.

Answer 162

A

All DNA polymerases (DNA pol) require a template and a primer (need 3’-OH). This is in contrast with RNA polymerases (RNA pol), which require a template, but do not require a primer.

The template strand is the sequence that directs synthesis (i.e., determines the sequence of newly synthesized chain). The primer is the initial segment of polymer that has to be extended.

Answer 163

A

Magnesium (Mg²⁺) is required.
Only correct base pairs have optimal interactions when oriented in the active site of DNA polymerase with the help of Mg²⁺ and Aspartate.
Mg²⁺ forms hydrogen bonds with minor groove of the growing chain
If wrong base pairing is formed, geometry will not be correct for proper H-bonds.
The DNA polymerase has a ‘lid’ that will close over the base-pair once the correct base pair has been formed.
All together, the error rate is lowered to to one mistake made by DNA pol only every 10⁴-10⁵ nucleotides.

Answer 164

A

DNA pol III has a clamp loading complex that ‘pops’ open the β-clamp, threads DNA through and closes the clamp.

This allows high processivity.

In comparison DNA pol I doesn’t have a clamp and can only add ~10 nt/sec.

Answer 165

A

Processivity Factor - β-clamp of DNA pol III
DNA ligase
Helicase (DNA B)
Topoisomerase (gyrase)
Single-stranded DNA binding (SSB) proteins
Primase
DNA polymerase I
DNA polymerase III

Answer 166

A

Many polymerases have a proof-reading function.

3’ → 5’ exonuclease activity
- Detects mismatched base pairs and hydrolyzes (removes) the incorrect nucleotide.
- This lowers the error rate to 1 in 10^6-8 nucleotides.
- Sometimes this activity hydrolyzes a correct pair - wasteful but improves accuracy.

Answer 167

A

DNA repair enzymes can scan the DNA after replication and repair the mismatched bases - error rate 1 in 10¹⁰ nucleotides.

Answer 168

A

oriC locus - specific site (1 per circular genome)

This is a segment of ~246bp that contains several (4-5) repeats of a sequence recognized by the DNA A protein.

3 tandem 13’mer AT rich sites.

Answer 169

A

Growth (replication) fork - the point at which new DNA is being synthesized.

Answer 170

A

False.

Replication at oriC happens bidirectionally.

Answer 171

A

DnaA protein binds at its binding sites within oriC and causes helix unwind at AT rich sites (aka DNA unwinding element - DUE)
DnaB protein binds to DNA with the help of DnaC that hydrolyzes ATP to load DnaB onto DNA
DnaB is a helicase and it will start unwinding and opening DNA using ATP hydrolysis (in opposite directions)
Opening the helix puts strain on the rest of the helix, so DNA gyrase (topoisomerase II) binds ahead of the replication fork and removes supercoils using ATP
ssDNA binding proteins (SSB) bind to ssDNA to prevent reannealing of DNA strands
- This is called preprimed complex and DNA replication can proceed.

Answer 172

A

Make one strand continuously (leading strand) and a second one discontinuously (lagging strand).

Answer 173

A

Leading: one continuous synthesis by DNApol

Lagging: series of short fragments (Okazaki fragments)

Answer 174

A

True.

The leading strand is made by continuous synthesis in the direction of DNA helix unwinding.

The lagging strand is made discontinuously in short stretches each made in the 5’→3’ direction as the DNA unwinds.

Answer 175

A

A 5’→3’ (forward) DNA-dependent DNA polymerase activity, requiring a 3’ primer site and a template strand
A 3’→5’ (reverse) exonuclease activity that mediates proofreading
A 5’→3’ (forward) exonuclease activity mediating nick translation during DNA repair.
A 5’→3’ (forward) RNA-dependent DNA polymerase activity. Pol I operates on RNA templates with considerably lower efficiency (0.1–0.4%) than it does DNA templates, and this activity is probably of only limited biological significance.

Answer 176

A

Activity: Synthesis of new DNA strand - the new strand synthesized by addition of nucleotides to the 3’ end will be complementary to the template strand

Requirements:

template strand
primer with free 3’-OH
dNTPs
Mg²⁺

Answer 177

A

3’→5’ exonuclease
5’→3’ exonuclease
- removes DNA fragments in ‘front’ of the enzyme
- removes bp in front of the moving enzyme, and as a result any ss break in DNA (nick) will migrate in front of the enzyme

Answer 178

A

The 5’→3’ exonuclease activity of DNA pol I will remove bp in front of the moving enzyme, and as a result any ss break in DNA (nick) will migrate in front of the enzyme.

Answer 179

A

Huge protein (900kDA and 10 polypeptides)
A current model of replication of DNA pol III implicates dual function.
- One part is involved in leading strand synthesis
- The other is involved in lagging strand synthesis
Different subunit compositions allow for this specialization
Has high fidelity and processivity

Answer 180

A

The ability to catalyze many reactions without releasing the substrate.

(i.e., DNA pol III has high processivity)

Answer 181

A

A primase (DnaG) synthesizes 4-5 nt primer (RNA) for both leading and lagging strands
DNA pol III replicates the leading and lagging strands at the same time (it has 2 core units)
- The lagging strand is looped in such a way that its synthesis goes in the same direction as the leading strand.
Every 1000bp the DNA pol III will release the lagging strand of DNA and then bind DNA further down the replication fork forming a new loop.
- This leads to discontinuous synthesis of lagging strand in short, 1000bp Okazaki fragments
Every time lagging strand synthesis restarts, primase (Dna G) needs to synthesize a fresh primer.
DNA pol I fills the gaps between Okazaki fragments and removes RNA primers (5’→3’ exonuclease activity) and replaces it with DNA.
DNA ligase connects breaks in newly synthesized strand of DNA by using ATP.

Answer 182

A

Connects breaks in newly synthesized strand of DNA by using ATP.

Answer 183

A

ter sites are short (20bp) sequences of DNA opposite of oriC that are bound by Tus proteins.
Ter-Tus complex stops a replication fork.
There are two sets of Ter-Tus complexes - one for each replication fork.
At the end of the DNA replication we will have catenated (intertwined) chromosomes that are resolved by special topoisomerases (Topoisomerase IV)

Answer 184

A

At the end of DNA replication we will have catenated (intertwined) chromosomes that are resolved by topoisomerase IV.

Answer 185

A

False.

In prokaryotes DNA replication and cell division happen at the same time (see image).

In eukaryotes these two processes are separated temporally.

Answer 186

A

More complex (more proteins and factors involved)
Less studied
Larger genomes
Chromosomes are linear rather than circular
- need to deal with ends (telomeres)
Several origins of replication on each chromosome
Replication fork movement is much slower
- ~50nt/sec vs. ~1000nt/sec in prokaryotes
Replication substrate is chromatin (i.e., histone bound DNA)
- Histone removal is rate-limiting
Okazaki fragments are shorter
- 100-200nt vs. 1000-2000 nt in prokaryotes
Primers are RNA-DNA made by Primase-pol-α
Main polyermase is DNA polymerase δ
- High processivity, associated with PCNA protein similar in function to β-clamp of prokaryotes
RNA primers are removed by combination of RNase H (degrades RNA in RNA-DNA duplexes) and MF1 (maturation factor - 5’→3’ exonuclease)

Answer 187

A

The ends of eukaryotic chromosomes - bound by proteins forming a loop.

Answer 188

A

DNA Polymerase δ

Answer 189

A

By combination of RNase H (degrades RNA in RNA-DNA duplexes) and MF1 (maturation factor - 5’→3’ exonuclease)

Answer 190

A

Eukaryotes

(100-200 nt vs. 1000-2000 nt)

Answer 191

A

Histone removal

Answer 192

A

Chromatin

Answer 193

A

By adding DNA sequences to the ends of chromosomes →catalyzed by telomerase.

Telomerase contains protein and RNA:

The RNA functions as the template
Complementary to the DNA sequence found in the telomeric repeat

Answer 194

A

DNA ends with repeated sequence (in humans AGGGTT) that can be recognized by the telomerase
- Telomerase RNA is complimentary to the AGGGTT).
Once bound to telomeres, proteins part of telomerase will extend (elongate) the telomere. This can be repeated several times to ensure correct length of the chromosomes
The complementary strand can be lengthened by action of primase and DNA pol.

Answer 195

A

DNA directed synthesis of RNA by RNA polymerase.

Answer 196

A

messenger RNA (mRNA) - template for translation
- ~5% of total RNA
- Each mRNA is usually unique (thousands - millions of unique sequences)
transfer RNA (tRNA) - carry activated amino acids to ribosomes
- ~15% of total RNA
- At least 1 tRNA per amino acid
- Usually smaller than mRNA
ribosomal RNA (rRNA) - major component of ribosomes
- Serve structural and catalytic roles
- ~80% of total RNA
- Few types: e.g., 23S, 16S, and 5S rRNAs
Other RNAs (<1% of total RNA)
- Examples: microRNA (miRNA), small nuclear RNA (snRNA), small interfering RNA (siRNA), long non-coding RNA (lnRNA), small nucleolar RNA (snoRNA), transfer-messenger RNA (tmRNA) - usually involved in regulation of gene expression.

Answer 197

A

Mg²⁺ similar to DNA pol, to coordinate NTPs in the active site.

Answer 198

A

Transcriptional control sequences: sequences other than promoter that control transcription
Promoter: sequence of DNA usually upstream (on the left) of transcription start (+1 site)
Transcript: RNA that is being transcribed
Termination sequences

Answer 199

A

Coding strand (non-template) - the strand of DNA that has the same sequence as RNA except T in place of U

Template strand (non-coding) - the strand of DNA that serves as a template during transcription (complimentary to coding strand)

Answer 200

A

People typically report a gene sequence as the DNA sequence that would be the mRNA if the T’s were written as U’s (the coding strand).

The template strand is complementary and antiparallel to the coding strand.

In eukaryotes, introns would also be present in the DNA.

Answer 201

A

Examples of σ proteins: σ subunit recognizes a promoter and initiates transcription. After transcription initiation σ subunit will dissociate from RNA pol.

Standard promoter: σ 70

Heat-shock promoter: σ 32

Nitrogen-starvation: σ 54

Answer 202

A

-10 sequence (Pribnow box)
- Consensus sequence is TATAAT
-35 sequence
- Consensus sequence is TTGACA
UP element
- Less common than the other elements - usually present in genes with strong transcription such as housekeeping genes.

Answer 203

A

True.

Strong promoters will initiate more effectively.

Answer 204

A

Sequence of promoter (e.g., sequence of -10 element, or presence of UP element)
Presence or absence of regulatory proteins
σ proteins

Answer 205

A

E.coli RNA pol consist of α2ββ’ω (core protein) and σ subunit. All together they form holoenzyme.

RNA pol holoenzyme slides along the DNA ‘looking’ for the promoter to initiate transcription.

σ subunit recognizes a promoter and initiates transcription. After transcription initiation σ subunit will dissociate from RNA pol.

Answer 206

A

Standard promoter: σ 70

Heat-shock promoter: σ 32

Nitrogen-starvation: σ 54

Answer 207

A

RNA pol holoenzyme scans the DNA.
σ protein determines specificity of what promoter will be recognized.
Once RNA pol holoenzyme is bound to the promoter (‘closed’ complex), it will unwind the DNA double helix (~17bp) forming ‘open’ complex.
Transcription will start (no primer is required)
σ protein will dissociate from RNA pol.
Only core RNA pol continues transcription promoter clearance and move to elongation phase.

Answer 208

A

RNA pol (core enzyme) synthesizes RNA using DNA as a template (non-coding strand)
RNA synthesis is in 5’→3’ direction
RNA pol has limited proof-reading capacity
Transcription occurs in transcription bubble - region of denatured DNA enclosed by RNA pol and containing the nascent (newly synthesized) RNA.
Transcription bubble contains 8 bp long DNA-RNA hybrid
DNA spontaneously rewinds back to double helix at the rear of the transcription bubble.

Answer 209

A

RNA is released from DNA and RNA pol
DNA helix completely rewind and RNA pol falls off DNA
There are two common mechanisms of termination:
- Rho independent
- Rho dependent

Answer 210

A

Nascent RNA has CG palindromic sequence that forms stem loop structure that forces RNA pol to pause
A/U region is only weakly bound to DNA so RNA will dissociate from DNA

Answer 211

A

Rho (ρ) is a protein that binds nascent RNA at its binding site and travels along the RNA using ATP hydrolysis
Once Rho catches up with RNA pol it will cause its dissociation from the DNA

Answer 212

A

Uses 3 different RNA polymerases: I, II, III
Chromatin (have to deal with histones)
mRNA is polyadenylated at 3’ end and has 5’ cap
RNA must be spliced - made as larger precursor with intervening sequences (introns) that are removed before mRNA is exported from nucleus
Activation of transcription can occur over long distances (enhancers)

Answer 213

A

RNA pol I: transcription of rRNAs (18S, 5.8S and 28S rRNAs)
RNA pol II: transcription of mRNAs, snRNA, miRNA
RNA pol III: transcription of tRNA, 5S rRNA, snRNAs

Answer 214

A

False - RNA pol II transcribes mRNA, snRNA and miRNA.

RNA pol I transcribes rRNAs (18S, 5.8S, and 28S rRNAs)

Answer 215

A

Chromatin remodelling - in order to be more accessible for RNA pol and other proteins.

Answer 216

A

In eukaryotes for transcription to occur chromatin must be remodelled by chromatin remodelling complex to be more accessible for RNA pol and other proteins.

E.g., histone acetyl transferase (HAT) adds acetyl groups to Lys of histones, making the histones less positively charged, which weakens the electrostatic interactions with DNA.

Answer 217

A

Regions of DNA that are less densely packed and more accessible are called euchromatin - active transcription.

Euchromatin regions are unique for each cell type.

Answer 218

A

Regions of DNA that are more densely packed - no transcription.

Heterochromatin regions are unique for each cell type.

Answer 219

A

Each RNA pol has a distinct type of promoter
Eukaryotes also have other sequences of DNA affecting the promoter - enhancers and insulators - with variable sequences and locations
Promoters, enhancers, insulators, and other regulatory elements are called ‘cis’ elements (sequences of DNA)
Eukaryotes have a number of proteins that bind promoters and enhancers, called transcription factors.

Answer 220

A

TATA box (aka Hogness box) - consensus sequence TATAAA: key cis-element, discovered first, between -30 and -100 nt.
CAAT box - sometimes present, usually -40 to -150 nt.

Note: These cis elements are not directly recognized by RNA pol II, it requires transcription factors to recognize the promoter and recruit RNA pol II.

Answer 221

A

A set of general transcription factors (TFIIs - involved in almost all RNA pol II transcription initiation) guide RNA pol II to the transcription start site:

TFIID binds to TATA box through TATA box binding protein (TBP) that unwinds and bends DNA
TFIIs + Pol II then sequentially or all at once bind to the promoter forming basal transcription complex
DNA unwinding changes the closed basal transcription complex into open basal transcription complex
The C-terminal domain of RNA Pol II is phosphorylated at multiple sites
RNA Pol II clears the promoter and transcription starts.

Note: This basal transcription complex does not initiate transcription by itself often. Gene specific transcription factors will bind to cis elements to increase transcription initiation rate (e.g., SREBP (TF) binds SRE (cis element) to increase transcription of HMG-CoA reductase.

Answer 222

A

False.

Unlike prokaryotic RNA, eukaryotic RNA is heavily processed, especially mRNA, preparing it for translation.

The unprocessed mRNA is known as pre-mRNA (i.e., not ready for translation)

Answer 223

A

False.

It is joined by an N-glycosidic linkage

Answer 224

A

False.

This is only true of free pentoses, not those that are components of nucleic acids.

Answer 225

A

False.

DNA contains deoxyribose, which lacks a hydroxyl group at the 2’ carbon.

Answer 226

A

Purine

Guanine

Answer 227

A

Purine

Adenine

Answer 228

A

Pyrimidine

Cytosine

Answer 229

A

False.

Cytosine makes three hydrogen bonds with guanine.

Answer 230

A

Ribose has -OH at 2’ carbon (component of RNA)

Deoxyribose has -H at 2’ carbon (component of DNA)

Answer 231

A

2 ATP used for conversion of R5P to PRPP

5 ATP used for synthesis of IMP from PRPP

2 ATP used for conversionIMP to GMP

1 ATP used for conversion of GMP to GDP

Net cost: - 10 ATP

Answer 232

A

1, 2 and 4

Answer 233

A

The synergistic effect of IMP, AMP, and GMP.

Answer 234

A

AMP will be a feedback inhibitor of the condensation of IMP with aspartate.

Answer 235

A

2 ATP used for conversion of R5P to PRPP

2 ATP used for synthesis of carbamoyl phosphate and UMP

2 ATP used for conversion of UMP to UTP

1 ATP used for conversion of UTP to CTP

Net cost: – 7 ATP

Answer 236

A

Orotic aciduria is caused by a deficiency of orotate phosphoribosyl transferase/orotidine-5’-decarboxylase (=UMP synthetase).

Since this enzyme is deficient UMP will not be produced in high enough quantities for pyrimidine synthesis.

UMP supplementation bypasses this block.

Answer 237

A

Allopurinol is a competitive inhibitor for xanthine oxidase.

It inhibits the enzyme to stop uric acid production.

Hypoxanthine (enol form) is another compound that can be used for gout treatment.

Answer 238

A

True, as long as HG-PRT is available, the bases will be salvaged to make AMP, GMP, and IMP, which inhibits the rate limiting step of the synthesis.

Answer 239

A

True.

Allopurinol directly inhibits the production of uric acid.

Answer 240

A

True.

Hypoxanthine is a substrate for the inhibited enzyme, and therefore, increases.

Answer 241

A

False.

As long as HG-PRT is available, the increasing xanthine, and hypoxanthine will consume the PRPP (unlike in Lesch-Nyhan syndrom)

Answer 242

A

True.

Xanthine is a substrate for the inhibited enzyme.

Answer 243

A

The incorporation of this analog into the growing strand would result in chain termination since there is no 3’-OH to form a phosphodiester bond with the next incoming nucleotide.

Answer 244

A

Alkaline hydrolysis of RNA.

At basic pH, the 2’-OH gets deprotonated and can perform a nucleophilic attack on the phosphate group in the phosphodiester bond of the sugar-phosphate backbone of the RNA.

Answer 245

A

False.

The rotation about the C-1’ to N-9 bond is not limited as purine can exist in both syn and anti conformations.

Answer 246

A

False.

Pyrimidines are only found in the anti- conformation.

Answer 247

A

False.

Anti- or syn- conformation in purines is determined by rotation about the C-1’ to N-9 bond.

Answer 248

A

The favoured form for both purines and pyrimidines is anti- because it limits steric hindrance with the sugar ring.

Answer 249

A

False.

The favoured form is anti because is limits steric hindrance with the sugar ring.

Answer 250

A

To join adjacent nucleotides in one strand.

Answer 251

A

To link nitrogenous base to pentose in nucleotide.

Answer 252

A

This is the difference between a nucleoside and a nucleotide.

Answer 253

A

To join complementary nucleotides between two strands.

Answer 254

A

False.

Left-handed helix is possible - less common.

Answer 255

A

False.

A-T base pairs are denatured more readily than G-C pairs.

Answer 256

A

False.

The sequence of bases will influence base stacking (pi-bonds) and folding of the DNA.

Answer 257

A

False.

The two strands are anti-parallel.

Answer 258

A

For the same strand:

[T] = I

[C] = I

[T] + [C] = 1 - 0.3 - 0.24 = 0.46

For the other strand:

[A] = I

[T] = 0.3

[A] + [T] = I

[G] = I

[C] = 0.24

[G] + [C] = I

Answer 259

A

Rotation about the C-1’ to N-9 bond.

Answer 260

A

GC base pairs have 3 hydrogen bonds as opposed to 2 hydrogen bonds between AT base pairs. Therefore, DNA with higher GC content requires a lot more energy to be broken down, leading to higher melting point temperature. The high T_m allows for survival of the thermophilic bacteria at termperatures in excess of 90^oC.

Answer 261

A

Thermophilic bacteria will have a higher melting temperature than E.Coli. Absorbance values (UV 260nm) increase when DNA is denatured as compared to intact double stranded DNA. If samples from both bacteria are heated to temperatures higher than 37°C but lower than 90°C, the bacteria from E.Coli will denature but the thermophilic bacteria will remain double stranded. By measuring the absorbance values of the two samples pre- and post- heating, we can identify the E. Coli bacteria as its absorbance value will be higher than its double stranded counterpart, while the absorbance of the thermophilic bacterial DNA will not change.

Answer 262

A

Yes, cations minimize repulsions between strands (phosphate backbones)

Answer 263

A

Yes, GC = 3-H bonds; AT = 2 H-bonds.

Answer 264

A

Yes, high or low pH changes nitrogenous base tautomers which affects H-bonds.

Answer 265

A

No, amount of DNA does not influence T_m.

Answer 266

A

Yes.

Whether the solvent is aqueous or organic influences the hydrophobic effect, which effects the stability of the DNA.

So organic solvent = less hydrophobic effect = less stable DNA = lower melting temperature.

Answer 267

A

False.

Relaxed DNA: 10.5 bp/turn → no supercoils and most energetically favourable state.

When DNA is underwould or overwound supercoiling helps to relieve some of the topological stress.

Supercoiling has higher or lower bp/turn overall compared to relaxed DNA.

Answer 268

A

False.

Any organism that has ds-DNA capable of replicating and transcription requires DNA to be underwound to allow for DNA separation.

Answer 269

A

False.
Negative DNA travel toward positive anode.

Separates based on size → supercoiled DNA is compacted in size and will appear smaller than relaxed DNA on a gel.

Answer 270

A

False.

Negative supercoiling (underwound) DNA makes strand separation easier for replication or transcription to occur.

Bacterial and eukaryotic DNA is typically underwound for this reason.

Overwinding is not usually seen.

Answer 271

A

True.

Supercoiling allows larger scale helical structure of relaxed DNA to be compacted so that it can fit the cellular environment.

Answer 272

A

False.

Topoisomerase II is coupled to ATP hydrolysis, not topoisomerase I.

Answer 273

A

Sample B.

Topoisomerase can underwind or overwind DNA.

The increase in size indicated by the banding on the gel for sample B pre- and post- treatment show that sample B is supercoiled.

Answer 274

A

A. Lk₀ = total bp/ number of base pairs per turn of relaxed DNA

Lk₀= 1700/10 = 170

B. DNA gyrase → topoisomerase II - changes Lk by 2 for every enzymatic turnover. When DNA is underwound Lk reduces.

Lk₁ = Lk₀ - (12 x 2) = 170 - 24 = 146

C. DNA in part B is underwound (> 10 bp per turn) so it needs to be overwound to relax it. Topoisomerase I changes Lk by 1 for every enzymatic turnover. When DNA is overwound Lk increases.

Lk₂ = Lk₁ + (8 x 1) = 146 + 8 = 154

Answer 275

A

False.

When DNA is overwound Lk increases.

Answer 276

A

Total number of base pairs per number of base pairs per turn.

Reduces when DNA is underwound.

Increases when DNA is overwound.

Answer 277

A

2: 0 for hybrid to light
1: 127 for hybrid to heavy

Answer 278

A

No. DNA polymerase catalyzes the incorporation of new dNTPs by cleaving off pyrophosphate, which have the β and γ phosphate groups. Only the α-groups are incorporated into the newly synthesized strand.

Answer 279

A

Top is lagging.

Bottom is leading.

Answer 280

A

False. 5’→3’ exonuclease performs a nick translation function.

Answer 281

A

False.

3’→5’ exonuclease performs a proof-reading function.

Answer 282

A

False.

DNA-A protein binds within the Ori-C site and causes helix unwinding at the AT rich sites (DUE region).

Answer 283

A

False.

DNA ligase seals gaps between DNA fragments.

DNA-B helicase helps to unwind and open up DNA.

There is no DNA-A helicase.

Answer 284

A

False.

DNA pol III extends the primer by the 3’ end, DNA primase places the primer.

Answer 285

A

Single-stranded binding proteins.

These proteins help to stabilize the separated DNA strands during the replication process.

Answer 286

A

Primer W = ACGTAGAGAC

Primer X = TACTCGGTCC

dNTP’s are added onto the 3’-OH end of the primers –chain elongation takes place in the 5’→3’ direction. Primers will have complementary sequences to their template strands. To amplify the region of interest: primer X will need to attach to the lagging (5’→3’) template strand and have a complementary sequence to the underlined region. Primer W will need to attach to the leading strand (3’→5’) template strand and have the same sequence as the underlined region.

Based on questions in tutorial: If the DNA was single stranded and you only need to amplify the region on the given 5’→3’ template, you would still need both primers to amplify the region of interest. Round 1 of PCR would only utilize primer X and produce the complementary 3’→5’ strand. All the following PCR rounds would utilize both primers to amplify the region of interest.

Answer 287

A

Telomeres are repeating sequences found at the ends of linear eukaryotic chromosomes. Due to the ‘primer problem’ chormosomes get shortened after each replication. Telomeres buffer the ends of chromosomes and allow for replication of all genetic information by lengthening the chromosome so that the essential part of the chromosome doesn’t get lost during replication.

Answer 288

A

It uses an RNA template to synthesize DNA telomeres. The enzyme has an RNA component and DNA polymerase activity.

Answer 289

A

58°C →annealing of primers

72°C →extension of primers –Taq polymerase adds dNTPs to 3’ end of primers.

Answer 290

A

Deamination of adenosines (A) on the template strand and mutation of ALL palindromic sequence formations in the mRNA.

Answer 291

A

5’ cap
Polyadenylation
Splicing
RNA editing

Answer 292

A

RNA 5’ Cap

Answer 293

A

RNA 5’ Cap

Answer 294

A

5’ to 5’ linkage to guanylate formed at the 5’ end during RNA maturation
The guanine is methylated at N-7
The 2’-OH of ribose is also methylated
Helps ribosome to recognize the 5’ end of mRNA
Improves stability and prevents degradation of mRNA
Present only on mRNAs

Answer 295

A

Addition of poly A tail at the 3’ end of mRNA
When most of mRNAs are transcribed they will have AAUAAA sequence at 3’ end
A special endonuclease recognizes this sequence and cuts mRNA downstream of the signal
Then PolyA polymerase adds 200-300 adenylate residues to the 3’-OH
Recognized by translation initiation factors
Improves stability of mRNA

Answer 296

A

Region of mRNA that is retained in mature mRNA (usually have coding sequence)

Answer 297

A

Regions of mRNA that are removed during maturation (usually non-coding sequence)

Answer 298

A

The removal of introns and linking exons together in mRNA.

In humans, there are on average 8 introns per mRNA.

Answer 299

A

β-globin has 3 coding sequences (exons) and 2 non-coding regions (introns).

Answer 300

A

Introns are considered ‘selfish’ elements of genome
Introns can include non-coding RNAs (miRNA, snRNA, etc.) that are involved in gene expression regulation
Intron-mediated enhancement of gene expression
Introns were first discovered as loops of DNA hybridized with corresponding mature mRNA.

Answer 301

A

True.

Alternative splicing - regions that are removed from mRNA will differ from cell to cell, as a result mature mRNA and proteins will have different sequences.

Answer 302

A

Insulin receptor

Exon 11 is retained in splicing
- IR-B (main metabolic receptor for insulin)
Exon 11 is removed during splicing
- IR-A (mainly transduces growth and anabolic effects of insulin, especially during development)

Answer 303

A

First 3’-OH of free guanosine acts as a nucleophile attacking the P in at the 5’ splice site and 5’ intron-exon border is cleaved
Then 3’-OH of 5’ exon acts as a nucleophile and attacks P at 3’ splice site and 3’ intron-exon border is cleaved
Self-catalyzing (a.k.a. ribozymes)

Answer 304

A

2’-OH Adenosine in the middle of the intron acts as a nucleophile attacking the P in the 5’ splice site and 5’ intron-exon border is cleaved
As a result, lariat structure with 2’-5’ phosphodiester bond is formed
The second step is the same as in Group 1: exons are joined and intron leaves as a lariat
Self-catalyzing (a.k.a. ribozyme)

Answer 305

A

Spliceosome mediated

Answer 306

A

Requires group snRNA and protein complex, called spliceosome (a.k.a. snRNP: small nuclear ribonuclear proteins)
Spliceosome recognizes introns and removes them
The mechanism involves formation of a lariat structure

Answer 307

A

rRNA: active site of peptidyl transferase in ribosome is part of the 23S rRNA in the large subunit.

Answer 308

A

Group I introns

Group II introns

rRNA: active site of peptidyl transferase in ribosome is part of the 23S rRNA in the large subunit

Answer 309

A

Change in nucleotide sequence in mRNA prior to translation, which leads to changes in protein sequences.
Rare - in humans less that 1% of all mRNA

Answer 310

A

Apolipoprotein B: 1 gene - 1 pre-mRNA, but 2 mature mRNAs:
- In the liver, no editing happens - so full length proteins are produced (ApoB100) and incorporated into LDL
- In the intestines, RNA editing is happening and C is edited to U, this introduces UAA stop codon and translation stops early, resulting in ApoB48, which is incorporated into chylomicrons.

Answer 311

A

False.

Cephalopods, such as octopus, edit more than 50% of mRNA in their brain.

RNA editing is rare in humans - less than 1% of all mRNA.

Answer 312

A

True

5’ capping and 3’ polyadenylation require specific enzymes and interactions found only in eukaryotes.

Answer 313

A

True.

Prokaryotic mRNA doesn’t require extensive post-transcriptional modification before translation.

Answer 314

A

True.

Ribosomes are found outside of the nucleus. mRNA must be modified (capping, polyA tail) and usually spliced before export + translation.

Answer 315

A

True.

Histone modifications can affect how DNA binds histone (e.g., Acetylation, methylation), and also how tightly DNA and histones are packed together.

Answer 316

A

False.

Eukaryotic mRNA requires processing. Prokaryotic mRNA can be directly translated during or after transcription.

Answer 317

A

A basal promoter (TATA box)

Answer 318

A

Splicing and removal of introns
Capping of the 5’ end
Conversion of a primary mRNA transcript into a mature mRNA transcript
Polyadenylation of the 3’ end

Note: Nucleotide proofreading function is not a type of RNA processing carried out by eukaryotes.

Answer 319

A

False.

They all recognize different promoter regions.

Answer 320

A

False.

RNA Pol II has similar structure to bacterial RNA polymerase.

Answer 321

A

False.

They recognize differen promoters.

Answer 322

A

False.

Eukaryotic transcription occurs in the nucleus.

Answer 323

A

TFIID is a transcription factor that is involved in binding the TATA box through TBP.

Answer 324

A

A. mRNA #2 is missing the poly-A tail. This is because the consensus sequence 5’-AAUAAA that is 10bp upstream is missing and without this sequence the poly-A tail is not added.

B. Presence of poly A tail suggests these mRNA came from eukaryotes and NOT prokaryotes (bacteria).

Answer 325

A

True.

2’OH in adenosine acts as the nucleophile.

Answer 326

A

False.

The adenine performs the first nucleophilic attack on the 5’ end, and not the 3’ hydroxyl.

Only after the lariat is released that the 3’-OH is available to perform a nucleophilic attack on the 5’ end and bring the exons together.

Answer 327

A

A guanine nucleoside.

Answer 328

A

A triplet (3 nts → 1AA) that DNA/RNA sequence to AA sequence.

Answer 329

A

It’s almost universal.
It’s sequential and unambiguous (non-overlapping)
It has no breaks
It has start and stop signals
It is degenerate (or redundant) - 64 codons but only 20 amino acids.

Answer 330

A

Codons that encode the same amino acid are called synonymous. Usually they only vary at the last nucleotide of the codon.

Answer 331

A

UAA

UAG

UGA

Normally these do not correspond to any amino acid and are not read by tRNA, but rather read by release factors.

Answer 332

A

AUG

This encodes Methionine.

Answer 333

A

Har Gobind Khorana (worked at UBC).

Answer 334

A

mRNA has purine-rich sequence (a.k.a. Shine-Dalgarno sequence) - GGAGG - usually located several nucleotides upstream of the AUG start codon.
The Shine-Dalgarno sequence forms H-bonds with 16S rRNA
The first amino acid is actually N-formyl methionine (fMet)
The N-formyl methionine tRNAi will bind methionine first, and then methionine is modified to fMet by a special enzyme.

Answer 335

A

mRNA does not have analog of SD sequence
Ribosome recognizes cap structure and then scans the mRNA from 5’ end until it encounters an AUG start codon.
Uses Met as the first amino acid (no fMet in eukaryotes)

Answer 336

A

No. fMet and Met are very often removed after translation.

Answer 337

A

formyl methionine

Answer 338

A

N-formylmethionine (fMet)

Answer 339

A

Sequence of non-overlapping nucleotide triplets.

Answer 340

A

DNA/RNA sequence with start and stop codons that encode a protein.

Answer 341

A

Recognition of anticodon loop and other secondary and tertiary structures of tRNA.

Answer 342

A

False.

There is only one aminoacyl tRNA synthetase per amino acid.

Answer 343

A

1st and 2nd base of the codon are specific to the 3rd and 2nd bases of anticodon (specific)
1st base of anticodon and 3rd base of codon are not always specific - this contributes to the degeneracy of the code.
Some tRNAs use hypoxanthine in the first position of the anticodon

NOTE: Need to know the table!

Answer 344

A

True.

Wobble pair

Answer 345

A

E.coli ribosome 70S (2700 kDA) consists of 2 parts:

Large subunit - 50S: 34 proteins, 23S rRNA and 5S rRNA
Small subunit - 30S: 21 proteins, 16S rRNA

2/3 of ribosome is RNA which is important for structure and function of ribosome.

Answer 346

A

tRNA has L-shaped 3D structure with CCA terminus at one end and anticodon loop at the opposite end.
Usually between 73-93 nucleotides long with some parts single stranded and some double stranded that form common structures:

Acceptor stem with amino acid attachment site
TΨC loop
Extra arm (not always present)
Anticodon
DHU loop
5’ end paired with G

Answer 347

A

tRNAs often contain unusual bases that prevent certain base pairing formation. These bases are:

Dihydrouridine (DHU)
Pseudouridine (Ψ)
5-methyl cytidine
Thymidylate

Answer 348

A

Cellular protein synthesis on ribosomes using mRNA as a template.

Translation of nucleic acid alphabet (4 letter) to amino acid alphabet (20)

Answer 349

A

To carry amino acids to the mRNA and ribosome.

Some parts are double stranded and some are single stranded.

Answer 350

A

Template recognition site containing 3 bases - anticodon
- Interacts with mRNA
- Anticodon base pairs with 3 bases on mRNA known as codon
- Each codon encodes a particular amino acid
Amino acid attachment site is at the 3’ end of the tRNA
- Usually ends with CCA
- Amino acid is attached to A via carboxylic acid group

Answer 351

A

5-Methylcytidine

(mC)

Answer 352

A

Dihydrouridine

(UH₂)

Answer 353

A

Aminoacyl-tRNA

Answer 354

A

They are the actual ‘translators’ that know both ‘languages’.

Attach the right amino acid to the right tRNA
Attach C-terminus of the amino acid to the 3’-OH of ribose in terminal adenosine in tRNA

Answer 355

A

Activation step: amino acid reacts with ATP to form aminoacyl-AMP (a.k.a. amino acyl adenylate) and PP_i.
Transfer of amino acid from aminoacyl-AMP to tRNA
1. Class 1: 2’-OH of ribose serves as a nucleophile attacking aminoacyl-AMP and forming aminoacyl-tRNA.
2. Class 2: 3’-OH of ribose serves as a nucleophile attacking aminoacyl-AMP and forming aminoacyl-tRNA.
Only for class 1 - aminoacyl is transesterified from 2’-OH to 3’-OH.

Overall:

Amino acid + ATP + tRNA → aminoacyl-tRNA + AMP + PP_i(→2P_i)

Answer 356

A

1,400 nucleotides

Answer 357

A

The initiation codon determines the correct reading frame, so it is more critical than the termination codon.

Prokaryotes determine the correct reading frame of a given mRNA sequence that is to be translated via the Shine-Dalgrano sequence. This sequence in the mRNA base pairs with a complementary sequence in the 16S RNA of the ribosome; this positions the correct start codon (AUG) on the 30S ribosomal subunit. Thus, the correct reading frame is distinguished by the proximity of the initiation codon (AUG) to the Shine-Dalgarno sequence.

Answer 358

A

mRNA:

5’-ACAUUUAUGCCUGAGCCGUAUAUCGAUGAAUGACACACC-3’

Amino acid:

N’-Met-Pro-Glu-Pro-Tyr-Ile-Asp-Glu-C’

Answer 359

A

Wherever Cys normally occurs.

It would be incorporated wherever the tRNA normally incorporates its amino acid (i.e., wherever Cys normally would be incorporated). The tRNA already read the codon on the mRNA and the amino acid itself does not play a role in this process.

Answer 360

A

120+80=200

200x3 nucleotides per amino acid =600bp

The DNA fragment is 400bp long, but the two proteins encoded a total of 200 amino acids.

Each amino acid residue requires a 3-letter codon, so at least 600 bps are required in mRNA to encode for the two proteins.

Answer 361

A

Polypeptide: Met-Arg-His-Asn

mRNA transcript: 5’-AUG-CGC-CAC-AAC-UGA-UGA-3’

Template strand: 3’-TAC-GCG-GTG-TTG-ACT-ACT-5’

Coding strand: 5’-ATG-CGC-CAC-AAC-TGA-TGA-3’

Yes, mRNA complementary to the given tRNA is CAC (His). If the 3rd position is changed so that the mRNA is now CAA, then it will encode Gln instead.

Answer 362

A

A.

Anticodon: 3’-UCG*-5’

Codon: 5’-AGA-3’

Amino acid: Arginine

B.

Anticodon: 3’-UCG-5’

Codon: 5’-AGC-3’

Amino acid: Serine

Answer 363

A

5’ → 3’ direction

Answer 364

A

mRNA is bound in 30S subunit, specifically 16S rRNA recognizes Shine-Dalgarno sequence.

Answer 365

A

Protein synthesis occurs in the 50S subunit.

50S has a ‘tunnel’ that allows the growing polypeptide chain to pass through and exit the ribosome.

Answer 366

A

E (exit), P (peptidyl), and A (aminoacyl)

Each site is usually occupied by tRNA.

Answer 367

A

Where aminoacyl-tRNA enters the ribosome.

Answer 368

A

Where the growing peptide chain is located.

Answer 369

A

Where the empty tRNA is located.

Answer 370

A

Initiation of protein synthesis.

Answer 371

A

Requires a set of translation initiation factors.

30S binds IF1 and IF3 which prevents 30S from binding 50S prematurely and directs IF2 to the P-site.
IF2 binds to fMet-tRNA_fMet
- IF2 is a G-protein, meaning it requires GTP hydrolysis for its function.
- IF2 bound to GTP will bind fMet-tRNA_fMet - ternary complex
Ternary complex binds to mRNA and 30S subunit with IF1 and IF3 forming 30S initiation complex
IF2 hydrolyzes GTP to GDP and Pi
- IF1, IF2-GDP, IF3 leave the initiation complex
- 50S subunit binds to the initiation complex forming 70S initiation complex
- Translation can start!

Note: fMet-tRNA_fMet is in the P-site

Answer 372

A

To prevent 30S from binding 50S prematurely, and also to direct IF2 to the P-site.

Answer 373

A

False.

IF2 is a G-protein.

Answer 374

A

False.

IF2 is a G-protein.

Answer 375

A

A G-protein (e.g., IF2, EF-Tu) requires GTP hydrolysis for its function.

Answer 376

A

IF2 bound to GTP will bind fMet-tRNA_fMet which forms the ternary complex.

Answer 377

A

Ternary complex binds to mRNA and 30s subunit with IF1 and IF3 forming the 30S initiation complex.

Answer 378

A

50S subunit binds to the initiation complex forming 70S initiation complex.

Answer 379

A

In the P-site.

Answer 380

A

tRNA with matching anticodon binds to the A-site.
The nitrogen from the amino group on the tRNA in the A-site attacks carbonyl group, attached to the tRNA in the P-site, catalyzed by 50S subunit rRNA (peptidyl transferase center)

Answer 381

A

N of amino group on the tRNA in A-site attacks the carbonyl group attached to the tRNA in the P-site.
Works via oxyanion tetrahedral intermediate
This results in separation of peptidyl from tRNA in the P-site and the formation of the peptide
With the formation of the peptide bond, the tRNA with peptidyl chain is in the P-site in the 50S subunit, but it is still in the A-site in the small subunit. Likewise, the empty tRNA is in the E-site in 50S, but it is still in the P-site in 30S

Answer 382

A

During elongation of protein synthesis in prokaryotes, EF (elongation factor) -Tu is a G-protein that can bind aminoacyl-tRNA, which protects the ester linkage and positions the aminoacyl-tRNA in the A-site of the ribosome.

Answer 383

A

EF-Tu hydrolyzes GTP to GDP once tRNA is in the A-site and codon/anticodon pairing is correct.
EF-Tu-GDP has low affinity to aminoacyl-tRNA and is released
EF-Ts is a GDP → GTP exchange factor that binds to EF-Tu-GDP and facilitates exchange of GDP to GTP
Now EF-Tu-GTP can deliver another aminoacyl-tRNA

Answer 384

A

False.

EF-Tu-GDP has low affinity to aminoacyl-tRNA.

Answer 385

A

EF-G bound to GTP binds the ribosome and hydrolyzes GTP causing conformational changes that force part of EF-G into the A-site.
This displaces the peptidyl tRNA into the P-site in the 30S subunit, dragging mRNA with it.
As a result mRNA will translocate by 3 bases.
Now, new codon is in the A-site, the empty tRNA is in the E-site.
Then, empty tRNA will leave the E-site and EF-G-GDP will leave the A-site.

Answer 386

A

Release factors - RF1 and RF2 recognize stop codons
One end of RF1 recognizes UAA and UAG, and one end of RF2 recognizes UAA and UGA
Opposite end of RF1/2 will interact with PTC coordinating water molecule with the help of rRNA
Water will attack the carbonyl group (ester linkage) and hydrolysis will occur.
Polypeptide is released from the ribosome
RF3 is a GTPase. It will hydrolyze GTP to help release RF1/2 from the ribosome

Answer 387

A

Release factors 1 and 2 (RF1 and RF2)

Answer 388

A

UAA and UAG

Answer 389

A

UAA and UGA

Answer 390

A

RF3 is a GTPase.

It will hydrolyze GTP to help release RF1 and RF2 from the ribosome during termination of translation elongation in prokaryotes.

Answer 391

A

Ribosomal recycling factor

RRF with the EF-G, IF-1, IF-3 will dissociate ribosomal subunits and mRNA to recycle the ribosomes for further use.

Answer 392

A

9 or more (called eIFs)

Answer 393

A

Several proteins bind both the 5’ and 3’ ends of the mRNA and bring the together in order to present them to the ribosome.
eIF4 complex recognizes cap (eIF4E) and polyA tail (PABP - poly-A binding protein)
eIF1 and eIF3 serve the same function as IF1 and IF3 in bacteria (homologs)
eIF2 is a homolog of IF2 in bacteria: delivers ternary complex
- eIF2 has regulatory subunit (alpha) that can be phosphorylated to slow down translation (during amino acid starvation, accumulation of missfolded proteins, viral infections)
43S pre-initiation complex (ternary complex + 40S) will scan mRNA (using ATP) to find start codon. When AUG is reached the large 60S subunit will bind and all eIF are released

Answer 394

A

During the initiation of translation in eukaryotes, eIF2 has a regulatory subunit (alpha) that can be phosphorylated to slow down translation (during amino acid starvation, accumulation of missfolded proteins, viral infections)

Answer 395

A

ternary complex and 40S

Answer 396

A

There is none!

The initiation codon is AUG (Methionine), which begins all eukaryotic polypeptides.

Answer 397

A

eEF1-alpha → EF-Tu
eEF1-beta → EF-Ts
eEF2 → EF-G

Answer 398

A

Eukaryotes only have 1 release factor (eRF1) that can recognize all stop codons.

eRF3 is GTPase similar to RF3 in bacteria.

Answer 399

A

Poorly understood.

ABCE1 protein acts similar to RRF in bacteria.

Answer 400

A

They are translated on the ribosomes associated with the ER rather than free cytosolic ribosomes.

This process involves special sequences at the start of mRNA/protein called signal peptide.

Answer 401

A

Stability
Immunogenicity
Delivery
Translation efficiency

Answer 402

A

inosine monophosphate (IMP)

Answer 403

A

5-phospho-ß-D-ribosylamine

Product of the committed step (step 2) of purine synthesis

Answer 404

A

Orotidine 5’-monophosphate (OMP)

Answer 405

A

Orotate, a pyrimidinecarboxylate anion

Answer 406

A

Carbamoyl phosphate

Answer 407

A

5-methylcytosine

Answer 408

A

5-methylcytosine

Answer 409

A

Hypoxanthine

Answer 410

A

cytidine-5’-diphosphate (CDP)

Answer 411

A

Uric acid

Answer 412

A

Uric acid

Answer 413

A

False.

NaNO₂ would greatly accelerate deamination of cytosine.

Answer 414

A

False.

NaNO₂ would greatly accelerate deamination of cytosine.

Answer 415

A

False.

NaNO₂ would greatly accelerate deamination of cytosine.

Answer 416

A

False.

NaNO₂ would greatly accelerate deamination of cytosine.

Answer 417

A

False.

Increased purine catabolism causes gout.

Answer 418

A

False.

Increased purine catabolism causes Gout.

Answer 419

A

False.

Increased purine catabolism causes Gout.

Answer 420

A

False.

Increased purine catabolism causes Gout.

Answer 421

A

Asynchronous with multiple origins of replication

Answer 422

A

Re-written as 3’-AGG-5’, codon is 5’-UCC-3’.

UCC codes for Serine.

Wobble base pairing allows 3’-AGG-5’ to base pair with 5’-UCC-3’ or 5’-UCU-3’.

C & D are correct.

Answer 423

A

False, it is the 30S.

Answer 424

A

False, this is IF-1.

Answer 425

A

False, IF-2 would be bound to the fMet-tRNA^fMet.

Answer 426

A

The codon is CGA, so this is the codon for Arg.

Argininyl-tRNA^Arg

Answer 427

A

False.

The codon is CGA, so this is the codon for Arg.

Argininyl-tRNA^Arg

Answer 428

A

To prevent peptide bond formation at the N-terminus.

Answer 429

A

UMP→UDP (nucleoside monophosphate kinase)

UDP→dUDP (nucleoside diphosphate reductase)

dUDP→dUTP (nucleoside diphosphate kinase)

dUTP→dUMP (dUTPase)

dUMP→dTMP (thymidylate synthase)

Answer 430

A

The phosphodiester backbone causes repulsion of the 2 strands.

Answer 431

A

Due to it’s 3’ to 5’ exonuclease activity.

Answer 432

A

X = G

Y = A

W = U

Z = G

The splice junctions are 5’-GU […] AG-3’

Answer 433

A

A TATA box present at the -25 region (part of DNA)
A poly (A) tail (mature mRNA not pre-mRNA)
A 5’ 7-methylguanosine cap (mature mRNA not pre-mRNA)
UP sequences (bacterial promoter)
Portions of the Inr sequence (yes, because Inr is around the +1 region)

Answer 434

A

There are no DNA polymerases that can synthesize DNA in the 3’→5’ direction.

Answer 435

A

This molecule has no 3’-OH present.

Answer 436

A

C3’ Endo

Answer 437

A

IMP would not contain a radiolabel.

Answer 438

A

Arrow #2 and #3

Answer 439

A

Thymine (aka 5’-methyluracil)

Answer 440

A

(T+C) = (A+G)

pyrimidines = purines

Answer 441

A

Hydrolysis of dNTPs to form dNMPs and PPi.

The spontaneous hydrolysis of PPi occurs after the formation of the phosphodiester bond and ensures that the DNA synthesis reaction is energetically favorable without the need for additional enzymes.

Answer 442

A

Topoisomerase Type II can induce supercoils whereas Topoisomerase Type I can only remove supercoils.
Topoisomerase Type II uses ATP and Topoisomerase Type I does not.

Answer 443

A

Magnesium ions act as co-factors.

Answer 444

A

Primer X = CCCACCGGGCT

Primer Y = GGGATGGCCAG

Answer 445

A

All of them!

Answer 446

A

Interaction with DnaB.

Answer 447

A

30S rRNA (16S rRNA is present here)

Answer 448

A

50S rRNA (23S rRNA is present here)

Answer 449

A

Coding strand

Answer 450

A

DNA polymerase I

Answer 451

A

Processivity

Answer 452

A

Uridine-5’-monophosphate, uridine monophosphate, uridylate
1 at the bottom nitrogen attached to the ribose sugar, then clockwise from there
Bypasses the blockage in the anabolic pathway

Answer 453

A

Folic acid is needed to produce tetrahydrofolate, which is involved in 1 carbon transfer of methyl group from uracil to make thymine. (i.e., dUMP→dTMP).
No production of dTMP means no dTTP, which is a substrate needed for DNA synthesis.
DNA synthesis is necessary for development of the baby (especially the neural tube formation stage!)

Answer 454

A

Asterisks should be at positions 4 and 5 on the ring.

Answer 455

A

B, further down the gel, smaller effective size
Lane 2 because there is still more DNA concentrated near the bottom of the gel, indicating the sample is still more supercoiled.

Answer 456

A

In the long peptide there are 250 codons followed by stop codon.
In the short peptide there are 220 codons followed by stop codon.
A single point mutation most likely occurred in the 221st codon of the long peptide, resulting in the change of a codon that codes for an amino acid to a STOP codon.
When a stop codon is introduced, this resulted in a truncated peptide (classified as a NONSENSE mutation).

Answer 457

A

Release factor
Finished peptide is hydrolyzed from the tRNA that is sitting in the peptidyl site. The ribosome complex dissociates. The 2 ribosomal subunits dissociate, mRNA is removed, tRNAs are released and finished peptide is released.
1^st, 2^nd, and 3^rd tRNA because they all have anticodons with 3^rd position I or G. I and G can wobble base pair. U can wobble base pair too, but it is not listed in any of the anticodons.
First: N-fMet-Ile-Cys-His-Ala-Glu-C | Second: N-fMet-Ile-Cys-Gln-Ala-Glu-C

Answer 458

A

Production of dCDP → dCTP activates production of dUDP → dTTP which activates production of dGDP → dGTP which activates production of dADP → dATP and the cycle continues

Answer 459

A

O6-Methylguanine

6-O-Methylguanine is a derivative of the nucleobase guanine in which a methyl group is attached to the oxygen atom.

It base-pairs to thymine rather than cytosine, causing a G:C to A:T transition in DNA.

Answer 460

A

AZT failed as a chemotherapy drug because human DNA pol has a higher affinity for dTTP than AZT, so it cannot prevent DNA strand from growing in cancerous cells in humans since human DNA pol will still incorporate dTTP instead of AZT when synthesizing DNA in those cancerous cells.

Answer 461

A

AZT prevents HIV because the HIVS reverse transcriptase has higher affinity for AZT than dTTP.

So, DNA synthesis will stop due to a lack of 3’-OH.

Answer 462

A

Riboucleotide reductase activated by rATP because a high level of rATP means that there are a lot more ribonucleotides so we should make more deoxynucleotides to balance the pool.

And on the other hand, riboucleotide reductase is deactivated by dATP and dADP because both high levels of dATP and dADP means that there are too much deoxynucleotides and we need more ribonucleotides.

Answer 463

A

Tu is needed for an amino acyl tRNA to enter the A site, hydrolysis of its GTP to GDP is needed so that the amino acid on the tRNA in the A site will attack the already existing chain in the P site.

Once the Tu-GDP comes off it must then be rephosphorylated by TS so that it can once again associate with an amino acyl tRNA and the peptide chain can be grown longer

Answer 464

A

N¹⁰-formyltetrahydrofolate

Answer 465

A

Aminopterin and methotrexate

Answer 466

A

Fluorouracil

Answer 467

A

Aspartate Transcarbamoylase (ATCase) catalyses the first committed step in pyrimidine biosynthesis. So for that reason it is a highly regulated enzyme.

Synthesis of carbamoyl phosphate by cytosolic carbamoyl phosphate synthetase II is considered an activation step prior to the committed step.

Answer 468

A

DNA pol I

Answer 469

A

ribosomal RNA

Answer 470

A

DNA pol III

Answer 471

A

ATP hydrolysis by DnaC

Answer 472

A

Pseudoruidine

Answer 473

A

Xanthylate

Answer 474

A

Dihydrouridine

Answer 475

A

Adenylosuccinate

Answer 476

A

Hypoxanthine

Answer 477

A

Carbamoyl phosphate

Answer 478

A

Amino groups of both Glycine and Aspartate are incorporated in purines during de novo synthesis. However for Glutamine, it is side chain amide that serves as a nitrogen source, not the amino group.

Answer 479

A

Adenine

Guanine

Hypoxanthine

Answer 480

A

False. CTP inhibits aspartate transcarbamoylase activity.

Answer 481

A

7-methylguanosine joined to the mRNA via a 5’ -> 5’ triphosphate linkage.

Answer 482

A

mRNA must be stored at such a low temperature because it is inherently unstable molecule, even when it is purified from cellular nucleases.

First of all, it is single stranded compared to DNA.

Second, mRNA doesn’t have stabilizing secondary structures like rRNA and tRNA.

Final and most important, RNA has 2’-OH in the ribose. This makes RNA susceptible to base-catalyzed hydrolysis. RNA hydrolysis will spontaneously occurs when the deprotonated 2’-OH of the ribose, acting as a nucleophile, attacks the adjacent phosphorus in the phosphodiester bond of the sugar-phosphate backbone of the RNA. This will lead to breakdown of RNA strand.

With long enough time RNA will be hydrolyzed all way to NMPs.

Answer 483

A

Rho binds to a rut site in the newly made RNA. However, the RNA polymerase may be far away down the template. How fast Rho catches RNA polymerase depends on their relative speed. Rho uses ATP hydrolysis to move along (5’ to 3’) the RNA, chasing the RNA polymerase. When it catches the polymerase, the RNA is pulled out of the DNA RNA-polymerase complex. If concentration of all 4 rNTPs are high, RNA polymerase moves at the maximum speed and can transcribe RNA all the way to the end until it reaches a Rho-independent terminator. This explains long RNA products.

However, when concentration of GTP + CTP + UTP is low RNA polymerase will move slow, whereas Rho will still move at the high speed along the newly made RNA because ATP concentration remain high. So very quickly Rh-factor will “catch” RNA polymerase and cause its dissociation from DNA template. This explains short RNA products.

Answer 484

A

Prokaryotes and eukaryotes have similar processes in translation initiation with small differences in proteins.

In translation initiation in both prokaryotes and eukaryotes, the small ribosomal subunit is bound by two proteins

Prokaryotes: 30S subunit bound by IF1 and IF3
Eukaryotes: 40S subunit bound by eIF1 and eIF3

The tRNA for the first amino acid is bound by a G-protein and delivered to the P site of the small subunit; the G-protein then detaches using GTP hydrolysis

Prokaryotes: fMet-tRNA^fMet bound by IF2
Eukaryotes: Met-tRNA^Met bound by eIF2

The large subunit binds the small subunit after initiation factors leave

Prokaryotes: 50S subunit binds after IF1, IF3, and IF2-GDP leave
Eukaryotes: 60S subunit binds after eIF1, eIF3, and eIF2-GDP leave

Answer 485

A

a) Most likely explanation is that intron 1 is a group I intron, and intron 2 is a group 2 intron. For group I introns external guanosine act as a nucleophile in the first step of splicing, so substituting adenosine with deoxyadenosine will not affect the splicing reaction. So mRNA will be successfully spliced and then translated. However, for group 2 introns 2’-OH of adenosine in the middle of an intron acts as a nucleophile. So if we are substituting adenosine with deoxyadenosine, there will be no 2’-OH to perform nucleophilic attack and splicing cannot happen. Intron will be retained in the mRNA, will be translated and this will change the sequence of the protein.
b) We would expect to observe Zoom phenotype, because cytidine does not act as a nucleophile in either type 1 or type 2 introns

Answer 486

A

UMP + ATP -> UDP + ADP by nucleoside monophosphate kinase (uridylate kinase)
UDP -> dUDP by ribonucleotide reductase
dUDP + ATP -> dUTP + ADP by nucleoside diphosphate kinase
dUTP -> dUMP by dUTPase (I don’t expect you to know the name of the enzyme for this reaction. I did talk about the reaction, but I didn’t name the enzyme in class. )
dUMP + THF ->dTMP + DHF by thymidylate synthase
dTMP + ATP -> dTDP + ADP by nucleoside monophosphate kinase (thymidylate kinase)
dTDP + ATP -> dTTP + ADP by nucleoside diphosphate kinase
dTTP -> DNA by DNA polymerase

UDP can also be converted into dCTP:

UDP phosphorylated by nucleoside diphosphate kinase, which uses ATP and yields UTP and ADP
UTP converted to CTP by glutamine-UTP amidotransferase, which uses glutamine and ATP, yielding CTP, glutamate, ADP, and Pi
Phosphate group in CTP is removed, yielding CDP
2’-OH group in CDP then removed by ribonucleotide reductase, which yields dCDP
dCDP can be phosphorylated by nucleoside diphosphate kinase, which uses ATP, yielding dCTP and ADP

Answer 487

A

a) coding strand

5’-CCTAGGAGGTTTGACCTATGGCGTTGAAAATCAATTGAATC -3’

b) mRNA

5’-CCUAGGAGGUUUGACCUAUGGCGUUGAAAAUCAAUUGAAUC -3’

c) protein

MALKIN

d) ATP equivalents required

Aminoacyl-tRNA synthesis (tRNA charging): 2 X 6 = 12

Initiation: 1 GTP by IF2 =1

Elongation (1 GTP by Tu + 1 GTP by EF-G) X 5 = 10

Termination 1 GTP by RF = 1

Total = 24

Answer 488

A

Beta-alanine

product of uracil degradation

Answer 489

A

Beta-alanine

Product of uracil degradation

Answer 490

A

Beta-alanine

Product of uracil degradation

Answer 491

A

Beta-aminoisobutyrate

Product of thymine degradation

Answer 492

A

Beta-aminoisobutyrate

Product of thymine degradation

Answer 493

A

Beta-aminoisobutyrate

Product of thymine degradation

Answer 494

A

Aminoacyl-tRNA synthesis (tRNA charging): 2 X 4 = 8

Initiation: 1 GTP by IF2 =1

Elongation (1 GTP by Tu + 1 GTP by EF-G) X 3 = 6

Termination 1 GTP by RF = 1

Total = 16

Answer 495

A

3’-5’ is the template strand

5’-3’ is the coding strand

Answer 496

A

1) bicarbonate–> carbamoyl phosphate –> 2 ATP
2) PRPP–> 2 ATP ( from purine pathway)
3) UMP–> UDP –> 1 ATP
4) UDP –> UTP–> 1 ATP

Total = 6 ATP

Answer 497

A

IMP -> rAMP/rGMP -> rADP/rGDP -> dADP/dGDP -> dATP/dGTP

rUMP -> rUDP -> rUTP -> rCTP -> rCDP -> dCDP -> dCTP

rUMP -> rUDP -> dUDP -> dUTP -> dUMP -> dTMP ->dTTP

Answer 498

A

DNApol III vs DNApol δ

DNApol I vs Rnase + MF1

DnaG vs Primase-Pol α

DNA ligase vs DNA ligase I

Answer 499

A

The bacterial homolog of the TATA box (i.e., the Hogness box) is called the Pribnow box which has a shorter consensus sequence.

Brainscape's Knowledge GenomeTM

MT2 Material Flashcards

Brainscape's Knowledge Genome^TM