Exam 2

Question 1

Deoxyribonucleotides are so named because they lack a hydroxyl at which position?

Hint: ribonucleotides have a hydroxyl at this position.

Answer 1

C2’

Question 2

What does N1 on adenine do?

Answer 2

Accepts H bond

Question 3

What is the difference between anti and syn?

Answer 3

Syn: indicates that both H and OH are present on the same side of the double bond. Anti: indicates that H and OH are present on the opposite sides of the double bond.

Question 4

What is the substrate for DNA polymerase?

Answer 4

dATP

Question 5

DNA synthesis occurs in which direction?

Answer 5

5’-to-3’ direction

Question 6

On DNA, the 5’ end is synonymous with ____, while the 3’ end is synonymous with ______.

Answer 6

The phosphate group, a hydroxyl

Question 7

What does nucleotide base tautomerization do?

Answer 7

It disrupts base pairing by changing hydrogen bond donor to acceptors (or vice versa).
Tautomerization changes hydrogen bond donors to acceptors . Since base pairing is based on the identity of the H-bond donor or H-bond acceptor, the base pairing is disrupted.

Question 8

Telomerase lengthens the ends of chromosomes. Telomerase is a ____ polymerase that is associated with a _____ template.

Answer 8

DNA; RNA To prevent the shortening of chromosomes during replication, telomerase extends the end of the DNA strand.

Question 9

What is telomerase? What does it do?

Answer 9

Telomerase is an enzyme that contains an RNA template within itself. During DNA replication, it uses this RNA template to bind DNA at the end of the chromosome. Once it binds to DNA, it polymerizes the end of the DNA strand using the RNA within itself as the template.

Question 10

DNA Polymerase fails to recognize that an incorrect nucleotide has been incorporated, resulting in a mismatch. Which repair pathway will correct this mistake?

Answer 10

Mismatch repair

Question 11

Base 5-methyl-cytosine -> Methylation of cytosine is mutagenic. True or false?

Answer 11

False. Methylation of cytosine at C5 does not disrupt any hydrogen bond donors or acceptors between nitrogenous bases.

Question 12

Which enzyme in base excision repair is not involved in depurinated nucleotide repair?

Answer 12

Glycosylase because it is not required to repair depurinated nucleotides. Glycosylase cleaves the glycosidic bond between the base and ribose sugar. When a nucleotide is depurinated, it lacks a base (specifically a purine base, either adenine pr guanine), therefore there is no glycosidic bond present. Therefore, glycosylase is not needed to repair depurinated nucleotides because there is no glycosidic bond to be cleaved by glycosylase.

Question 13

Can UV light produce “dimers” as a type of DNA damage for pyrimidine nucleotides? Can it for purines?

Answer 13

Yes; No

Question 14

Which of the following scenarios will result in gene activation?
A. A ligand promotes binding of an activator to DNA.
B. An activator is degraded.
C. Expression of a repressor increases.
D. A ligand promotes binding of a repressor to DNA.

Answer 14

A. A ligand promotes binding of an activator to DNA.
An activator stimulates gene expression. Therefore, a ligand promoting the binding of an activator to DNA will result in activation of gene expression.

Question 15

Which amino acid is able to form hydrogen bonds with DNA bases?

Answer 15

Asparagine because its R-group has hydrogen bond donors and acceptors.

Question 16

Cytosine is methylated resulting in 5-methylcytosine. What is the most likely outcome?

Answer 16

Gene expression will decrease because methylation of cytosine in DNA is an example of DNA methylation. DNA methylation decreases gene expression. Methylation can affect transcription factor binding; transcription factors bind to specific DNA sequences, and methylation in the CpG regions upstream of the promoter can decrease that bonding affinity

Question 17

RNA polymerase binds directly to a/an:

Answer 17

Promoter in euchromatin. The promoter is a sequence of nucleotides upstream of a gene. To begin transcription, RNA polymerase binds to the promoter. When DNA is coiled in the heterochromatin form, it is not very accessible, and transcription is limited. RNA polymerase binds and is able to start transcription when DNA is in the uncoiled euchromatin form so that the promoter and the gene of interest are accessible.

Question 18

RNA polymerase requires a primer to initiate polynucleotide synthesis. True or false?

Answer 18

False

Question 19

Consider the following coding strand. What mRNA molecule will be synthesized from this sequence?

5’-CCGTTAAACGCTA-3’

Answer 19

5’-CCGUUAAACGCUA-3’

Question 20

Rho helicase terminates transcription by:

Answer 20

melting the RNA-DNA duplex, which causes RNA polymerase to fall off. Rho helicase binds to an exposed Rho site on the forming RNA. It will move up the RNA strand until it reaches the RNA polymerase where it melts the RNA-DNA duplex. This will cause the RNA strand to be released and for RNA polymerase to dismantle.

Question 21

The 5’ end of a gene corresponds to the 5’ end of the mRNA and the:

Answer 21

5’ end of the coding strand and 3’ end of the template strand.
This is because the DNA template (non-coding strand) and DNA coding strand are completely complimentary and antiparallel to one another. The term template strand refers to the sequence of DNA that is copied during the synthesis of the mRNA. The opposite strand, (the one with a base sequence directly corresponding to the mRNA sequence) is called the coding strand because it corresponds to the sequence being transcribed. The only difference between the mRNA and coding strand is thymine from the coding strand of DNA will be replaced by uracil for mRNA. The direction will be the same for both the mRNA and the coding strand (5’ to 3’).

Question 22

Proteins are synthesized:

Answer 22

Amino to carboxy terminal. Proteins are synthesized in this direction. The mRNA is read by the ribosome in the 5’-3’ direction and the first amino acid to be produced makes up the amino termini and will have a free amino group. The last amino acid added to the polypeptide will contain a free carboxylic acid, making it the carboxy termini.

Question 23

Deaminated cytosine forms __________ and deaminated adenine forms ____________.

a. thymine; guanine

b. uracil; guanine

c. uracil; hypoxanthine (the nucleoside form is called inosine)

d. thymine; hypoxanthine (the nucleoside form is called inosine)

Answer 23

c. Uracil, hyproxanthine

Question 24

During protein synthesis, a peptide bond forms when the on the incoming amino acid attacks the on the polypeptide chain.

Answer 24

Amine, carbonyl

Question 25

Which of the following statements regarding translation is INCORRECT?
Inosine in the tRNA anticodon loop forms a wobble base pair with mRNA codon nucleotides.

The 3’ and 5’ untranslated regions (UTRs) of an mRNA are not removed by splicing (the UTRs are part of the exons).

Every organism uses the same code. This means that an mRNA from fruit flies can be translated in a human cell.

Eukaryotic ribosomes always start translation at the first nucleotide at the 5’ end of an mRNA.

Answer 25

d. Eukaryotic ribosomes always start translation at the first nucleotide at the 5’ end of an mRNA.

Eukaryotic ribosomes begin translation at the first AUG start codon(which is not the first nucleotide). The nucleotides at the 5’ end prior to the AUG start codon are part of the 5’ UTR (untranslated region) and do not code for the protein and are not translated.