Practice Exam #2

Question 1

RNA polymerase recognizes which of the following DNA regions in transcription

Answer 1

promoter

Question 2

in mutant DNA polymerase without 5’–>3’ exonuclease function, what property is expected to have? it will:

Answer 2

likely to keep the RNA primers

Question 3

While prokaryotes have a single transcription factor (_), eukaryotes have several TF-II.

Answer 3

sigma only

Question 4

All tRNA molecules:

Answer 4

have a cloverleaf shape for their secondary structure.

Question 5

What is the most likely result from mutating a 3’-5’ exonuclease activity in the DNA polymerase?

Answer 5

unable to proof reading

Question 6

Before replication start with DNA polymerase, a set of molecules (__________) recognize and bind the origin of replication.

Answer 6

initiator proteins

Question 7

Telomerase is a unique enzyme in that:

Answer 7

has a template.

Question 8

In eukaryotes, what initially binds to the TATA box on the DNA template?

Answer 8

TBP

Question 9

A rho-independent transcription terminator includes:

Answer 9

• inverted repeats which when transcribed fold into a hairpin
• a region in the RNA molecule that causes RNA polymerase to pause

Question 10

Glycosylase and apurinic base are usually associated with:

Answer 10

BER

Question 11

The _______ is an expensive protein in the sense that spends ATP to rotate around the single strand of DNA template for the lagging strand.

Answer 11

helicase

Question 12

The following terms best associate with: Rec and BRCA

Answer 12

HR

Question 13

Eukaryotic transcription factors assist the ______ in binding to the _________.

Answer 13

RNA polymerase; promoter

Question 14

Which protein is observed exclusively in association with eukaryotic DNA replication?

Answer 14

telomerase

Question 15

In eukaryotes, enzyme responsible for transcribing all protein-coding genes is

Answer 15

RNA polymerase II

Question 16

____________ bind to enhancers in eukaryotic cells to increase the rate of transcription.

Answer 16

Transcription activators

Question 17

Introns within primary eukaryotic transcripts

Answer 17

Vary considerably in size and number among different genes

Question 18

What kind of RNA functions in splicing and is associated with the spliceosome?

Answer 18

snRNA

Question 19

A cell of the pancreas would specialize in producing insulin, among other functions, while a muscle cell would specialize in the production of the protein myosin. This means that

Answer 19

Pancreas cells activate the gene that would produce insulin, while muscle cells activate the gene that would produce myosin.

Question 20

Which enzyme seals nicks and joins Okazaki fragments?

Answer 20

DNA ligase

Question 21

Is the XP patients, the patients lacking a DNA repair mechanisms are likely to have more or fewer variations between individuals than a species that possesses repair mechanisms?

Answer 21

More, because more spontaneous mutations will go uncorrected.

Question 22

true or false: Alternative splicing increases variability

Answer 22

true

Question 23

RNA in eukaryotes is produced in the

Answer 23

nucleus

Question 24

Which nuclease cuts in the middle of the DNA strand?

Answer 24

endonuclease

Question 25

All DNA polymerases require a primer with a 3’ OH group to begin DNA synthesis. The primer is:

Answer 25

a short stretch of RNA nucleotides.

Question 26

true or false: The production of polycistronic mRNA in prokaryotes serve as a mechanism of gene regulation because polycistronic mRNA contains the coding sequences for several proteins on a single transcript. These proteins are often the enzymes needed for a common metabolic pathway. This is an economical way for the cell to regulate the synthesis of related proteins in a coordinate manner by a single signal so that appropriate quantities of each are made at the same time. This is termed coordinate regulation.

Answer 26

true

Question 27

In prokaryotes, the strength of the binding of RNA polymerase to different promoters

Answer 27

Depends on the promoter sequences

Question 28

true or false: RNA forms a helix.

Answer 28

false

Question 29

In rho-dependent transcription termination, the rho factor binds to ___________.

Answer 29

mRNA

Question 30

Which of the following features is common to both DNA replication and transcription?

Answer 30

A sugar-phosphate bond is formed between the 3’ hydroxyl and the 5’ phosphate

Question 31

Which of the following is unique to eukaryotes?

Answer 31

removal of introns

Question 32

When RNA is transcribed from a gene, which strand of DNA is used?

Answer 32

the template strand

Question 33

If the dispersive model of DNA replication had been correct, Meselson and Stahl would have observed that DNA extracted from bacterial cells following a second round of DNA replication in 14N would have been:

Answer 33

only of hybrid density, and somewhat lighter than after one round of replication.

Question 34

Which of the following enzymes is responsible for specifically transcribing tRNA in eukaryotic cells?

Answer 34

RNA polymerase III

Question 35

true or false: The core promoter includes elements such as the TATA box which are usually located downstream from the transcription initiation site. Several transcription factors are required to start initiation in eukaryotes to bind and recognize the promoter region (TBP, BRE and DPE).

Answer 35

false

Question 36

Which of the following is unique to prokaryotes?

Answer 36

Coupled transcription-translation

Question 37

DNA replication in eukaryotes differs from replication in prokaryotes in that:

Answer 37

eukaryotic chromosomes have many separate origins of replication, whereas prokaryotic chromosomes have a single origin of replication

Question 38

true or false: You analyzed two new mutations in the chicken albumin gene. The comparison of the amino acid sequences of the wildtype and mutated proteins showed that one mutation resulted from substitution of Ala in the position 69 by Pro, and in the second mutant, Asn in the position 70 has been replaced by Ser. However, when you cloned and sequenced mutated genes you found that altered codons were not adjacent, in fact they were 318 base pairs apart. An explanation to these findings is that the first mutation has changed the last codon of one exon, while the second mutation has changed the first codon of the next exon. 318 base pairs represent the intron separating two exons.

Answer 38

true

Question 39

true or false: Every division, lagging strand gets shorter by ~150 nucleotides and there is a minimum length of DNA needed for initiation of an Okazaki fragment. The enzyme telomerase is able to base pair with the last telomeric repeat of the imperfectly replicated DNA strand driving the addition of a hexanucleotide repeat. Telomerase is a DNA polymerase that carries its own RNA molecule to use as a primer at the end of the lagging strand.

Answer 39

true

Question 40

Single strand binding proteins are usually associated with only one strands of DNA:

Answer 40

lagging

Question 41

The following terms best associate with: replication error and 3’>5’ exonuclease activity

Answer 41

proofreading

Question 42

In replication, the termitation of the prokaryotes involves a ____________ while eukaryotes involves a ______________.

Answer 42

harpin, cleavage

Question 43

The simbol (-) on a DNA nucelotide refers to the ____________ direction from the origin of replication.

Answer 43

upstream

Question 44

Okazaki fragments are associated with synthesis of:

Answer 44

lagging strand

Question 45

Which mRNA processing event adds stability to the mRNA?

Answer 45

• addition of a 5’ cap

- addition of a poly(a) tail to the 3’ end

Question 46

If the DNA molecule below is transcribed from left to right, what is mRNA sequence 5’-TCGACTTGCAGGACCCTGCGGTAC-3’ 3’-AGCTGAACGTCCTGGGACGCCATG-5’

Answer 46

5’-UCGACUUGCACCACCCUGCGGUAC-3’.

Question 47

You have generated a mouse strain in which the mice cannot make functional telomerase. You predict that the mutant mice will:

Answer 47

undergo progressive chromosome shortening each generation.