Genetics MCQs Flashcards

Question 1

Q

Nucleosomes inhibit…

A) activators

B) RNA polymerase

C) translation

D) assembly of transcription factors

Answer

A

D) assembly of transcription factors

Feedback: Nucleosomes inhibit the formation of transcription factor complexes but not activators or RNA polymerase.

Question 2

Q

Unlike prokaryotes, the control of transcription by eukaryotes is designed to react to change by…

A) changing

B) ignoring change

C) remaining constant

D) changing the environment

Answer

A

C) remaining constant

Feedback: Eukaryotes seek to maintain homeostasis, remaining stable in the face of changing environments.

Question 3

Q

A form of binding motif containing a nearly identical sequence of 60 amino acids in many eukaryotes is the…

A) homeodomain motif

B) zinc finger motif

C) leucine zipper motif

D) universal motif

Answer

A

A) homeodomain motif

Feedback: The homeodomain motif, common to many eukaryotes, contains a nearly identical sequence of 60 amino acids.

Question 4

Q

Which of the following does not occur in the function of the catabolite activator protein (CAP) of E. coli?

A) Cyclic-AMP binds to the CAP protein.

B) The protein changes shape.

C) Space is increased by the binding of tryptophan.

D) Helix-turn-helix motifs are enabled to bind to the DNA.

Answer

A

C) Space is increased by the binding of tryptophan.

Feedback: The binding of cAMP to the CAP protein causes the protein to change shape, allowing helix-turn-helix motifs to bind to the DNA.

Question 5

Q

All of the following can be found in a human transcription complex except…

A) activator

B) RNA

C) enhancer

D) silencer

Answer

A

B) RNA

Feedback: The human transcription complex is highly complex, containing activators, enhancers, TATA binding proteins, silencers, basal factors, and coactivators.

Question 6

Q

Enhancers are…

A) proteins located adjacent to promoters

B) distant sites where regulatory proteins bind

C) expediters of RNA polymerase capture

D) proteins that bind with repressors, deactivating them

E) a bacterial form of promoters

Answer

A

B) distant sites where regulatory proteins bind

Feedback: In genetics, an enhancer is a short (50-1500 bp) region of DNA that can be bound with proteins (activators) to activate transcription of a gene or genes. These proteins are usually referred to as transcription factors. Enhancers are generally cis-acting, located up to 1 Mbp (1,000,000 bp) away from the gene and can be upstream or downstream from the start site, and either in the forward or backward direction.

Question 7

Q

When tryptophan is present in the medium, the transcription of tryptophan producing genes in E. coli is stopped by a helix-turn-helix regulator binding to the…

A) trp repressor

B) trp operon

C) trp promoter

D) trp operator

E) trp polymerase

Answer

A

C) trp promoter

Question 8

Q

When a homeodomain binds to DNA, the actual binding portion of the homeodomain is…

A) a leucine zipper

B) the operon

C) zinc fingers

D) the histine

E) a helix-turn-helix motif

Answer

A

E) a helix-turn-helix motif

Feedback: In proteins, the helix-turn-helix (HTH) is a major structural motif (a supersecondary structure in a chain-like molecule) capable of binding DNA. It is composed of two α helices joined by a short strand of amino acids and is found in many proteins that regulate gene expression. It should not be confused with the helix-loop-helix domain

Question 9

Q

The assembly of transcription factors on a promoter begins some 25 nucleotides upstream where it binds to a start _______________ sequence.

A) ATAT

B) AATT

C) TTAA

D) TAAT

E) TATA

Answer

A

E) TATA

Feedback: The TATA box is a DNA sequence (cis-regulatory element) found in the promoter region of genes in archaea and eukaryotes.

Considered to be the core promoter sequence, it is the binding site of either general transcription factors or histones (the binding of a transcription factor blocks the binding of a histone and vice versa) and is involved in the process of transcription by RNA polymerase.

Question 10

Q

When tryptophan is present in the environment of E. coli, the tryptophan binds to the…

A) trp operon

B) trp promoter

C) trp operator

D) trp repressor

E) trp polymerase

Answer

A

D) trp repressor

Feedback: In the bacterium E. Coli, a group of five genes code for enzymes required to synthesise the amino acid tryptophan. All five genes are transcribed together as a unit called an operon.

An operon is a group of genes that is under the control of a single operator site. A regulatory protein called a repressor can bind to the operator site and prevent transcription. When typrophan is lacking in the environment, the repressor is inactive.

RNA polymerase binds to the promoter site and then proceeds down the DNA, transcribing the genes for the tryptophan biosynthesis enzymes.

When tryptophan is present in the environment, the organism no longer needs to make tryptophan. Tryptophan binds to the repressor and activates it.

The activated repressor now binds to the operator, located withing the tryptophan promoter and blocks transcription.

The tryptophan repressor is a helix-turn-helix regulatory protein. When tryptophan is absent from the environment, the repressor is in an inactive conformation and cannot bind to the DNA to prevent transcription.

When tryptophan is abundant, two molecules of tryptophan bind to the repressor.

This alters the orientation of the helix-turn-helix motifs in the repressor and causes their recognition helices to fit into adjacent major grooves of the DNA.

Thus the synthesis of tryptophan occurs when it is needed, but is repressed when tryptophan is available.

Question 11

Q

Transcription factors appear to be unable to bind to a nucleosome because…

A) activators are inhibited by the configuration

B) of inhibition of RNA polymerase

C) of histones positioned over promoters

D) nucleosomes are especially vulnerable to repressors

E) operators are placed in an inaccessible position

Answer

A

C) of histones positioned over promoters

Question 12

Q

In the zinc fingers motif, the spacing of the helical segments is performed by…

A) beta sheets

B) helical clusters

C) zinc atoms

D) gamma helices

E) an alpha helix

Answer

A

A) beta sheets

Question 13

Q

Translation repressor proteins may shut down translation of processed mRNA transcripts by…

A) binding with the poly-A tail

B) resetting the reading frame

C) reinserting introns into the transcript

D) excising a short sequence of nucleotides

Answer

A

D) excising a short sequence of nucleotides

Question 14

Q

In many animals, the genes that regulate the development of stem cells are activated…

A) once

B) only twice

C) up to 10 times

D) over a hundred times

E) not at all

Question 15

Q

The leucine zipper motif involves the cooperation of two _______________ subunits.

A) leucine

B) protein

C) RNA

D) polymerase

E) histone

Answer

A

B) protein

Question 16

Q

Regulatory domains of most activators interact with

A) the transcription factor complex

B) RNA polymerase

C) repressors

D) the regulatory factor complex

E) the DNA binding domain

Answer

A

A) the transcription factor complex

Question 17

Q

The operon that controls tryptophan producing genes in E. coli consists of _______________ .

A) only one gene

B) two genes

C) three genes

D) four genes

E) five genes

Answer

A

E) five genes

Question 18

Q

In order for a gene to be transcribed, RNA polymerase must have access to the DNA helix and be able to bind to the genes

A) activator

B) regulator

C) promoter

D) operator

E) repressor

Answer

A

C) promoter

Question 19

Q

In the function of the lac operon in E. coli, the lac genes are transcribed in the presence of lactose because

A) RNA polymerase binds to the operator

B) the repressor cannot bind to the promoter

C) an isomer of lactose binds to the repressor

D) CAP does not bind to the operator

E) of the absence of cAMP

Answer

A

C) an isomer of lactose binds to the repressor

Question 20

Q

The role of methylation of DNA is now viewed as…

A) interfering with DNA transcription by blocking base pairing between cytosine and guanine

B) complexing with enhancers to prevent transcription

C) prevention of mutation

D) insuring that genes that are turned off, stay off

E) irrelevant to gene transcription

Answer

A

D) insuring that genes that are turned off, stay off.

Question 21

Q

In order for the helix-turn-helix motif to bind to DNA, the _______________ must fit into the major groove of the DNA.

A) homeotic switches

B) zinc fingers

C) operator

D) recognition helix

E) protein link

Answer

A

D) recognition helix

Question 22

Q

The most common form of gene expression regulation in both bacteria and eukaryotes is…

A) translational control

B) transcriptional control

C) post-transcriptional control

D) post-translational control

E) control of passage from the nucleus

Answer

A

B) transcriptional control.

Question 23

Q

In eukaryotes, many genes may have to interact with each other, requiring more interacting elements than can fit around a single promoter. This physical limitation is overcome by…

A) alternating promoters and operators

B) placing promoters on both sides of each gene

C) the use of very long promoters

D) distant sites in a chromosome controlling transcription of a gene

E) having factors on one chromosome control genes on another gene

Answer

A

D) distant sites in a chromosome controlling transcription of a gene

Question 24

Q

E. coli is able to use foods other than glucose in the absence of available glucose, because falling levels of glucose cause an increase of…

A) cAMP

B) CAP

C) lactase

D) glu operons

E) tRNA

Question 25

Q

In the absence of glucose, E. coli can import lactose to change into glucose and galactose because CAP binds to the…

A) cAMP

B) DNA

C) lac operon

D) operator

E) repressor

Question 26

Q

Which is not part of the lac operon?

A) repressor

B) activator protein

C) operator

D) promotor

E) structural gene

Answer

A

B) activator protein

Question 27

Q

In an operon the location of the regulatory region occurs ________ the structural genes.

A) after

B) within

C) before

Answer

A

C) before

Question 28

Q

In eukaryotic gene regulation, the location of the promoter is always before and the location of the enhancer always after the gene being regulated.

A) True

B) False

Question 29

Q

Proteins that block the passage of RNA polymerase are called:

A) operons

B) activators

C) repressors

D) enhancers

E) promoters

Answer

A

C) repressors

Question 30

Q

Which of the following is part of an operon?

A) structural genes

B) an operator

C) a promoter

D) a CAP binding site

E) all of the above

Answer

A

E) all of the above

Question 31

Q

Which of the following are not matched correctly?

A) exon splicing-occurs in nucleus

B) post-translational modifications-phosphorylation

C) snRNA-splice out exons from transcript

D) activated enhancers-trigger transcription

E) all are matched correctly

Answer

A

C) snRNA-splice out exons from transcript

Question 32

Q

A single gene may use a regulatory site to control the expression of that gene, but genes rarely have multiple regulatory sites.

A) True

B) False

Question 33

Q

If the genetic code consisted of four bases per codon rather than three, the maximum number of unique amino acids that could be encoded would be…

A) 16
B) 64
C) 128
D) 256
E) 512

Question 34

Q

An RNA-dependent RNA polymerase is likely to be present in the virion of a…

A) DNA virus that multiplies in the cytoplasm

B) DNA virus that multiplies in the nucleus

C) minus-strand RNA virus

D) plus-strand RNA virus

E) transforming virus

Answer

A

C) minus-strand RNA virus

Question 35

Q

In E. coli, the inability of the lac repressor to bind an inducer would result in…

A) no substantial synthesis of ß-galactosidase

B) constitutive synthesis of ß-galactosidase

(C) inducible synthesis of ß-galactosidase

(D) synthesis of inactive ß-galactosidase

(E) synthesis of ß-galactosidase only in the absence of lactose

Answer

A

A) no substantial synthesis of ß-galactosidase

Question 36

Q

Which of the following statements about retrotransposons is correct?

A) They transpose via an RNA intermediate.

B) They contain genes for ribosomal proteins.

C) They possess a gene for RNA-dependent RNA polymerase.

D) They possess genes that encode proteins that integrate RNA into chromosomes.

E) They are found only in bacteria.

Answer

A

A) They transpose via an RNA intermediate.

Question 37

Q

A mutation deleting an upstream activating sequence for a single gene would be expected to be…

A) polar

B) trans-dominant

C) cis-dominant

D) silent

E) revertible

Answer

A

C) cis-dominant

Feedback: cis dominant- mutations (eg of an operator) that alter the functioning of genes on that same piece of DNA.

Question 38

Q

Proline disrupts α-helical structure in proteins because it is…

A) an acidic amino acid

B) an aromatic amino acid

C) an imino acid

D) a basic amino acid

E) a sulfur-containing amino acid

Answer

A

C) an imino acid

Feedback: Proline is unique in that it is the only amino acid where the side chain is connected to the protein backbone twice, forming a five-membered nitrogen-containing ring. Strictly speaking, this makes Proline an imino acid. Proline is unable to occupy many of the main chain conformations easily adopted by all other amino acids.

Proline can introduce kinks into alpha helices, since it is unable to adopt a normal helical conformation.

Question 39

Q

All of the following are proteins within the core nucleosome particle EXCEPT

A) H1

B) H2A

C) H2B

D) H3

E) H4

Answer

A

A) H1

Feedback: The nucleosome core particle consists of approximately 147 base pairs of DNA wrapped in 1.67 left-handed superhelical turns around a histone octamer consisting of 2 copies each of the core histones H2A, H2B, H3, and H4

Linker histones such as H1 and its isoforms are involved in chromatin compaction and sit at the base of the nucleosome near the DNA entry and exit binding to the linker region of the DNA.

Question 40

Q

In the context of prokaryotic gene expression, which of the following is the most appropriate definition of an operator?

A) A cluster of genes that are regulated by a single promoter.

B) A DNA-binding protein that regulates gene expression.

C) A non-coding, regulatory DNA sequence that is bound by RNA polymerase.

D) A non-coding, regulatory DNA sequence that is bound by a repressor protein.

Answer

A

D) A non-coding, regulatory DNA sequence that is bound by a repressor protein.

Feedback:
In prokaryotes it is common to find a group of genes encoding enzymes required for a single metabolic pathway clustered together in the genome, so that they can be transcribed as a single polycistronic mRNA. This allows coordinated regulation of their expression by a single promoter, The term used for such a cluster of genes and their promoter is an operon. Using the E. coli lac operon as an example, the three co-regulated genes encoding β-galactosidase, lactose permease, and transacetylase are called the lacZ, lacY and lacA genes, respectively. There is also a separate I gene (I for inducibility) that codes for a protein called the lac repressor, and there is a stretch of DNA called the operator region to which the lac repressor protein can bind. The promoter where RNA polymerase binds to begin transcription is flanked by two regulatory DNA sequences, each of which is recognized and bound by a different regulatory protein. The lac repressor binds to the operator sequence, which lies just downstream of (and partially overlapping) the promoter.

Question 41

Q

In terms of lac operon regulation, what happens when E. coli is grown in medium containing both glucose and lactose?

A) Both CAP and the lac repressor are bound to the DNA.

B) CAP is bound to the DNA but the lac repressor is not.

C) Lac repressor is bound to the DNA but CAP is not.

D) Neither CAP nor the lac repressor are bound to the DNA.

Answer

A

D) Neither CAP nor the lac repressor are bound to the DNA.

Feedback:
In the presence of high glucose there is no cyclic AMP (cAMP) to cause catabolite gene-activator protein (CAP) to bind to the DNA, and this binding is necessary for the attachment of RNA polymerase to the promoter. With no lactose the lac repressor is bound to the operator. With low glucose but no lactose, although CAP binds and assists the RNA polymerase to bind, transcription still does not occur because the lac repressor is bound to the operator, blocking polymerase movement. In the presence of lactose, a small amount is converted to allolactose, which acts as the inducer. Allolactose binds to the repressor, causing its release from the operator and transcription can proceed.

Question 42

Q

Which of the following can be described as ‘a sequence that can be several thousand base pairs upstream or downstream of a eukaryotic promoter and which increases gene expression as much as 200-fold.’?

A) CAAT box

B) Enhancer

C) Insulator

D) TATA box

Answer

A

B) Enhancer

Feedback:
Eukaryotic genes often have control elements, called enhancers, which increase gene expression and which can be thousands of base pairs away from the promoter. These can be tissue-specific, enhancing transcription in only certain tissues. The enhancer is brought into proximity with the promoter by looping of the DNA that lies between them.

Question 43

Q

Nuclear receptors belong to which class of transcription factor?

A) Helix-loop-helix proteins

B) Helix-turn-helix proteins

C) Leucine zipper proteins

D) Zinc finger proteins

Answer

A

D) Zinc finger proteins

Feedback: 
An important class of regulatory protein that contains zinc fingers with four cysteines are the steroid receptor family. Steroid receptors are members of the nuclear receptor family of zinc finger transcription factors, which also includes receptors for thyroid hormone, vitamin D, and retinoic acid. (Retinoic acid is important in embryonic development). Nuclear receptors contain two zinc fingers, one of which is actually involved in protein-protein rather than protein-DNA interactions. The receptors homodimerize or heterodimerize through protein interactions of one of their two zinc fingers, and then bind to their DNA recognition sequences by the other.

A zinc finger is a small protein structural motif that is characterized by the coordination of one or more zinc ions in order to stabilize the fold.

Question 44

Q

What best describes the mechanism by which the coactivator CREB-binding protein (CBP) activates transcription?

A) CBP has DNA methyltransferase activity.

B) CBP has histone acetyl transferase activity.

C) CBP interacts with the basal transcription complex.

D) CBP interacts with the basal transcription complex and also has histone acetyl transferase activity.

Answer

A

C) CBP interacts with the basal transcription complex.

Feedback:
A well-studied example of a coactivator that works with a number of different transcription factors (TFs) is CBP. CBP stands for CREB-binding protein, CREB being the first TF found to have CBP as a coactivator. CREB in turn stands for cAMP-response-element-binding protein. CREB is TF that activates specific genes in response to cAMP signalling. When CBP is recruited to a promoter by binding to CREB or another TF, it then activates transcription by more than one mechanism. It can bind to components of the basal initiation complex. However, it also has histone acetyl transferase activity (HAT), which catalyses acetylation of lysine residues of histone proteins. Histone acetylation reduces the positive charge on histone proteins and is generally associated with ‘loosely wound’ chromatin and therefore with transcriptional activation.

Question 45

Q

Which of the following statements, concerning regulation of trp operon expression by attenuation, is correct?

A) The leader peptide sequence encodes enzymes required for tryptophan synthesis.

B) The leader peptide sequence contains no tryptophan residues.

C) Rapid translation of the leader peptide allows completion of the mRNA transcript.

D) Rapid translation of the leader peptide prevents completion of the mRNA transcript.

Answer

A

D) Rapid translation of the leader peptide prevents completion of the mRNA transcript.

Feedback:
Translation of the trp operon encoded mRNA begins with the synthesis, not of one of the enzymes, but of a short ‘leader’ peptide encoded by the 5’ end of the message. The leader peptide contains codons encoding tryptophan. If there is tryptophan in the cell, then tRNA charged with tryptophan will be available and ribosomes will move rapidly along the sequence encoding the leader peptide. The effect of this is to leave part of the mRNA sequence free to base pair with itself, forming a stem-loop transcription termination signal so the mRNA is never completed and the enzymes needed for tryptophan synthesis are not made. The leader peptide has no other function but to play this regulatory role by its translation, and is rapidly degraded.

Question 46

Q

Which of the following statements regarding regulation of transferrin-receptor protein synthesis is correct?

A) The iron-responsive element is an iron-binding sequence in the mRNA that encodes transferrin-receptor protein.

B) The iron-responsive element is in the 5’ untranslated region of the mRNA that encodes transferrin-receptor protein.

C) When iron is abundant the IRE-binding protein binds to and stabilizes the mRNA that encodes transferrin-receptor protein.

D) When iron is scarce the IRE-binding protein binds to and stabilizes the mRNA that encodes transferrin-receptor protein.

Answer

A

D) When iron is scarce the IRE-binding protein binds to and stabilizes the mRNA that encodes transferrin-receptor protein.

Feedback:
The synthesis of the transferrin-receptor protein is a case where mRNA stability regulates synthesis of the protein and is itself regulated. The receptor is responsible for the transport of iron into cells by endocytosis and more receptor protein is therefore required when iron is scarce, to increase the efficiency of iron import. In the 3’ untranslated region of the mRNA is a group of five stem loops called an iron-responsive element (IRE). In the absence of iron, an IRE-binding protein (IRP) attaches to the IRE and stabilizes the mRNA, thus increasing receptor synthesis with a consequent increase in the import of iron into the cell. In iron abundance, iron complexes with the IRP, which then no longer binds to the IRE. The IRE is thus exposed to attack by a specific endonuclease which cleaves the message, leading to its degradation. The result is a reduced import of iron into the cell.

Question 47

Q

‘RNAi’ stands for which of the following?

A) RNA inducer.

B) RNA insertion.

C) RNA interference.

D) RNA intron.

Answer

A

C) RNA interference.

Feedback:
RNAi stands for RNA interference and refers to a natural mechanism for destroying selected RNA molecules, silencing gene expression and also protecting the cell from foreign RNA molecules, such as viral RNA.

Question 48

Q

Suppose a certain gene contains the double-stranded sequence:
5’- ATGTTTAGCGCC -3’
3’- TACAAATCGCGG -5’
If the top strand is the sense strand and codes for an mRNA whose sequence begins ‘ATG’, which of the following would be the sequence of the corresponding segment of antisense RNA?

A) 5’-AUGUUUAGCGCC-3’

B) 5’-CCGCGAUUUGUA-3’

C) 5’-GGCGCUAAACAU-3’

D) 5’-UACAAAUCGCGG-3’

Answer

A

C) 5’-GGCGCUAAACAU-3’

Feedback:
An antisense RNA has the complementary sequence to mRNA. In the example, the coding sequence in the DNA is the top strand starting ATG (which is the start codon) so the complementary sequence would be the RNA equivalent of the bottom DNA strand. U takes the place of T in RNA and all single-stranded molecules are written 5’ - 3’ so the sequence of the complementary RNA would be 5’-GGCGCUAAACAU-3’.

Question 49

Q

The effectors of gene silencing are short double-stranded RNA molecules produced by the action of the enzyme dicer. Approximately what size are these molecules?

A) 11 bp

B) 22 bp

C) 75 bp

D) 100 bp

Answer

A

B) 22 bp

Feedback:
Large double-stranded RNA molecules or microRNA hairpin molecules are processed into pieces of about 22 base pairs long by the enzyme dicer. These 22-base pair molecules initiate the process of RNA interference.

Question 50

Q

Which of the following is true of the lac operon in E. coli?

A) The operon is only switched on in the absence of lactose in the growth medium.

B) The lac operon messenger RNA is a polycistronic mRNA (it carries information for synthesis of several proteins)

C) The enzyme β-galactosidase is only produced in large quantities when the lac repressor is bound to the operator.

D) The promoter is the binding site for the lac repressor.

Answer

A

B) The lac operon messenger RNA is a polycistronic mRNA (it carries information for synthesis of several proteins)

Feedback:
A feature of operons in bacteria is that a single mRNA is made that spans several genes coding for proteins with related functions. The expression of these genes is coordinated by control of transcription of the polycistronic message.
The lac operon is only switched on when there is lactose in the growth medium. The enzymes encoded by the lac operon are needed to transport lactose into the cell and to break it down into glucose and galactose. These enzymes are only needed when lactose is the sole carbon source in the growth medium.
Binding of the lac repressor to the operator region blocks transcription. So β-galactosidase is only produced in large quantities when the lac repressor is removed from the operator in the presence of lactose.

Question 51

Q

Which of the following statements about mRNA stability is correct? Please select all that apply.

A) Prokaryotic mRNAs have a half-life of only a few minutes.

B) Regulation of mRNA stability is a way of regulating gene expression.

C) It is thought that polyA tails stabilise eukaryotic mRNAs.

D) Histone mRNAs have especially long polyA tails and are especially stable.

Answer

A

A) Prokaryotic mRNAs have a half-life of only a few minutes.

B) Regulation of mRNA stability is a way of regulating gene expression.

C) It is thought that polyA tails stabilise eukaryotic mRNAs.

Feedback:
Most prokaryotic mRNAs have half-lives of 2-3 minutes. Eukaryotic mRNAs have half-lives of a few minutes to many hours or even days. PolyA tails are thought to stabilise eukaryotic mRNAs and polyA binding protein protects that mRNA from exonuclease activity. Histone mRNAs do not have polyA tails and their stability is determined by a stem-loop structure at the 3’ end of the mRNA.

Question 52

Q

An epigenetic change in gene expression is an inherited change that does not involve any change in the nucleotide sequence of the gene. True or false?

A) True

B) False

Answer

A

A) True

Feedback:
An epigenetic change in gene expression is a change can persist through many cell divisions or for many generations. There are several mechanisms for epigenetic changes including modifications of histones, methylation of DNA and RNA interference. All of these can cause long-term, heritable, changes in gene activity.

Question 53

Q

Which of the following statements is true of RNA interference?

A) RNA interference is a normal way for organisms to regulate gene expression.

B) RNA interference is a mechanism for combating virus infection in plants.

C) RNA interference occurs only in vertebrates.

D) RNA interference is already used therapeutically for many disorders.

Answer

A

A) RNA interference is a normal way for organisms to regulate gene expression.

B) RNA interference is a mechanism for combating virus infection in plants.

Feedback:
RNA interference is a normal method of gene regulation widely used in plants, animals (both vertebrates and invertebrates), and viruses. It is also thought to be a method for protecting plants and other organisms from virus attack. Though it has great therapeutic potential, there are barriers to be overcome for its development, including developing delivery systems to deliver the microRNAs into cells and tissues.

Question 54

Q

Which of the following is true of RNA synthesis (transcription)?

A) RNA synthesis is always in the 5’ - 3’ direction.

B) RNA polymerase needs a primer to initiate transcription.

C) In transcription, U is inserted opposite T.

D) New nucleotides are added on to the 2’ OH of the ribose sugar.

Answer

A

A) RNA synthesis is always in the 5’ - 3’ direction.

Feedback:
In transcription, as in DNA replication, nucleotides are added on to the 3’ OH of the growing chain so RNA synthesis is always in the 5’ - 3’ direction. Unlike DNA synthesis, a primer is not needed for initiating a new strand. RNA contains U instead of T, and U base pairs with A and is inserted opposite A (not T) in the template.

Question 55

Q

In bacterial promoters, which of the following describes the ‘Pribnow box’?

A) The 5’ untranslated region

B) The -10 box

C) The -35 box

D) The termination sequence

Answer

A

B) The -10 box

Feedback:
Initiation of transcription takes place when RNA polymerase locks on to the gene at the promoter. The polymerase binds to short stretches of base sequences that are similar in many promoters. These are called ‘elements’ or ‘boxes’. The Pribnow box is named after David Pribnow, who discovered it, and it is found around 10 base pairs upstream of the transcriptional start site, i.e. at -10. It is often referred to as the -10 box.

Question 56

Q

The role of the sigma factor in bacterial RNA polymerase is:

A) to catalyse RNA synthesis:

B) to position RNA polymerase correctly on the template DNA.

C) to terminate RNA synthesis.

D) to unwind the DNA template.

Answer

A

B) to position RNA polymerase correctly on the template DNA.

Feedback:
RNA polymerase of E. coli is a large complex of several protein subunits. The ‘core’ enzyme (two copies of an α subunit, two related subunits termed β and β’, and an ω subunit) has an affinity for DNA, can unwind the DNA template, and can even catalyse RNA synthesis from the template strand, but it cannot recognize the correct initiation site until it is joined by another protein from the cytosol, the sigma subunit (σ) or sigma factor. With this attached, the polymerase binds to the -35 and Pribnow boxes and initiation of transcription can start.

Question 57

Q

Which of the following statements regarding termination of transcription in prokaryotes is correct?

A) In Rho dependent termination the Rho factor moves along the DNA template ahead of the RNA polymerase.

B) Rho factor has topoisomerase activity for relieving supercoiling.

C) Termination often involves a stem-loop structure forming in the RNA transcript.

D) Termination often involves a stem-loop structure forming in the DNA template.

Answer

A

C) Termination often involves a stem-loop structure forming in the RNA transcript.

Feedback:
 At the end of many transcribed prokaryote genes are sequences that result in the transcribed RNA having a stem loop structure which somehow disrupts the elongation process. Immediately following the stem loop structure in the mRNA transcript is a string of around 8 U residues, giving weak bonding of the RNA to DNA because of weaker double A-U hydrogen bonding. This facilitates complete detachment of the mRNA and hence terminates transcription. An alternative method of termination of transcription in many prokaryote genes requires an additional protein called the Rho factor, which attaches to the newly transcribed mRNA and moves along it behind the RNA polymerase. At the termination site, the polymerase pauses, possibly because of a difficult to separate G-C-rich sections of the DNA, and this pause allows the Rho factor to catch up with the polymerase. The Rho factor has an unwinding (helicase) activity for unwinding the RNA-DNA duplex formed by transcription. ATP breakdown is involved. Unwinding releases mRNA and terminates transcription

Question 58

Q

What is the role of eukaryotic RNA polymerase I?

A) Transcription of mRNA only.

B) Transcription of mRNA, rRNA and tRNA.

C) Transcription of ‘small’ RNAs including TRNAs, 5S RNAs and snRNAs.

D) Transcription of the major rRNA transcript.

Answer

A

D) Transcription of the major rRNA transcript.

Feedback:
 The basic enzymic reaction by which RNA is synthesized in eukaryotes is the same as in prokaryotes. However, in eukaryotes there are three different RNA polymerases, designated I, II, and III, which are responsible for transcribing different classes of gene.  RNA polymerase I (Pol I) transcribes the major rRNA transcript. RNA polymerase II (Pol II) transcribes mRNA.  RNA polymerase III (Pol III) transcribes 'small' RNAs: tRNAs, 5S rRNA and small nuclear RNAs (snRNAs), a class of nonprotein-coding RNA molecules with varied functions such as RNA splicing.

Question 59

Q

Which of the following does the abbreviation TBP stand for?

A) TATA-box binding protein

B) Transcription associated factor

C) Transcription factor binding protein

D) TATA box polymerase

Answer

A

A) TATA-box binding protein

Feedback:
TBP is a protein that attaches the transcription factor TFIID to the TATA box, in eukaryotes, and its full name is ‘TATA-box binding protein’.

Question 60

Q

Which of the following best describes the ‘cap’ modification of eukaryotic mRNA?

A) A modified guanine nucleotide added to the 3’ end of the transcript.

B) A modified guanine nucleotide added to the 5’ end of the transcript.

C) A string of adenine nucleotides added to the 3’ end of the transcript.

D) A string of adenine nucleotides added to the 5’ end of the transcript.

Answer

A

B) A modified guanine nucleotide added to the 5’ end of the transcript.

Feedback:
 The RNA of the eukaryotic gene primary transcript immediately undergoes a modification at its 5’ end, called capping. At the 5’ end of the primary RNA transcript there is a triphosphate group because the first nucleotide triphosphate simply accepts a nucleotide on its 3’-OH. The terminal phosphate of this is removed and a GMP residue is added from GTP (Fig. 24.11). The 5’-5’ triphosphate linkage is unusual. The G is then methylated in the N-7 position as is also the 2’-OH of the second nucleotide. The cap protects the end of the mRNA from exonuclease attack and it is involved in initiation of translation as described in the next chapter. Termination of transcription in eukaryotic cells is less understood than the prokaryotic mechanism. Most, though not all, eukaryotic mRNAs end in a string of up to 250 adenine residues known as a 3’ polyA tail. The polyA tail is not directly encoded by the gene, but its position is directed by a polyadenylation signal (AAUAAA) that is encoded by the gene and transcribed in the primary transcript.

Question 61

Q

Which of the following statements regarding splicing of eukaryotic mRNA transcripts is correct?

A) Exons are spliced out and introns are retained in the mature mRNA transcript.

B) Several reactions in the splicing process involve hydrolysis of ATP.

C) Small nuclear RNAs are retained in the mature mRNA transcript.

D) Splicing takes place in the cytosol.

Answer

A

B) Several reactions in the splicing process involve hydrolysis of ATP.

Feedback:
 Splicing involves removal of the unwanted RNA introns of a primary transcript and joining up the exons into mRNA. The key to it is the transesterification reaction. In this, a phosphodiester bond is transferred to a different -OH group. There is no hydrolysis and no significant energy change during the bond rearrangements. In most cases, the splicing reaction in the nucleus is catalysed by very complex protein-RNA structures called spliceosomes. They are complexes of about 300 different proteins and also five RNA molecules, 100-300 bases long in higher eukaryotes, called small nuclear RNAs (snRNAs). These are associated with proteins in structures known as small ribonucleoprotein particles (snRNPs, pronounced ‘snurps’), each containing multiple protein subunits. The spliceosome undergoes various rearrangements during different phases of the splicing reaction. These involve changes in the base pairing of snRNAs with the transcript and with each other. Several of the interactions and rearrangements require ATP hydrolysis, so although the transesterification itself does not involve energy changes, splicing is an energy requiring process.

Question 62

Q

Which of the following statements about mRNA splicing is true?

A) The existence of split genes has no advantage

B) Self-splicing introns do not require the help of any protein for splicing to occur accurately.

C) β-thalassaemia results from a genetic defect in the spliceosome.

D) Splicing occurs in the cytosol.

Answer

A

B) Self-splicing introns do not require the help of any protein for splicing to occur accurately.

Feedback:
 Self-splicing introns do not require a spliceosome or any other proteins for the splicing reaction. Instead, the RNA transcript catalyses its own splicing reaction and is, therefore, an example of a ‘ribozyme’ (an RNA enzyme). Split genes have the advantage that they allow alternative splicing so that a single gene can code for variant forms of the same protein. These variant forms may have slightly different properties suited to different tissues. β-thalassaemia results from faulty splicing of the β-globin transcript. However, the spliceosome is normal and the reason for the fault is a mutation in the β-globin gene itself. Splicing occurs in the nucleus and mature transcripts are then transported into the cytosol for translation (protein synthesis) to occur.

Question 63

Q

Is the following statement true or false? ‘Eukaryotic mRNA is transcribed by RNA polymerase II.’

A) True

B) False

Answer

A

A) True

Feedback:
 In eukaryotes, RNA polymerase II transcribes all messenger RNAs (mRNAs). RNA polymerase I and III transcribe non-coding RNAs including ribosomal RNA (rRNA) and transfer RNA (tRNA).

Question 64

Q

The polyA tail on eukaryotic mRNA is encoded by a long string of Ts at the end of the gene. Is this statement true or false?

A) True

B) False

Answer

A

B) False

Feedback:
 There is no coding sequence for polyA. The polyA tail is added on to the mRNA after transcription. After transcription, the transcript is cleaved near to the polyadenylation signal, AAUAAA, and as many as 200 adenine nucleotides are added, using ATP as the source of these adenines.

Answer 59

A

C) Promoter.

D) DNA-dependent RNA polymerase.
Feedback:
 DNA-dependent RNA polymerases are the enzymes responsible for transcription in both eukaryotes and bacteria, and both have promoters for initiation of transcription. However, only eukaryotic messages are capped at the 5’ end with methylated GMP, and only eukaryotic messages have a polyA tail (a string of up to 200 adenines that are added to the 3’ end).

Answer 60

A

A) Protecting bacteria by cleaving the DNA of infecting viruses.

Feedback:
Bacteria have restriction enzymes as protection against bacterial DNA viruses (bacteriophages). When these viruses infect a cell, they insert their own DNA, which takes over the targeted cell. The cell’s restriction enzymes cut the invading DNA, which destroys the phage. There will be large numbers of short base sequences in the infected cell’s own genome identical to the sequence targeted by the restriction enzyme. The bacteria protect themselves by methylating A or C bases in the target sites in their own DNA; the enzymes no longer recognize the methylated sites.

Answer 61

A

C) Separation of DNA fragments by electrophoresis followed by transfer to a membrane and then hybridization with a labelled probe sequence.

Feedback:
In Southern blotting DNA fragments are first separated by electrophoresis. The DNA in the gel is then made single stranded by exposure to dilute alkali, and transferred by capillary action to the nylon membrane, which is laid over it. The relative positions of the different DNA fragments are thus preserved on the membrane. The membrane is then soaked in buffer containing the labelled probe, which base pairs with the DNA fragment of interest. The position of that fragment is then determined by autoradiographic (exposure of the membrane to X-ray film), or fluorescent detection of the hybridized probe.

Answer 62

A

D) ddNTPs prevent further DNA synthesis once they are incorporated into the DNA sequence.

Feedback:
Dideoxy derivatives of nucleoside triphosphates (ddNTPs) are used in DNA sequencing, because if one of these molecules is added to a growing DNA chain, synthesis stops. This is because DNA polymerase adds a nucleotide to the 3’-OH of a growing DNA chain and ddNTPs lack the 3’-OH group. Hence although they can be added to a chain via their 5’-phosphate, the chain is then terminated. In a sequencing reaction the fraction of newly synthesized DNA chains that are terminated will depend on the relative proportions of dNTP and ddNTP. Four reactions are set up, one with a mixture of dATP and ddATP, another with a mixture of dCTP and ddCTP and so on for each of the four bases, In each reaction a set of DNA fragments of different lengths are synthesized, each one terminating where there is an incident of the relevant base in the sequence. The lengths of these DNA fragments are then determined by electrophoresis.

Answer 63

A

D) Approximately 1 billion

Feedback:
30 cycles will produce approximately 230 PCR products, which is 1.07 billion.

Answer 64

A

B) Dideoxy-dNTPs (ddNTPs)

Feedback:
Dideoxy-dNTPs, which lack a 3’ OH on the deoxyribose, cause chain termination if they are present in a DNA synthesis reaction so they are not used in PCR. They are used in DNA sequencing. A PCR reaction requires a template, dNTPs, a thermostable (heat-tolerant) DNA polymerase and primers.

Answer 65

A

B) It allows selection of E. coli host cells that contain plasmid in which the insert has been ligated.

Feedback:
A typical plasmid vector has a gene for ampicillin resistance and a gene for the N-terminal 146 amino acids of the enzyme β-galactosidase. Restriction enzyme sites, which can be used as cloning sites, are situated within the gene for the N-terminal fragment of β-galactosidase. Following transformation, only cells with a plasmid will form colonies on the ampicillin containing agar plate, but most of the incorporated plasmids will not have received the DNA insert. This is where the gene for the N-terminal fragment of β-galactosidase comes in. If the part enzyme it encodes joins to the missing C-terminal portion of the enzyme the two reconstitute the active enzyme. An engineered E. coli strain is used for the cloning into whose main chromosome the gene for the missing C-terminal section of β-galactosidase has been inserted. Any plasmid that had the DNA fragment inserted into it cannot synthesize the N-terminal β-galactosidase fragment because the insertion disrupts the coding region. A plasmid with a successful insert will therefore not give rise to active β-galactosidase, while one without an insert will.

Answer 66

A

D) analysis of mRNA expression.

Feedback:
RT-PCR is a common technique for detecting the presence of a particular mRNA transcript in a cell or tissue type. mRNA cannot be amplified directly by PCR, so here the RT stands for ‘reverse transcription’, as the mRNA is first copied to make just a single cDNA strand, which is then subjected to PCR using primers specific for the transcript of interest. If the mRNA was present in the cell an amplified double-stranded DNA PCR ‘product’ of the expected size will be made, and if it was absent no PCR product will be seen. RT-PCR methodology has been successfully adapted to make it quantitative (qPCR or Real-Time PCR) so that the level of gene expression, not just the presence or absence of a transcript, is measured.

Answer 67

A

C) PCR is used for DNA profiling (DNA fingerprinting).

Feedback:
DNA fingerprinting, or DNA profiling, is used for forensic analysis of crime scenes and for paternity testing. The method that is now widely used is the analysis of short tandem repeats in non-coding DNA. The number of these repeats varies from one individual to another and PCR is used to analyse the numbers of repeats. The chance of two individuals having an identical profile is reckoned to be about 1 in 1 billion. Nevertheless it can happen, especially in related individuals, and identical twins have the same profiles. SNPs (single nucleotide polymorphisms in coding or non-coding DNA) are not used for forensic analysis.

Answer 68

A

B) To allow negative selection of cells in which integration of the targeting sequence has occurred by random insertion into the genome.

Feedback:
In creating a knockout mouse cultured ES cells are transfected with a targeting vector in which the target gene is replaced by a neomycin resistance gene flanked by sequences homologous to the normal gene. Homologous recombination replaces the target gene with the NRG construct in a small number of cases. Stably transfected cells are now resistant to the neomycin due to expression of the NRG so that these cells can be selected by growing them in the presence of the antibiotic. Cells without the NRG fail to grow. After that, it is necessary to select those cells in which integration has occurred by homologous recombination rather than by random insertion. In constructing the targeting vector, a gene for thymidine kinase is inserted at one end of the construct, well outside the region of homology to the target gene. If the DNA is incorporated randomly, the thymidine kinase gene goes in with it and is expressed; if the incorporation is by homologous recombination the kinase gene is not incorporated (because it is outside the homology region). If the kinase gene is expressed, it makes the cell sensitive to gancyclovir, which can be used as a negative selection method to kill cells in which random integration has occurred.

Answer 69

A

A) A gene ‘gun’

Feedback:
Electroporation and transformation of competent cells are methods commonly used for introducing DNA into bacteria. Microinjection is used for animal cells and oocytes, which are bigger and do not have cell walls so microinjection using a fine glass needle is possible. One way of introducing DNA into plant cells is to bombard them with DNA-coated particles fired at high speed from a ‘gene gun’.
Page reference: 471

Answer 70

A

A) True

Feedback:
SCID is a genetic disorder in which the enzyme adenosine deaminase is deficient in lymphocytes, causing a toxic build-up of ATP and loss of function of the lymphocytes. Gene therapy aims to replace the faulty gene in bone marrow cells which give rise to the lymphocytes. In clinical trials there have been problems because the therapy has triggered oncogenes and caused leukaemia in a number of patients.

Answer 71

A

B) Restriction enzymes are endonucleases.

D) Restriction enzymes are part of the defence system of bacteria against attack by bacteriophages.

Feedback:
Restriction enzymes are endonucleases (they cut within a DNA molecule, not at the end) and they are part of the natural defences of bacteria against foreign DNA, such as bacterial viruses (bacteriophages). There are many restriction enzymes, in different bacteria, with many different cut sites. Some restriction enzymes produce sticky ends with a short single-stranded overhang; others cut straight across the DNA to produce blunt ends.

Answer 72

A

A) Hepatitis B vaccine.

B) Human insulin for treatment of diabetes.

C) Human growth hormone.

D) Tissue plasminogen activator, used for blood clot removal.

Feedback:
All of these are produced by gene cloning. Hepatitis B vaccine is simply a Hepatitis B virus coat protein produced in yeast from the cloned coat protein gene. The other proteins are human proteins produced from the cloned gene or, in the case of insulin, a synthetic copy of the human gene.

Answer 73

A

D) topoisomerases

Feedback:
A group of enzymes, known as topoisomerases catalyse the process of removing supercoils ahead of the replication fork. They act on the DNA and isomerize or change its topology. The roles of the other enzymes in DNA replication are: helicase - separating the two strands of the DNA double helix, DNA polymerases - addition of new nucleotides to the replicating strand, primase - polymerization of RNA primers.

Answer 74

A

D) Formation of a phosphodiester bond between the 3’-OH of one Okazaki fragment and the 5’-phosphate of the next on the lagging strand.

Feedback:
In the case of the leading strand, primase lays down a single RNA primer at the origin of initiation and the DNA polymerase proceeds from there until replication is complete, but the same will not suffice for the lagging strand. The solution is that, as the DNA unwinds, there is repeated initiation of the lagging strand by primase, each primer being extended by the polymerase into a short stretch of DNA synthesis, 1,000-2,000 bases long in E. coli and 100-200 in eukaryotes. The net result is illustrated in Fig. 23.11. The short stretches of DNA attached to RNA primers on the lagging stand are called Okazaki fragments. DNA ligase catalyses formation of a phosphodiester bond between the 3’-OH of one Okazaki fragment and the 5’-phosphate of the next, a process requiring energy. In some prokaryotes and all eukaryotes, ATP supplies this.

Answer 75

A

B) Human chromosomes

Feedback:
Telomeres are found at the ends of the linear double-stranded DNA molecules in human chromosomes. They protect chromosome ends from nucleases and they also provide a special mechanism for replication of chromosome ends, using the enzyme telomerase. Circular molecules such as bacterial chromosomes and most mitochondrial genomes do not need these specialised DNA ends.

Answer 76

A

A) 3’ - 5’ exonuclease activity

Feedback:
The proofreading activity possessed by many DNA polymerases is an exonuclease activity that degrades mismatched bases that have been wrongly incorporated into the growing chain. This is, therefore, an exonuclease activity (exonucleases digest from the end of a DNA chain) and it operates ‘backwards’ from the 3’ growing end, i.e. 3’ - 5’.

Answer 77

A

D) Guanine in the new strand of the helix is methylated at GATC.

Feedback:
Mismatch repair is a system that repairs mismatches that have slipped and evaded proofreading during DNA replication. DNA becomes methylated at the G of GATC sequences after replication; however, this does not occur immediately so the new strand, containing the error, can be distinguished from the methylated parental strand. In bacteria, the protein MutS recognises the distortion of the helix that results from the mismatch. Then MutH selects the nearest GATC on the unmethylated strand and nicks the DNA. This strand is then degraded past the mismatch and replaced with new DNA, thus replacing the faulty strand with fresh DNA.

Answer 78

A

C) Nucleotide excision repair

Feedback:
Nucleotide excision repair is an almost universal repair mechanism in which a section of damaged DNA is removed and replaced with new DNA by a DNA polymerase. It is used to repair photoproducts caused by UV damage and bulky DNA lesions caused by a variety of mutagens.

Answer 79

A

B) Xeroderma pigmentosum (XP)

Feedback:
People born with the disorder, xeroderma pigmentosum, have a mutation in one of the genes coding for nucleotide excision repair enzymes. Therefore they are unable to carry out efficient repair on sunlight damage and they are hypersensitive to sunlight. They have to protect their skin from daylight or risk getting skin cancer.

Answer 80

A

B) S phase.

Feedback:
S stands for DNA ‘Synthesis’. It takes about 8 hours of the 24-hour human cell cycle.
G1 stands for ‘Gap 1’ and is the period of cell growth after cell division.
G2 stands for ‘Gap 2’ and is the period between the end of S-phase and the beginning of mitosis.
M stands for ‘Mitosis’ and is the period during which the chromosomes condense and the cell divides.

Answer 81

A

A) True

Feedback:
Unwinding of the double helix during DNA replication causes overwinding in front of the replication fork and the build-up of positive supercoils. These supercoils are released by the enzyme, topoisomerase I, which nicks the DNA, allows rotation of the helix, and then reseals the DNA duplex.

Answer 82

A

B) False

Feedback:
Nucleotides are always added on to the 3’ OH end of the growing DNA strand during DNA replication. There are no DNA polymerases that can add nucleotides to the 5’ end.

Answer 83

A

A) The E. coli chromosome is a single replicon.

B) Replication begins at oriC.

Feedback:
The E. coli chromosome is a circular, double-stranded DNA molecule. It replicates as a single replicon, beginning at the origin, oriC, with two replication forks travelling in opposite directions (bidirectional replication) until the entire chromosome has been replicated. Replication takes place at about 1000 base pairs per second and the entire chromosome takes about 40 minutes to be completely replicated.

Answer 84

A

A) Telomeres become shorter as cells age.

B) Somatic cells have very little telomerase.

Feedback:
Telomerase is the enzyme that synthesises chromosome ends, or telomeres. Telomerase occurs in germ cells which are continuously dividing. However, most cell lines need only to go through a certain number of cell divisions in their life time and the stock of telomerase is used up as they age. Most differentiated somatic cells have little or no telomerase and their telomeres get shorter as they divide. Many (but not all) cancer cell lines have high levels of telomerase, which allows them to carry on dividing unchecked.

Answer 85

A

a. Inhibit expression of a specific gene product.

Answer 86

A

d: pEGFP uncut

Answer 87

A

b. The production of double stranded RNA.

Answer 88

A

c. A nuclease.

Feedback: The Dicer enzyme is able to snip a double-stranded form of RNA into segments that can attach themselves to genes and block their activity.

Dicer cleaves dsRNA into smaller fragments called short interfering RNAs (siRNAs) and microRNAs (miRNAs). Dicer then helps load these fragments into a large multiprotein complex called RISC.

Dicer is an endoribonuclease.

Answer 89

A

a. The RISC complex.

Feedback: The RNA-induced silencing complex, or RISC, is a multiprotein complex that incorporates one strand of a small interfering RNA (siRNA) or microRNA (miRNA). RISC uses the siRNA or miRNA as a template for recognizing complementary mRNA.

Ago, Argonaute protein family plays a central role in RNA silencing processes, as essential catalytic components of the RISC.

RISC complex is responsible for the gene silencing phenomenon known as RNA interference (RNAi). AGO proteins bind different classes of small non-coding RNAs, including microRNAs (miRNAs), small interfering RNAs (siRNAs) and Piwi-interacting RNAs (piRNAs). Small RNAs guide Argonaute proteins to their specific targets through sequence complementarity (base pairing), which then leads to mRNA cleavage or translation inhibition.

Answer 90

A

d. The template DNA is amplified in an exponential manner as the reaction proceeds.

Answer 91

A

a. -35 sequence, -10 sequence, RBS, ATG.

Answer 92

A

b. all proteins in an organism

Feedback: The proteome is the entire set of proteins, produced or modified by an organism or system. Proteomics is the large-scale study of proteins, particularly their structures and functions.

Answer 93

A

d. A sequence that could encode methionine.

Answer 94

A

d. An antibody

Feedback: The Western blot is a widely used to detect specific proteins in a sample of tissue homogenate or extract.

It uses gel electrophoresis to separate native proteins by 3-D structure or denatured proteins by the length of the polypeptide. The proteins are then transferred to a membrane (typically nitrocellulose or PVDF), where they are stained with antibodies specific to the target protein.

The gel electrophoresis step is included in Western blot analysis to resolve the issue of the cross-reactivity of antibodies.

Answer 95

A

c. neomycin (neo) resistance

Answer 96

A

b. Terminal transferase

Feedback: Terminal transferase (TdT) is a template independent polymerase that catalyzes the addition of deoxynucleotides to the 3’ hydroxyl terminus of DNA molecules. Protruding, recessed or blunt-ended double or single-stranded DNA molecules serve as a substrate for TdT.

Answer 97

A

b. reverse transcriptase (RT)

Answer 98

A

c. The 1970s.

Answer 99

A

c. 4.5 months

Answer 100

A

a. inverted terminal repeat (ITR)

Answer 101

A

b. A restriction endonuclease.

Feedback: Type IIM restriction endonucleases, such as DpnI, are able to recognize and cut methylated DNA

Answer 102

A

d. The His tag

Feedback: Affinity tags are peptide sequences genetically grafted onto a recombinant protein. Often these tags are removable by chemical agents or by enzymatic means, such as proteolysis or intein splicing. Tags are attached to proteins for various purposes.

The ideal affinity tag should be small in size and as inert as possible to limit any potential interaction with the recombinant protein or proteins that might be present in culture media.

The poly(His) tag is a widely used protein tag; it binds to metal matrices.

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