H2 Biology COPEG Extra Lesson ACJC

1 With reference to prokaryotic and eukaryotic genomes, which of the following statement is not true?

A They both have extrachromosomal DNA. B They both have non-coding regions.C They both have regulatory sequences.D They are both associated with histones.

2 Within a cell, the amount of polypeptide made using a given mRNA molecule depends partly on

A the degree of DNA methylationB the rate of which the mRNA is degradedC the presence of transcription factorsD the types of ribosomes present in the cytoplasm

3 The diagram below shows how acetylation of histones promotes loose chromatin structure. Recent evidence has shown that chemical modification of histones play a direct role in regulation of gene expression.

Which of the following best explains how acetylation regulates gene expression?

A Helicase action is enhanced by acetylation.

B Acetylation of histones neutralizes their negative charges and encourages binding to DNA polymerase.

C When nucleosomes are highly acetylated, chromatin becomes less compact and DNA is more accessible for transcription.

D RNA polymerase works better by binding with acetyl groups.

4 Which of the following is an example of a possible step in the post-transcriptional control of gene expression?

A The addition of methyl groups to cytosine bases of DNA.B The binding of transcription factors to a promoter.C The removal of introns and splicing together of exons.D Gene amplification during a stage in development.

DNA

histone acetylase

histone deacetylase

histone tail

0

50

100

0 2 4 6 8 0 2 4 6 8 10 12

Age (months)

% of Globin Chains

Prenatal Postnatal

Birth

H2 Biology COPEG Extra Lesson ACJC

5 The globin gene family in humans consists of the α, β and γ genes. These genes code for the globin chains that make up haemoglobin and are expressed at different levels during different developmental stages. The graph shows the expression of the various globin chains during the prenatal (fetal) and postnatal (after birth) periods.

Which of the following cannot account for the differences in the levels of expression of globin chains?

A Methyl groups are added to regulatory sequences of γ-globin genes during the postnatal period, allowing for some proteins to bind.

B Alternative splicing has occurred to form the mature mRNA of the α-globin and β-globin genes, resulting in differences in the rate of expression of globin chains during the prenatal period.

C A growth factor triggers the expression of a transcription factor that increases the rate of β-globin gene expression during the postnatal period.

D The shortening of poly(A) tail in the mRNA of γ –globin genes reduces its stability, resulting in a decrease in the rate of expression of γ-globin chains during the postnatal period.

6 Which of the following experimental procedures is most likely to hasten mRNA degradation in a eukaryotic cell?

A enzymatic shortening of the poly(A) tail B methylation of C nucleotides C removal of the 5' capD removal of one or more exons

H2 Biology COPEG Extra Lesson ACJC

7 A geneticist introduces a transgene into yeast cells and isolates five independent cell lines in which the transgene has integrated into the yeast genome. In four of the lines, the transgene is expressed strongly, but in the fifth there is no expression at all. Which is a likely explanation for the lack of transgene expression in the fifth cell line?

A A transgene integrated into a heterochromatic region of the genome. B A transgene integrated into a euchromatic region of the genomeC The transgene was mutated during the process of integration into the host cell

genomeD A transgene integrated into a region of the genome characterized by high

histone acetylation

8 Which of the following about eukaryotic control elements are correct?

1 attachment of RNA polymerase to promoter requires interaction with activator proteins

2 attachment of activator proteins to enhancers increases the basal activity of the promoter

3 attachment of activator proteins to silencers suppresses the basal activity of the promoter

4 enhancers and silencers are DNA sequences usually found upstream of structural genes and can regulate gene expression

A 2 and 4B 3 and 4C 1, 2 and 4D 1, 3 and 4

9 The percentage of the human genome that is transcribed is larger than that predicted based on the range of proteins made by the cell. Which of the following accounts for the difference?

A Alternative splicing can result in more than one kind of protein produced from one gene.

B Some genes are transcribed to produce RNA that is not meant to serve as a template for protein synthesis.

C The enhancers present in the human genome are also transcribed to bring about an increase in the transcription of protein-coding genes.

D The telomeric regions are also transcribed to give telomerase, which helps to maintain the telomere length.

H2 Biology COPEG Extra Lesson ACJC

10 The following events occur in the extension of telomeres.

1 Further extension of 3’ end of telomere2 Translocation of telomerase to end of telomere3 Reverse transcription to extend the 3’ end of the telomere4 Complementary base-pairing of RNA template with single-stranded end

of telomere5 DNA polymerase catalyses formation of complementary strand to form

double –stranded DNA.

In which order do these events take place?

A 3 2 4 5 1B 3 4 2 1 5C 4 2 3 5 1D 4 3 2 1 5

11 Which of the following methods of regulating gene expression is common to both prokaryotes and eukaryotes?A Binding of proteins to control elementsB DNA methylation C Histone acetylationD Post-transcriptional modification of RNA

12 Which of the following about cancerous cells is false?

A They are able to divide for a certain number of times and then stop dividing.B They are able to divide further when in contact with neighbouring cells.C They are not able to differentiate properly.D They are not able to exhibit anchorage dependence.

13 Which of the following statements about cancer is / are true?

1 Cancer is a result of increased cell division which promotes the mutation of a proto-oncogene.

2 Individuals who inherit one inactive copy of tumour suppressor gene are more likely to develop cancer than individuals with two non-mutant copies.

3 Mutagenic activation of a single oncogene is sufficient to convert a normal cell into a cancer cell.

A 2 only B 1 and 3C 2 and 3D All of the above

H2 Biology COPEG Extra Lesson ACJC

1 Fig. 1 below shows the various parts of a gene.

E = exon, I = intronFig. 1

a) Describe the role of regulatory sequence 1 in causing the gene to be expressed. [4]

1) Specific transcription factors activators bind to the enhancers ;2) This recruits a DNA-bending protein which causes the DNA to bend to bring the

activator close to the promoter;3) Mediator proteins will bind to the bound activators, recruiting RNA polymerase and

general transcription factors to the promoter;4) forming the transcription initiation complex on the promoter;5) transcription of gene at a high rate; @ 1 mark

b) Explain what could happen to gene expression if a short sequence of DNA was inserted into or near the

(i) Promoter [2]1. RNA polymerase may not be able to bind to the disrupted promoter2. due to sequence not complementary to binding site of RNA polymerase / TATA box

binding protein);3. and so gene becomes transcriptionally inactive / silenced ;

(ii) Poly(A) addition signal [2]

With the poly(A) addition signal disrupted, the mRNA formed does not have a poly(A) tail;This decreases the stability of the mRNA/decreases its half-life/causes mRNA to be degraded ;

Hence gene is less expressed ; @ 1 mark

(c) Describe the role of tumour suppressor genes in normal cells. [2]Tumour suppressor genes code for proteins which1. are involved in DNA repair/fixing damaged or broken DNA2. cause (mutated) cells to undergo apoptosis.3. control the adhesion of cells to each other or to an extracellular matrix.4. are components of cell-signalling pathways that inhibit the cell cycle.@ ½ mk

(d) Predict whether a mutation in the tumour suppressor gene will be a dominant or recessive

mutation. Give reasons for your answer. [2]

1. Recessive mutation.

I3I2I1

Regulatory sequence 1

Regulatory sequence 2

Poly(A) addition signal

Termination sequencePromoter

E4E3E2E1

H2 Biology COPEG Extra Lesson ACJC

2. Loss of function mutation, which means that the normal, non-mutated gene can still function and used for synthesis of tumour suppressor protein;3. Both genes must be mutated for tumour suppressor protein to be non-functional;@ 1mk

(e) Describe one other way in which a cell’s ability to proliferate can become uncontrolled.[4]