SSP COPEG

H2 Biology COPEG Extra LessonACJC1 Within a cell, the amount of polypeptide made using a given mRNA molecule depends partly on

Athe degree of DNA methylationBthe rate of which the mRNA is degradedCthe presence of transcription factorsDthe types of ribosomes present in the cytoplasm

Note: The rate of which the mRNA degrades is dependent and directly proportional to the length of poly A tail present at the 3 end of the mRNA. It is degraded by exoribonucleasespresent in the cytoplasm. Hence a longer poly A tail will take a longer time for degradation allowing more time for translation. DNA methylation will cause the chromatin to be tightly packed which prevents transcription of the gene. The presence of transcription factors (general of specific) will bring about transcription of genes while the type of ribosomes (70/80s) is not relevant in this question. On another note, the location of the ribosome, free ribosome or bound onto the RER will determine the destination of the protein.

2 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?

histone tailhistone deacetylasehistone acetylaseDNA

AHelicase action is enhanced by acetylation.BAcetylation of histones neutralizes their negative charges and encourages binding to DNA polymerase.CWhen nucleosomes are highly acetylated, chromatin becomes less compact and DNA is more accessible for transcription.DRNA polymerase works better by binding with acetyl groups.

Note: Acetylation removes the positive charge of the lysine R group hence causing the lysine of the histone to bind loosely to the sugar phosphate backbone of the DNA. Recall that helicase serves to break the hydrogen bond between the DNA strands in semi conservative replication.

3 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.

Note: removal of introns and splicing together of exons are brought about by splicesosome and they are part of post transcriptional modification of the primary RNA. The other post transcriptional mod is the addition of the 5 guanosine cap and the addition of the poly A tail at the 3 end. After these modifications, the primary RNA becomes a mature mRNA. Addition of methyl groups affects DNA packing and the binding of transcription factor (general / specific) determines if transcription takes places or not. Gene amplification results in multiple template of the gene so that more than one template can be used for transcription at the same time. This result in high amount of mRNA produced in a short time.

4 The globin gene family in humans consists of the , and 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.

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Age (months)% of Globin ChainsPrenatalPostnatalBirth

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

AMethyl groups are added to regulatory sequences of -globin genes during the postnatal period, allowing for some proteins to bind. BAlternative splicing has occurred to form the mature mRNA of the -globin and -globin genes, resulting in differences in the rate of expression of globin chains during the prenatal period.CA growth factor triggers the expression of a transcription factor that increases the rate of -globin gene expression during the postnatal period.DThe shortening of poly(A) tail in the mRNA of globin genes reduces its stability, resulting in a decrease in the rate of expression of -globin chains during the postnatal period.

Note: The , and genes indicates that 3 different genes codes for the 3 polypeptides. Hence the only possible reason that cant account for the graph is alternative splicing as it implies 3 protein being coded for by 1 gene. The 3 protein would consist of different combination of exons (from the same gene) spliced together.

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

Aenzymatic shortening of the poly(A) tail Bmethylation of C nucleotides Cremoval of the 5' capDremoval of one or more exons

Note: Normally removal of the 5cap takes place after the poly A tail have been shortened to a critical length. Upon removal of the 5 cap the 5 exonucleases will rapidly degrade the mRNA at the 5end. Enzymatic shortening of the poly A tail will take a while depending on its length. Hence removal of the 5 cap will have a drastic effect than the shortening of the poly A tail.

6 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?

AA transgene integrated into a heterochromatic region of the genome. BA transgene integrated into a euchromatic region of the genomeCThe transgene was mutated during the process of integration into the host cell genomeDA transgene integrated into a region of the genome characterized by high histone acetylation

Note: A transgene is a gene from a foreign species. In the heterochromatic region of the genome, the DNA is tightly packed hence the promoter regions of gene cannot be accessed by general transcription factors and RNA polymerase. This would repress gene transcription. Histone acetylation will lead to looser packing of the chromatin.

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

1 attachment of RNA polymerase to promoter requires interaction with activator proteins2 attachment of activator proteins to enhancers increases the basal activity of the promoter3 attachment of activator proteins to silencers suppresses the basal activity of the promoter4 enhancers and silencers are DNA sequences usually found upstream of structural genes and can regulate gene expression

A2 and 4B3 and 4C1, 2 and 4D1, 3 and 4

Note: Attachment of activator proteins to enhancers stabilizes the formation of the transcription initiation complex, hence increase the transcriptional rate of the gene.

8 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?

AAlternative splicing can result in more than one kind of protein produced from one gene.BSome genes are transcribed to produce RNA that is not meant to serve as a template for protein synthesis.CThe enhancers present in the human genome are also transcribed to bring about an increase in the transcription of protein-coding genes. DThe telomeric regions are also transcribed to give telomerase, which helps to maintain the telomere length.

Note: Genes can also codes for rRNA and tRNA which serves a structural or enzymatic function. rRNA can serve a structural function in the formation of the ribosome or an enzymatic function eg peptidyl transferase in the ribosome. tRNA serves a structural function in transporting specific amino acid in translation. Alternative splicing would lead to the observation that more proteins are formed that the no. of genes (for polypeptide) present. Enhancers and silencers (control elements) are never transcribed. Telomerase are coded for by the gene for it. Telomerase functions to extend the ends of the DNA molecules.

9 Which of the following methods of regulating gene expression is common to both prokaryotes and eukaryotes?ABinding of proteins to control elementsBDNA methylation C Histone acetylationDPost-transcriptional modification of RNA

Note: Some example of control elements in prokaryotes are promoter, operator, activator binding regions. DNA methylation is present in prokaryotes but not for the purpose of gene control. Rather they serve to protect its own DNA from degradation by distinguishing it from a viral DNA. Exonuclease are present in bacteria to degrade viral DNA during infection. Prokaryote have non histone proteins that functions in DNA packing. There is no post transcriptional mod of RNA as transcription and translation takes place simultaneously due to a lack of nuclear envelope.

1Fig. 1 below shows the various parts of a gene.E1E2E3E4PromoterTermination sequencePoly(A) addition signalRegulatory sequence 2Regulatory sequence 1I1I2I3

E = exon, I = intronFig. 1

a) State the terms used to describe regulatory sequence 1 and 2. [1]Regulatory sequence 1enhancer/distal control element ;

Regulatory sequence 2proximal control element ; @ mark

b) State the functions of the following: [2]Promotersite where RNA polymerase binds ;

Exoncodes for a sequence of amino acids in a portion of a polypeptide; @ 1 mark

c) Describe the role of regulatory sequence 1 in causing the gene to be expressed. [3]1) Specific transcription factors (activators) bind to the enhancers ;

2) This recruits a DNA-bending protein which causes the DNA to bend;

3) Mediator proteins will bind to the bound activators, recruiting RNA polymerase and general transcription factors;

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

Note: Point 1 to 4 details the immediate role of the enhance while point 5 describes its ultimate role.

d) Explain what could happen to gene expression if a short sequence of DNA was inserted into or near the(i) Promoter [2]RNA polymerase may not be able to bind to the disrupted promoter;

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

Note: Scope of both questions relates to gene expression, hence answers must end by addressing if the gene is transcribed or not and its rate of transcription if applicable.