3g1 mock exam crib

7/26/2019 3G1 Mock Exam Crib

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Exam duration 90 minutes

Answer three questions out of the four available.

Question 1

a) Given a single-stranded DNA molecule dissolved in water, what needs to be added to allowsynthesis of a complementary strand?

buffer/salts, complementary DNA primer, dNTPs, DNA polymerase.

b) What two reagents made it possible for the polymerase chain reaction to be easily carried out andwidely used.

cheaply available DNA primers, pure thermostable DNA polymerases

c) A researcher has amplified the open reading frame of a Green Fluorescent protein gene and clonedit next to a promotor in E. coli. However it does not seem to have worked and the bacteria do not

fluoresce.

i) What are the different types of mutation that might have occurred during the PCR of this gene?

synonymous, non-synonymous codon changes, insertions/deletions, introduction of a stop codon

ii) Which is most likely to be responsible for the inactivity of the clone and why?

insertion/deletion likely to change the reading frame so most likely to cause deleterious mutation. (or

other reasonable argument)

iii) How would the researcher show that it was indeed a mutation that had caused the inactivity, ratherthan a problem with the host strain itself?

Either: Purify the plasmid and transform it into a different strain (indirect) or sequence the construct

(direct evidence).

d) DNA is extracted from a human blood sample, PCR is used to amplify ten genes from chromosome22, and the purified PCR products are sequenced using the original Sanger dideoxy sequencingmethod. At certain locations, specific bases in the sequences generated are found to be ambiguous.

i) Explain why this is the case.

Apart from the sex chromosomes, blood cells have two copies of each chromosome. Where thereare sequence variations (most commonly single nucleotide polymorphisms) the sequencing reactionwill be made up of equal quantities of each variant, and so the sequence at that location will beambiguous.

ii) The same amplified samples are then sequenced on the Illumina platform but there is no sign ofambiguity at the above positions in the individual sequencing reads.

Explain why this is expected and what computational test you would carry out on the sequences to

confirm your explanation.


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On the Illumina platform, a single molecule seeds each sequencing reaction so, at this level, the

variation is lost. By aligning the sequences generated, it should be possible to see the variationpresent at the previously observed locations in the form of different bases in different reads.

Question 2

a)

i) what is a cDNA?

An mRNA molecular that has been reverse transcribed into DNA

ii) what is a cDNA library?

A population of cDNA molecules or clones, representing the mRNA present in a biological sample.

iii) what enzyme is essential to making cDNAs?

Reverse transcriptase

b) You have sequenced the genome of a recently discovered bacterium and now want to annotate thesequence with the location of the genes.

Briefly outline two possible approaches you might take and name a commonly used algorithm that

underpins them.

Two of: Align cDNA (mRNA) sequences, align known protein sequences, ab initiomethods. All can usedynamic programming algorithms.

c) An affine gap penalty scheme produces biologically better alignments of protein sequences than alinear gap penalty: explain what affine gap penalties are and why they give better alignments.

Gaps not tolerated well in cores of protein, but well tolerated on surface loops. So, if it is possible toinsert a single residue it is likely to be possible to insert several. Therefore paying a fixed penalty tostart a gap and then a reduced per-residue penalty to extend it is appropriate, and this is the affine

scheme.

d) A pigment-producing gene is cloned into a plasmid and several thousand colonies are examined.Only one is coloured and when the plasmid it contains is sequenced it is found to contain a mutationthat introduces a stop codon in the 5th codon of the pigment gene. Puzzled, the researchers sequenceplasmids from several of the non-pigmented colonies and find exactly the same mutation.

Provide a simple explanation for the rare pigmented colony.

Various answers are possible though the best would be a mutation in a tRNA gene which allows,through wobble recognition, for it to recognise the stop codon and introduce an amino-acid at thatlocation thus allowing active protein to be generated.


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12/01/2011

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1a. Define catabolismand anabolism. Are they linked? How?

1b. Net result of complete catabolism of glucose by the yeast Saccharomyces cerevisiae

in absence of oxygen.

1c. Would S. cerevisiae under these conditions (i.e. 100% alcoholic fermentation) be a good system

for the synthesis of coupled-to-growth products (*), such as heterologous proteins?

(*)products whose synthesis requires high biomass yield (i.e. high production of ATP per glucose consumed).

1d. What alternative conditions (and/or organisms) could be used for the production of coupled-to-

growth heterologous proteins with minimum accumulation of byproducts?

3G1 course. Question

3G1 course. Answers

1a. Catabolism. Biochemical processes involved in the breakdown of organic compounds, usually leading to

the production of energy (ATP, GTP).

Anabolism. Metabolic processes involved in the synthesis of cell constituents from simpler molecules

such as organic and/or inorganic precursors. An anabolic process usually requires energy (ATP, GTP).

Yes. They are linked. Activated energy carriers (e.g. ATP, -main energy carrier-, NADH, NADPH,

FADH2

) link generation of energy (catabolism) to biosynthetic (anabolic pathways) and growth.


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12/01/2011

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1b. Yeast in absence of oxygen -- > (100% alcoholic fermentation) -- >-- > Glycolysis (glucose to pyruvate) + pyruvate to ethanol.

[1] Glycolysis. From glucose (6C) to pyruvate (3C):Glucose + 2NAD++ 2ADP + 2Pi -- > 2Pyruvate + 2NADH

++ 2ATP

[5] From pyruvate (3C) to ethanol (2C)

[5a] Pyruvate -- > CO2+ CH3CHO (enzyme pyruvate decarboxylase, PDC)

[5b] CH3CHO + NADH+ -- > ethanol + NAD+ (enzyme alcohol dehydrogenase, ADH)

Net result: [1] + 2 [5] -- > Glucose + 2ADP + 2Pi -- > 2 ethanol + 2ATP + 2CO2

(Reactions [1] and [5] in grey --> optional)

3G1 course. Answers

1c. No. Alcoholic fermentation results in, a) low biomass yields (due to low amounts of ATP procuded perglucose consumed, -much lower than through the tricarboxylic acid (TCA) cycle (i.e. respiratory

pathway)-, and b) accumulation of toxic by-products (e.g. acetate, ethanol, organic acids), which will

compromise the quality of the product and the downstream purification process.

1d. Conditions in which S. cerevisiae exhibits a predominant respiratory metabolism (ca. 100% respiration)would be preferable, but they are difficult to implement (e.g. need for high aeration levels, continuouscultures with low feed supplementation, etc). Alternative organisms with a predominant respiratory

metabolism would result in higher biomass yields with lower costs of aeration and purification.

Mammalian cells under controlled conditions leading to respiratory metabolism could be used as well,

especially for the production of complex heterologous proteins (e.g. mammalian proteins such as

antibodies, etc). The final decision should be based on the comparison of the different alternatives, theirproductivity and costs.

3G1 course. Answers


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Question

Antibody therapeutics are projected to dominate the pharmaceutical market in the nextdecade.

a) Draw and label the basic structural features of an antibody.

b) How do the basic structural features allow the antibody to function?c) Using a heavy chain gene as an example, describe the generation of antibody diversity.d) What is meant by humanising an antibody?

Answer

a)

b) The function of antibodies is the highly specific recognition of foreign antigens as a

means to neutralize and target these antigens for destruction.1. The bivalent (two antigen recognition sites on one antibody) structure assists inmaking antigen-antibody complexes.

2. The disulfide bridges keep the four peptide chains in proper, robust structure3. The light and heavy chain combinations allow for greater combinatorial diversity4. The variable regions allow for very high antigen specificity5. The constant regions determine isotype (e.g. IgG) and specify higher order

structure (e.g. IgA=dimer; IgM=pentamer) and cellular functions.c)

A similar calculation can be made for light chain genes (kappa and lambda) so the totalpossibilities are on the order of 108.

d) Humanising antibodies is important for monoclonal therapeutics to limit human immunereaction to the recombinant therapeutic antibody. Humanising antibodies can beachieved by:

1. the placement of rodent complementarity determining regions into a human antibody.

2. engineering recombinant mice with human antibody genes3. Using phage display of a human variable region for antigen recognition and subsequent

fusion with a human constant region.

3g1 mock exam crib

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