Questions on recombinant DNA technology

Question 1

A regulator is asked to approve the farming of a genetically modified salmon which carries a growth hormone gene from a chinook salmon, placed under a promoter that runs all year rather than only in summer. Evaluate the case for approving it.

Accepted Answer

B1 in favour: the fish reaches market size in roughly half the usual time and on less feed, so a given mass of protein is produced from less feed, less water and less space, and the farm supplies a market that would otherwise be supplied by fishing wild stocks B1 against: an escaped transgenic fish could breed with wild salmon and pass the gene into a wild population, where it could not be recalled, and a fast-growing fish may compete with wild fish for food and for spawning grounds B1 against, financially and socially: the stock is patented, so a farmer buys new stock rather than breeding it and is in a different economic relationship with the supplier, and consumers who object to eating a modified animal need labelling if their choice is to mean anything B1 the conditions attached matter as much as the gene itself: the approved fish are reared as predominantly sterile females in tanks on land, which meets the escape argument twice over, by removing the route to the sea and by removing the ability to breed if one escapes B1 a judgement supported by those points, for example that approval is defensible so long as the containment conditions are enforced and the product is labelled, and that the case would be much weaker for the same fish reared in sea cages, where containment depends on a net; an objection that the fish is unnatural names no mechanism and carries no weight on its own

Question 2

Describe how a knockout mouse is produced from embryonic stem cells, and explain how it can be used to investigate the function of a gene.

Accepted Answer

B1 a length of DNA matching the target gene is made with a marker gene, such as one for antibiotic resistance, inserted into it so the gene is disrupted B1 this is put into mouse embryonic stem cells, where in a few cells it replaces the normal gene by homologous recombination; cells carrying it are selected with the antibiotic B1 the selected cells are injected into a blastocyst, which is implanted into a surrogate mother, giving chimaeric offspring B1 chimaeras are bred, and heterozygous offspring are crossed so that a quarter of their young have both alleles inactivated B1 the knockout mice are compared with normal mice of the same strain, and whatever structure or process differs indicates the function of the gene

Question 3

Explain why the same restriction endonuclease must be used to cut both the gene and the plasmid, and explain what DNA ligase then contributes.

Accepted Answer

B1 a restriction endonuclease cuts only at its own recognition sequence, and an offset cut through the two strands leaves each fragment with a short single-stranded overhang, a sticky end B1 using one enzyme on both means every fragment carries the same overhang, so the exposed bases of the gene are complementary to those of the cut plasmid B1 the overhangs pair by hydrogen bonding between complementary bases, which holds the pieces together but leaves the sugar-phosphate backbones unjoined B1 DNA ligase catalyses the formation of phosphodiester bonds along both strands, which makes the join permanent

Question 4

A polymerase chain reaction begins with 5 molecules of double-stranded target DNA and runs for 20 cycles. Calculate the number of molecules present at the end, and calculate how long the run takes if each cycle lasts 90 seconds.

Accepted Answer

M1 the number of molecules doubles each cycle, so the number at the end is the starting number multiplied by 2 raised to the number of cycles M1 5 × 2²⁰ = 5 × 1 048 576 A1 5 242 880 molecules A1 20 × 90 = 1800 seconds, which is 30 minutes

Question 5

Well under one bacterium in a hundred takes up a plasmid, and many plasmids close on their own sticky ends without accepting an insert. Suggest why a marker gene is needed at all, and suggest how placing the gene of interest inside a gene for a fluorescent protein allows the recombinant cells to be picked out.

Accepted Answer

B1 after transformation the culture holds three populations: cells with no plasmid, cells with a self-closed plasmid and cells with the recombinant plasmid. They look identical B1 a marker gene gives an observable difference, so the very small proportion of cells that took up the plasmid wanted can be found without testing every colony B1 a plasmid that closed on itself has an intact fluorescent protein gene, so it makes the protein and its colony glows under ultraviolet light B1 a plasmid carrying the insert has that gene interrupted in the middle, so no functional protein is made and the colonies wanted are the ones that do not glow, but only among cells first selected for carrying a plasmid, for example on antibiotic agar, since cells with no plasmid do not glow either

Question 6

Compare obtaining a gene by treating mRNA with reverse transcriptase with obtaining the same gene by cutting it out of chromosomal DNA using a restriction endonuclease, referring to the material each route starts from, to introns, and to the host in which each product can be expressed.

Accepted Answer

B1 the reverse transcriptase route starts from the mRNA of a cell that makes a great deal of the protein wanted, such as a β cell of the pancreas for insulin, whereas the restriction route starts from the chromosomal DNA of any cell at all, since every cell carries the gene whether it transcribes it or not B1 reverse transcriptase builds a DNA strand against the RNA template and DNA polymerase then adds the second strand, whereas a restriction endonuclease cuts the double-stranded gene out directly, which works only if its recognition sequence happens to lie on each side of the gene B1 the mRNA has already been spliced, so the complementary DNA carries no introns, whereas the fragment cut from the chromosome carries every intron the gene has B1 the complementary DNA can therefore be expressed in a prokaryote such as Escherichia coli, which has no spliceosomes, whereas the cut fragment yields a functional protein only in a host that can splice, such as a yeast or a mammalian cell culture

Question 7

Describe how a recombinant plasmid is introduced into bacterial cells, and describe what the plasmid must carry if the inserted gene is to be copied as the culture grows.

Accepted Answer

B1 the bacteria are made permeable to DNA, the standard method being to suspend them in an ice-cold calcium chloride solution B1 a brief heat shock at about 42 °C then drives plasmids into the cells; electroporation, a short high-voltage pulse, is the alternative, and either way the process is called transformation B1 uptake is inefficient: only a small fraction of the cells take up a plasmid at all, and many plasmids have re-closed on their own sticky ends without an insert, which is why a marker gene is needed afterwards to find the cells wanted B1 the plasmid must carry an origin of replication, so the host copies it and passes copies to both daughter cells at every division, and the gene is amplified along with the culture; that is amplification in vivo, as against the in vitro amplification of the polymerase chain reaction

Question 8

Different organisms use the synonymous codons for an amino acid at different frequencies, and a gene written in the codons its host uses most often is usually translated faster, giving a higher yield of protein. Explain why the gene machine method allows a gene's codons to be chosen for a particular host, while the reverse transcriptase method does not.

Accepted Answer

B1 the gene machine builds the gene from scratch, a nucleotide at a time, from a sequence designed in advance, so the codon at every position can be chosen B1 the genetic code is degenerate, so most amino acids have more than one codon, which is what makes a choice between codons possible at all B1 synonymous codons give the same amino acid sequence, so the primary structure of the protein, and so its folding and function, is unchanged by the choice B1 the reverse transcriptase method copies an existing mRNA by complementary base pairing, so the gene carries exactly the codons the original cell used, and they cannot be altered without further engineering

Question 9

Compare in vivo amplification of a gene by a plasmid replicating inside host cells with in vitro amplification by the polymerase chain reaction, referring to speed, scale and what is produced.

Accepted Answer

B1 in vivo amplification by dividing cells is slower to reach a large number of copies than PCR, but the culture can be scaled up to industrial volumes, whereas PCR is very fast but is normally run at laboratory scale B1 PCR produces DNA only, whereas replication inside host cells can be followed by transcription and translation, so protein is also produced if the gene is expressed B1 PCR needs no living cells, only a thermostable polymerase, primers, nucleotides and a thermal cycler, whereas in vivo amplification needs a living, transformed host culture kept alive and fed B1 a diagnostic test that needs enough DNA to sequence or run on a gel uses PCR, while production of a protein such as human insulin in bulk uses transformed cells in culture

Question 10

Bt cotton carries a bacterial gene coding for a protein toxic to the caterpillars that eat the crop. Suggest how this reduces insecticide use, and suggest one way its widespread use could undermine its own long-term effectiveness.

Accepted Answer

B1 the gene from Bacillus thuringiensis makes the toxin inside the plant's own tissues, so every plant in the field carries its own protection against the caterpillars continuously B1 because the toxin is always present, farmers do not need to spray insecticide against those caterpillars as often, cutting insecticide use and its cost B1 constant exposure to the same toxin in every plant selects strongly for any caterpillar that happens to carry resistance to it, more strongly than occasional spraying would B1 resistant caterpillars then survive and breed, so resistance can spread through the pest population and reduce the crop's protection over time, unless non-Bt refuge areas are planted to slow the process

Question 11

Describe how a gene gun is used to produce a genetically modified crop plant.

Accepted Answer

B1 microscopic particles of gold or tungsten are coated with copies of the DNA, which includes the wanted gene and a marker gene B1 the particles are fired at high speed, for example by high-pressure helium, into plant tissue such as embryos or callus, passing through the cellulose cell walls B1 in a few cells the DNA reaches the nucleus and is inserted into a chromosome; these cells are selected by growing the tissue on a medium containing the antibiotic or herbicide the marker gene resists B1 plant cells are totipotent, so surviving cells are grown with plant hormones into whole plants, every cell of which carries the gene, including those that form pollen and ovules

Question 12

Mice heterozygous for a knockout of gene G were crossed with each other. The table shows the genotypes of 152 pups examined at birth; the figures are constructed for this question. Calculate the number of pups of each genotype expected if all genotypes survive equally, and suggest what the results show about gene G and how its role in adult mice could still be studied.

Accepted Answer

M1 a cross between two heterozygotes gives a 1 : 2 : 1 ratio, so a quarter, a half and a quarter of 152 are expected A1 38 with both alleles working, 76 heterozygous and 38 with both alleles knocked out B1 no homozygous knockout pups were born, and the other two classes are close to a 1 : 2 ratio, so mice lacking gene G die before birth: the gene is essential for embryonic development B1 its adult role could be studied with a conditional knockout, in which the gene is inactivated only in one tissue or only after a chosen age

Question 13

Ordinary soya oil goes rancid quickly and is unstable for frying. Explain how genetic modification of soya beans has produced an oil that resists oxidation.

Accepted Answer

B1 ordinary soya oil contains a high proportion of polyunsaturated fatty acids, linoleic and linolenic acid, with two and three carbon-carbon double bonds B1 oxygen reacts at the double bonds, so the more double bonds a fatty acid has, the faster it oxidises and the oil turns rancid B1 in high-oleic soya the gene for the desaturase enzyme FAD2, which converts oleic acid into linoleic acid, is silenced B1 so oleic acid, with only one double bond, accumulates and makes up most of the oil, which therefore oxidises more slowly and needs no hydrogenation, so no trans fats are formed

Question 14

State the three temperatures used in a cycle of the polymerase chain reaction, and state what each of them is for.

Accepted Answer

B1 95 °C, at which the hydrogen bonds between paired bases break and the two strands separate B1 50 to 65 °C, at which the primers bind to their complementary sequences at each end of the target region B1 72 °C, which is the optimum for Taq polymerase as it extends each primer to build the new strand

Question 15

Explain why DNA fragments move towards the positive electrode during gel electrophoresis, and explain why the shortest fragments travel furthest.

Accepted Answer

B1 every nucleotide carries a phosphate group and phosphate groups are negatively charged, so every fragment is attracted towards the positive electrode B1 charge does not separate the fragments, because a longer fragment carries proportionately more of it B1 the agarose gel is a mesh, and a short fragment threads through it more easily than a long one, so in a given time the short fragments travel further

Question 16

Describe the three ways of obtaining a gene for genetic engineering, naming a situation that suits each one.

Accepted Answer

B1 reverse transcriptase acting on mRNA suits producing an intron-free gene when a cell that expresses it heavily, such as a pancreatic β cell for insulin, is available B1 a restriction enzyme cutting the gene directly from the chromosome suits a quick route when a suitable recognition sequence flanks the gene, though the product still carries its introns B1 the gene machine builds the gene chemically from known sequence data, needing no biological starting material at all and allowing the codons to be chosen to suit the host

Question 17

Explain why a plasmid vector needs a promoter the host recognises, in addition to an origin of replication.

Accepted Answer

B1 a promoter is the region of DNA to which RNA polymerase binds to start transcription, and a host's RNA polymerase recognises only its own kind of promoter sequence B1 a human gene's own promoter, carried across with the gene, would not be recognised by a bacterium's RNA polymerase, so the gene would not be transcribed without a host promoter placed in front of it B1 an origin of replication only ensures the plasmid itself is copied and passed to daughter cells; it does nothing to make the inserted gene transcribed, which is a separate requirement met by the promoter

Question 18

Explain why Taq polymerase, rather than a human DNA polymerase, is used in the polymerase chain reaction.

Accepted Answer

B1 the reaction is repeatedly heated to 95 °C to separate the DNA strands, and a human DNA polymerase's tertiary structure would be disrupted at that temperature, so the enzyme would be denatured B1 Taq polymerase comes from Thermus aquaticus, a bacterium of hot springs, and its tertiary structure is not disrupted at 95 °C B1 because Taq survives every denaturation step, one addition of it at the start works for all the cycles, instead of fresh enzyme having to be added after every one

Question 19

Explain why a virus used as a vector to carry a gene into human cells is first modified, and explain why a gene gun rather than a viral vector or a heat shock is often used to transform cereal plants.

Accepted Answer

B1 the genes the virus needs to replicate and to cause disease are removed, so that it can deliver the gene into a cell but cannot make new viruses or harm the host B1 the removed genes are replaced by the gene to be delivered, using the virus's natural ability to bind to cells and get its nucleic acid inside B1 plant cells have a cellulose cell wall, and the gene gun forces DNA-coated particles through it, while heat shock works on bacteria and few suitable biological vectors infect cereals

Question 20

A DNA ladder was run on a gel beside a sample. Ladder fragments of 6000, 3000, 1500 and 750 base pairs moved 20, 31, 42 and 53 mm from the well, and a band in the sample moved 36.5 mm; the results are invented for this question. Distance moved falls in a straight line with log₁₀ of fragment length. Calculate the length of the fragment in the sample.

Accepted Answer

M1 log₁₀ 3000 = 3.48 and log₁₀ 1500 = 3.18, the markers either side of the band M1 36.5 mm is halfway between 31 and 42 mm, so log₁₀ length = (3.48 + 3.18) ÷ 2 = 3.33 A1 length = 10^3.33 = about 2100 base pairs; accept 2000 to 2200

Question 21

When running a gel to separate DNA fragments, a student loaded the samples, mixed with loading dye, into wells at the end of the gel next to the negative electrode, and ran a DNA ladder in one lane of the same gel. Explain each of these three decisions.

Accepted Answer

B1 DNA is negatively charged because of its phosphate groups, so it moves towards the positive electrode and must start at the negative end to cross the gel B1 the loading dye is dense, so the sample sinks into the well instead of spreading into the buffer, and it is coloured, so the progress of the run can be seen B1 the ladder's fragments are of known length and are run under exactly the same conditions, so the lengths of the sample's fragments can be read against them

Question 22

In Southern blotting, DNA fragments separated by gel electrophoresis are made single-stranded and transferred from the gel onto a nylon membrane, keeping their positions. A labelled probe, a short single strand of DNA whose base sequence is complementary to part of the target gene, is then added, and probe that has not bound is washed off. Explain why the fragments must be single-stranded for the probe to find its target, and explain why it matters that they keep their positions during the transfer.

Accepted Answer

B1 the probe binds by complementary base pairing, which needs the target's bases to be unpaired; in double-stranded DNA they are already paired with the other strand B1 a fragment's position shows how far it moved in the gel, which depends on its length, so keeping the positions lets the band the probe labels be matched to a fragment of known length

Genotype of pups at birth	Number of pups
both alleles working	51
heterozygous	101
both alleles knocked out	0