S6 Biology Paper: Gene Expression, Mutation And Biotechnology (100 Marks)

S6 Biology Paper: Gene Expression, Mutation and Biotechnology (100 marks)

2026-11-11

S6 Biology – gene expression and biotechnology

S6 Biology paper: transcription and translation, mutation, base composition and hydrogen bonds, recombinant DNA and gel electrophoresis, PCR and GM crops.

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Paper details　S6 Biology gene expression practice paper with full marking scheme

Question 1 Transcription and translation (18 marks)

The mature mRNA of a gene contains 900 nucleotides, of which 120 lie in untranslated regions. The mRNA is translated in the cytoplasm and the stop codon does not code for an amino acid. Assume the average molecular mass of an amino acid is 110. Answer the following.

#	Question	Answer and marking notes
1	Find the number of nucleotides in the coding region and the number of codons.	Answer: 900 minus 120 gives 780 nucleotides, and 780 divided by 3 gives 260 codons. (Marking: 2 marks for the subtraction, 1 for the value, 1 for dividing by three.)
2	Find the number of amino acids in the protein.	Answer: one of the 260 codons is a stop codon, so the number of amino acids is 260 minus 1, which gives 259. (Marking: 3 marks for identifying the stop codon, 2 for the answer.)
3	Estimate the molecular mass of the protein in daltons.	Answer: 259 multiplied by 110 gives 28490 daltons, which is about 28.5 kDa. (Marking: 2 marks for the method, 2 for the value with unit.)
4	Explain why the protein isolated in an experiment usually has a lower molecular mass than the estimate above.	Answer: post-translational modification changes the mass, for example the removal of a signal peptide, protein splicing or the addition of modifying groups; some peptide segments are cut out during processing and the sample may be partly degraded by proteases during purification, so the measured mass is usually lower. (Marking: 2 marks for post-translational modification, 2 for removal of peptide segments, 1 for degradation or modifying groups.)

Question 2 Gene mutation and frameshift effects (16 marks)

The coding region of a structural gene contains 600 base pairs. In strain A an extra base pair is inserted after the 150th base of this coding region. In strain B a single base substitution occurs at the 90th base. Answer the following.

#	Question	Answer and marking notes
5	Find the number of codons and the number of amino acids coded by the normal coding region.	Answer: 600 divided by 3 gives 200 codons, and excluding the stop codon the region codes for 199 amino acids. (Marking: 2 marks for the codon number, 2 for the amino acid number.)
6	In strain A, how many codons before the insertion point are unaffected, and how many codons after it are shifted out of frame?	Answer: 150 divided by 3 gives 50 codons that are unaffected, and the remaining 200 minus 50, that is 150 codons, are all shifted out of frame. (Marking: 2 marks for the unaffected number, 3 for the shifted number.)
7	Explain why a frameshift mutation usually makes the protein non-functional.	Answer: after the frame shift the codons are regrouped and the whole amino acid sequence from the insertion point onwards changes; a stop codon very likely appears early, so translation ends prematurely and the shortened protein loses its normal three-dimensional shape and active site, which usually destroys its function. (Marking: 2 marks for the changed sequence, 1 for premature termination, 1 for loss of structure or function.)
8	If the substitution in strain B occurs at the third base of a codon, explain why its effect is usually smaller.	Answer: the genetic code is degenerate, so several different codons can code for the same amino acid; a change at the third base therefore often gives a synonymous mutation, the amino acid sequence stays the same and protein function is unaffected. (Marking: 2 marks for degeneracy, 1 for synonymous mutation or unchanged sequence.)

Question 3 Base composition and hydrogen bonding in DNA (18 marks)

Sample A is a double-stranded DNA molecule of 2000 bases in which adenine makes up thirty per cent of all bases. Sample B also contains 2000 bases, of which adenine makes up twenty-five per cent. There are 2 hydrogen bonds between A and T and 3 hydrogen bonds between G and C. Answer the following.

#	Question	Answer and marking notes
9	Find the number of each of the four bases in sample A.	Answer: A is 2000 times 0.30, that is 600, and T equals A, so T is also 600; G and C together make up 2000 minus 1200, that is 800, so G is 400 and C is 400. (Marking: 1 mark each for A and T, 1 for the combined G and C, 1 each for G and C.)
10	Find the total number of hydrogen bonds in sample A.	Answer: the A to T pairs give 600 times 2, that is 1200 bonds, and the G to C pairs give 400 times 3, that is 1200 bonds, so the total is 2400 hydrogen bonds. (Marking: 2 marks for the A and T part, 2 for the G and C part, 1 for the total.)
11	Find the number of G and C in sample B and the percentage of all bases that each represents.	Answer: A and T are each 2000 times 0.25, that is 500; G and C together make up 1000, so G and C are each 500, which is twenty-five per cent each and fifty per cent for the two together. (Marking: 1 mark for A and T, 2 for the G and C numbers, 1 for the percentages.)
12	Predict which sample has the higher melting temperature and explain your answer.	Answer: sample B has the higher melting temperature. Its G and C content is fifty per cent against forty per cent in sample A, and since G pairs with C by 3 hydrogen bonds while A pairs with T by only 2, more energy is needed to separate the two strands of sample B, so its melting temperature is higher. (Marking: 1 mark for the choice, 1 for comparing the content, 2 for the number of hydrogen bonds.)

Question 4 Recombinant DNA and gel electrophoresis (18 marks)

A plasmid of 4000 bp contains one EcoRI site. A gene of interest of 1200 bp is inserted at that site to form a recombinant plasmid. The inserted fragment contains a second EcoRI site 400 bp from its own start. Answer the following.

#	Question	Answer and marking notes
13	Find the total length of the recombinant plasmid.	Answer: 4000 plus 1200 gives 5200 bp. (Marking: 2 marks for the method, 2 for the value with unit.)
14	After complete digestion with EcoRI, find the length of each fragment obtained.	Answer: the recombinant plasmid has two EcoRI sites, so digestion gives 2 fragments: one runs from the start of the insert to the cut and is 400 bp long, the other follows the rest of the circular molecule and is 5200 minus 400, that is 4800 bp. (Marking: 2 marks for identifying two sites, 1 for the number of fragments, 1 for 400 bp, 2 for 4800 bp.)
15	In gel electrophoresis, which fragment moves faster? Explain.	Answer: the 400 bp fragment moves faster because it is shorter and meets less resistance when passing through the mesh of the gel, so it travels further in the same time. (Marking: 1 mark for the choice, 2 for length and resistance, 1 for the distance travelled.)
16	If the original plasmid without the insert is also run as a control, explain why its band appears between the two fragments.	Answer: the original plasmid has only one EcoRI site, so digestion gives a single linear molecule of 4000 bp; this is shorter than 4800 bp but longer than 400 bp, so its band lies between the two fragments. (Marking: 2 marks for identifying 4000 bp, 2 for comparing the sizes.)

Question 5 Amplification in the polymerase chain reaction (15 marks)

Each cycle of the polymerase chain reaction doubles the number of copies of the target fragment. The reaction starts from one double-stranded DNA molecule and every cycle is assumed to be one hundred per cent efficient. Answer the following.

#	Question	Answer and marking notes
17	How many copies of the target fragment are present after 10 cycles?	Answer: 2 raised to the power 10 gives 1024 copies. (Marking: 2 marks for the exponential form, 2 for the value.)
18	After 30 cycles, express the number of copies in scientific notation to three significant figures.	Answer: 2 raised to the power 30 gives 1073741824 copies, that is 1.07 times 10 to the power 9 copies. (Marking: 2 marks for the value, 2 for scientific notation with significant figures.)
19	Find the minimum number of cycles needed for the number of copies to exceed one million.	Answer: 2 raised to the power 19 gives 524288 copies, which is below one million, while 2 raised to the power 20 gives 1048576 copies, which is above it, so at least 20 cycles are needed. (Marking: 2 marks for testing 19 cycles, 2 for the conclusion.)
20	State two reasons why the actual yield is usually lower than the theoretical value.	Answer: the primers and the deoxynucleoside triphosphates become exhausted late in the reaction so the efficiency of amplification falls; the DNA polymerase gradually loses activity during the high temperature denaturation steps; and the accumulating products compete for primers, which also lowers the efficiency. (Marking: 1.5 marks for each valid reason, 3 marks for two.)

Question 6 Genetically modified crops: uses and risks (15 marks)

In a country the area planted with genetically modified maize rose from 0.5 million hectares in 2010 to 3.2 million hectares in 2025, while the total area of maize in the country is 4 million hectares. Answer the following.

#	Question	Answer and marking notes
21	Find the increase in the planted area expressed as a multiple.	Answer: 320 minus 50 gives 270, and 270 divided by 50 gives 5.4, so the area increased by a factor of 5.4, which is five hundred and forty per cent. (Marking: 2 marks for the difference, 1 for dividing by the original value, 1 for the multiple.)
22	Find the proportion of the national maize area planted with genetically modified maize in 2025.	Answer: 320 divided by 400 gives 0.8, so genetically modified maize covers eighty per cent of the national maize area. (Marking: 2 marks for the method, 2 for the proportion.)
23	State two benefits of growing genetically modified maize.	Answer: some varieties carry insect resistance or herbicide tolerance, so fewer insecticides and herbicides need to be applied; insect resistant varieties suffer less feeding damage, which raises yield per hectare and lowers production costs; and drought or salt tolerant varieties extend the area that can be cultivated. (Marking: 2 marks for each benefit, 4 marks for two.)
24	State two risks associated with gene flow and the evolution of resistance.	Answer: pollen can be carried by wind or insects to nearby crops of the same species or to wild relatives, so the introduced gene may spread into other populations; and long term large scale planting of insect resistant varieties places selection pressure on the pest population, so resistant individuals survive and multiply quickly and the resistance trait loses its value. (Marking: 1.5 marks for gene flow, 1.5 for the evolution of resistance.)

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