IGCSE Biology 0610 Paper 4: Genetics, DNA and Biotechnology

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 75 minutes | Total: 100 marks | Parts: 31

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (16 marks) – DNA structure, genes and protein synthesis

(a) Describe the double-helix structure of a DNA molecule and state how the four bases pair. (3 marks)

(b) A DNA molecule contains 1200 bases, of which 22% are adenine. Calculate the number of guanine bases. (3 marks)

(c) Define the terms gene and allele. (3 marks)

(d) Explain how the 64 codons determine the amino acid sequence of a protein. (3 marks)

(e) State where transcription and translation take place in a cell. (2 marks)

(f) Explain why some gene mutations do not change the function of the protein produced. (2 marks)

Question 2 (17 marks) – Monohybrid crosses and inheritance ratios

(a) Two heterozygous (Bb) plants are crossed. State the gametes produced and the genotypic ratio of the offspring. (3 marks)

(b) The cross produces 800 offspring plants. Calculate the numbers showing the dominant and recessive phenotypes and explain the ratio. (4 marks)

(c) Use a Punnett square to show the cross between Bb and bb, and state the genotypic and phenotypic ratios of the offspring. (4 marks)

(d) Explain how a test cross can be used to find the genotype of a plant showing the dominant phenotype. (3 marks)

(e) Two heterozygous parents have two children. Calculate the probability that both children show the recessive phenotype, giving your answer as a fraction and a percentage. (3 marks)

Question 3 (17 marks) – Codominance, ABO blood groups and sex linkage

(a) Distinguish between codominance and incomplete dominance, using the ABO blood group as an example of codominance. (3 marks)

(b) A man of blood group AB and a woman of blood group B (genotype I-B i) have 400 children. Calculate the number with each blood group. (4 marks)

(c) Explain why colour blindness is more common in males than in females. (3 marks)

(d) A carrier mother (XCXc) and an unaffected father (XCY) have 260 sons. Calculate how many sons are colour blind and describe the daughters. (4 marks)

(e) Explain why a sex-linked condition such as colour blindness cannot pass from a father to his son. (3 marks)

Question 4 (16 marks) – Mutation, variation and natural selection

(a) Define mutation and name two factors that increase the rate of mutation. (3 marks)

(b) Explain why most mutations are harmful and usually recessive. (3 marks)

(c) Explain how antibiotic-resistant bacteria become common in a bacterial population. (4 marks)

(d) A population of 500 insects is 12% resistant. If all resistant insects survive but only 40% of the non-resistant insects survive, calculate the population at the start of the next generation. (3 marks)

(e) Explain why using the same pesticide over large areas for a long time makes resistance rise quickly, and suggest one way to reduce this. (3 marks)

Question 5 (17 marks) – Genetic engineering and biotechnology

(a) Outline the three main steps in genetic engineering used to make a human protein. (3 marks)

(b) Human insulin consists of an A chain of 21 amino acids and a B chain of 30 amino acids. Calculate the total and give two advantages of making insulin with bacteria rather than extracting it from pigs and cattle. (4 marks)

(c) State two features that make a plasmid suitable as a vector. (3 marks)

(d) A circular plasmid is 4800 base pairs long and is cut by one restriction enzyme at 3 sites. State the number of fragments produced and explain how gel electrophoresis separates them. (4 marks)

(e) A polymerase chain reaction starts with 5 DNA molecules. How many molecules are present after 10 cycles? (3 marks)

Question 6 (17 marks) – Cloning, asexual reproduction and bioethics

(a) Compare the genetic variation produced by sexual and asexual reproduction. (3 marks)

(b) Give two advantages of tissue culture (micropropagation) in agriculture. (3 marks)

(c) A single cell divides by mitosis to form a group of 64 cells. How many divisions are needed, and how does mitosis keep the chromosome number constant? (4 marks)

(d) Explain how embryonic stem cells can be used to treat disease and state one ethical objection. (4 marks)

(e) Explain why keeping genetic diversity in crop varieties matters for resisting pests and diseases. (3 marks)

Answers and marking notes

Question 1 – DNA structure, genes and protein synthesis

(a) Two complementary polynucleotide strands are joined by hydrogen bonds and twist into a double helix (one mark); adenine pairs with thymine and guanine pairs with cytosine (one mark); the sugar-phosphate backbones lie on the outside with the bases on the inside (one mark). [3 marks]

(b) Thymine is also 22%, that is 264 bases (one mark); the remaining 672 bases are shared equally between guanine and cytosine (one mark); so there are 336 guanine bases (one mark). [3 marks]

(c) A gene is a length of DNA that codes for one polypeptide or protein (one mark); an allele is a different version of the same gene, found at the same position on homologous chromosomes (one mark); alleles may be dominant or recessive (one mark). [3 marks]

(d) Three bases form one codon (one mark); 61 of the 64 codons code for the 20 amino acids and the other 3 are stop codons (one mark); the order of the codons fixes the order of amino acids and therefore the shape and function of the protein (one mark). [3 marks]

(e) Transcription takes place in the nucleus (one mark); translation takes place on ribosomes in the cytoplasm (one mark). [2 marks]

(f) The genetic code is degenerate, so the mutated codon may still code for the same amino acid (one mark); even if an amino acid is replaced, the protein still works if the active site is not affected (one mark). [2 marks]

Question 2 – Monohybrid crosses and inheritance ratios

(a) Each parent produces B and b gametes in equal numbers (one mark); the offspring genotypes are BB, Bb and bb (one mark); the ratio is 1:2:1 (one mark). [3 marks]

(b) The genotypic ratio is 1:2:1 and both BB and Bb show the dominant phenotype (one mark); the dominant phenotype therefore makes up three quarters, which is 600 plants (one mark); the recessive phenotype bb makes up one quarter, which is 200 plants (one mark); the phenotypic ratio is therefore 3:1 (one mark). [4 marks]

(c) The gametes are B, b and b, b (one mark); the combinations give two Bb and two bb (one mark); the genotypic ratio is 1:1 (one mark); phenotypically half are dominant and half recessive, a ratio of 1:1 (one mark). [4 marks]

(d) Cross the plant with a homozygous recessive (bb) plant (one mark); if all the offspring show the dominant phenotype the parent is BB (one mark); if about half the offspring are recessive the parent is Bb (one mark). [3 marks]

(e) The probability that one child is recessive is one quarter (one mark); the two events are independent, so the probabilities are multiplied to give one sixteenth (one mark); that is 6.25%, about 6.25% (one mark). [3 marks]

Question 3 – Codominance, ABO blood groups and sex linkage

(a) In codominance both alleles are fully expressed (one mark); in incomplete dominance the offspring show an intermediate phenotype (one mark); a person with blood group AB carries both A and B antigens, so the A and B alleles are codominant (one mark). [3 marks]

(b) The father gives I-A or I-B gametes and the mother gives I-B or i gametes (one mark); the offspring genotypes I-A I-B, I-A i, I-B I-B and I-B i each occur with probability one quarter (one mark); 100 children are group AB and 100 are group A (one mark); group B totals 200 children, one quarter plus one quarter (one mark). [4 marks]

(c) The allele for red-green colour blindness is carried on the X chromosome (one mark); a male has only one X chromosome, so one copy of the allele is enough to cause the condition (one mark); a female has two X chromosomes and must inherit the recessive allele on both to be affected (one mark). [3 marks]

(d) Each son has an equal chance of receiving XC or Xc from his mother (one mark); so 130 of the 260 sons are colour blind (one mark); every daughter receives XC from her father and is therefore not colour blind (one mark); half of the daughters are carriers (XCXc) (one mark). [4 marks]

(e) A son inherits his Y chromosome from his father (one mark); all of a son X chromosomes come from his mother (one mark); so an allele on the father X chromosome can only be passed to a daughter (one mark). [3 marks]

Question 4 – Mutation, variation and natural selection

(a) A mutation is a change in the base sequence of DNA (one mark); ionising radiation such as X-rays or ultraviolet light (one mark); chemical mutagens such as tobacco tar or certain drugs (one mark). [3 marks]

(b) A mutation changes the amino acid sequence, so an enzyme or structural protein may lose its function (one mark); random changes usually reduce how well the organism is adapted (one mark); a recessive allele is masked by a normal dominant allele and so is not expressed at once (one mark). [3 marks]

(c) Random mutation produces a resistance allele (one mark); the antibiotic kills individuals without resistance (one mark); resistant individuals survive and reproduce, passing the allele to their offspring (one mark); the frequency of the resistance allele rises from generation to generation (one mark). [4 marks]

(d) There are 60 resistant insects (one mark); of the 440 non-resistant insects, 176 survive (one mark); the next generation therefore starts with 236 insects (one mark). [3 marks]

(e) The pesticide imposes strong selection so that only resistant individuals reproduce (one mark); resistance alleles accumulate quickly and the population becomes mainly resistant (one mark); rotate pesticides with different modes of action and combine them with biological control (one mark). [3 marks]

Question 5 – Genetic engineering and biotechnology

(a) A restriction enzyme cuts out the required gene (one mark); DNA ligase inserts the gene into a plasmid vector (one mark); the vector is put into host cells such as bacteria and the cells that took it up are selected (one mark). [3 marks]

(b) The total is 51 amino acids (one mark); bacteria produce it quickly in large amounts so supply is reliable (one mark); the product is pure with exactly the human sequence and carries no animal pathogens (one mark); production does not depend on the number of animals slaughtered (one mark). [4 marks]

(c) It has a recognition site that a restriction enzyme can cut (one mark); it carries a marker gene such as an antibiotic-resistance gene so transformed cells can be selected (one mark); it can replicate on its own so the gene is copied many times inside the host cell (one mark). [3 marks]

(d) Each cut in a circular DNA molecule adds one fragment, so 3 cuts give 3 fragments (one mark); DNA is negatively charged and moves towards the positive electrode (one mark); shorter fragments move faster and longer ones move more slowly (one mark); fragments of different length therefore stop at different positions and form separate bands (one mark). [4 marks]

(e) Each cycle doubles the number of DNA molecules (one mark); after 10 cycles each starting molecule has become 2 to the power 10, which is 1024 molecules (one mark); starting with 5 molecules gives 5120 molecules in total (one mark). [3 marks]

Question 6 – Cloning, asexual reproduction and bioethics

(a) Sexual reproduction involves meiosis and fertilisation, so each offspring has a different combination of alleles (one mark); asexual reproduction such as vegetative propagation or tissue culture produces offspring genetically identical to the parent (one mark); in asexual reproduction variation comes only from mutation (one mark). [3 marks]

(b) Large numbers of plants can be produced in a short time (one mark); the offspring keep the desirable features of the parent and are uniform (one mark); virus-free plants can be produced without being limited by seed supply or season (one mark). [3 marks]

(c) The number of cells doubles at each division and 2 to the power 6 is 64 (one mark); so 6 divisions are needed (one mark); the chromosomes are copied before division (one mark); the copies are shared equally between the two daughter cells, so each has the same chromosome number as the parent (one mark). [4 marks]

(d) Stem cells are undifferentiated and can develop into many cell types (one mark); introducing them into damaged tissue replaces dead or faulty cells (one mark); embryonic stem cells are taken from an early embryo, which is destroyed in the process (one mark); this raises the ethical objection about the status of the embryo and requires strict regulation and informed consent (one mark). [4 marks]

(e) Different varieties carry different alleles for resistance (one mark); when a single variety is grown over large areas one disease can destroy the whole crop (one mark); genetic diversity is the raw material for breeding and natural selection, allowing new resistant varieties to be produced (one mark). [3 marks]

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