ESAT Biology: Genetics, DNA and inheritance, set 1

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• Aim to spend about 90 seconds on each question.

1. 11 mark

   A student examines a stained smear of human blood cells under a microscope and notices that mature red blood cells (erythrocytes) appear to have no nucleus. Which statement is correct?

   1. A Mature red blood cells still contain genetic material in their cytoplasm, since DNA is not confined to the nucleus in eukaryotic cells.
   2. B Mature red blood cells have lost their nucleus during development, so they contain no genetic material and cannot carry out protein synthesis from a nuclear template.
   3. C Mature red blood cells never had a nucleus, because red blood cells are not eukaryotic cells.
   4. D Mature red blood cells still contain a nucleus, but it is too small to be seen under a light microscope.
2. 21 mark

   In pea plants, the allele for tall stem (T) is dominant over the allele for short stem (t). A plant has the genotype Tt. Which of these correctly describes this plant?

   1. A Homozygous dominant, with a tall phenotype.
   2. B Heterozygous, with a short phenotype.
   3. C Homozygous recessive, with a short phenotype.
   4. D Heterozygous, with a tall phenotype.
3. 31 mark

   In pea plants, tall (T) is dominant to short (t). A gardener crosses a tall plant of unknown genotype with a homozygous recessive short plant (tt) to find out whether the tall plant is TT or Tt. Of the offspring produced, 8 are tall and 9 are short. What is the most likely genotype of the original tall plant?

   1. A Tt, since a heterozygous parent crossed with tt is expected to produce tall and short offspring in a ratio close to 1:1, matching the observed 8:9 split.
   2. B TT, since a homozygous dominant parent crossed with tt can still occasionally produce short offspring by chance.
   3. C tt, since the presence of short offspring shows the tall parent must carry two copies of the recessive allele.
   4. D The cross cannot determine the unknown genotype, because both TT and Tt parents can produce some short offspring when crossed with tt.
4. 41 mark

   Cystic fibrosis is an inherited condition caused by a recessive allele (f) of a gene on an autosome; the dominant allele (F) does not cause the condition. Two unaffected parents have a child with cystic fibrosis. What is the probability that their next child will also have cystic fibrosis?

   1. A 0, because both parents are unaffected and so cannot pass on the condition.
   2. B 1/4
   3. C 3/4
   4. D 1/2
5. 51 mark

   Which of the following best defines the term 'genome'?

   1. A The complete set of genetic material (DNA) of an organism, contained within its chromosomes.
   2. B The complete set of proteins produced by an organism's cells.
   3. C A single chromosome that carries all the genes needed for one characteristic.
   4. D The complete set of alleles present across a population of one species.
6. 61 mark

   A sample of double-stranded DNA is analysed and found to contain 30 per cent adenine bases. What percentage of the bases in this sample are guanine?

   1. A 30 per cent
   2. B 20 per cent
   3. C 40 per cent
   4. D 70 per cent
7. 71 mark

   Which statement correctly describes the structure of a single DNA nucleotide?

   1. A A nucleotide is a double helix made of two nucleotide strands held together by hydrogen bonds.
   2. B A nucleotide consists of a sugar and a phosphate group, plus one of four nitrogenous bases.
   3. C A nucleotide consists of a phosphate group and a nitrogenous base only, with no sugar component.
   4. D A nucleotide consists of two sugar groups joined to a single phosphate group and a nitrogenous base.
8. 81 mark

   A section of a gene contains 300 nucleotide bases on the coding strand, ignoring any start or stop signals. How many amino acids could this section code for?

   1. A 300
   2. B 900
   3. C 100
   4. D 150
9. 91 mark

   Which statement best explains why changing the sequence of amino acids in a polypeptide can change how the resulting protein functions?

   1. A The sequence of amino acids determines the DNA base sequence of the gene, which then determines the protein's function.
   2. B The order of amino acids has no effect on protein function; only the total number of amino acids matters.
   3. C Amino acids only affect a protein's colour, not its shape or function.
   4. D The sequence of amino acids determines the protein's three-dimensional shape, and a protein's shape determines how it functions.
10. 101 mark

    Which statement about gene mutations is correct?

    1. A Every mutation that occurs in an organism's DNA changes its phenotype, since any change to the nucleotide sequence must alter the resulting protein.
    2. B Most gene mutations have no effect on phenotype, some have a small effect, and occasionally a mutation determines the phenotype.
    3. C Gene mutations only ever occur in reproductive cells, so mutations in other body cells cannot affect an organism at all.
    4. D A mutation always changes the number of chromosomes in a cell, altering how genes are inherited.
11. 111 mark

    In a genetic engineering procedure, a gene for human insulin is inserted into a bacterial plasmid so that bacteria can produce insulin. Which pair of enzymes is used, and what is the role of each?

    1. A Ligase cuts the DNA at specific sequences to remove the gene and open the plasmid, and a restriction enzyme joins the gene into the cut plasmid.
    2. B Restriction enzymes join the gene into the plasmid, and ligase is only used afterwards to help the bacteria take up the plasmid.
    3. C Restriction enzymes cut the DNA at specific sequences to remove the gene and open the plasmid, and ligase joins the gene into the cut plasmid.
    4. D A single restriction enzyme both cuts out the gene and reseals the plasmid, so no other enzyme is required.
12. 121 mark

    Which of these correctly ranks totipotent, pluripotent and multipotent stem cells by how many different cell types each can differentiate into, from greatest to least?

    1. A Totipotent, then pluripotent, then multipotent.
    2. B Multipotent, then pluripotent, then totipotent.
    3. C Totipotent, then multipotent, then pluripotent.
    4. D Pluripotent, then totipotent, then multipotent.
13. 131 mark

    Which statement correctly distinguishes selective breeding from natural selection?

    1. A Selective breeding and natural selection are the same process, since both result in populations becoming better adapted over time.
    2. B In selective breeding, humans choose which organisms reproduce based on desired characteristics, whereas in natural selection the environment determines which organisms survive and reproduce.
    3. C In natural selection, humans choose parents with desirable characteristics to breed, whereas selective breeding happens without any human input.
    4. D Selective breeding always increases genetic variation within a population, while natural selection always reduces it.
14. 141 mark

    A population of bacteria is exposed to an antibiotic. Most bacteria are killed, but a small number survive because they already carried a mutation giving them antibiotic resistance. These survivors reproduce, and the resulting population is mostly resistant to the antibiotic. Which of these best explains this outcome in terms of natural selection?

    1. A The antibiotic caused the surviving bacteria to develop resistance in response to the treatment, and they then passed this newly acquired resistance to their offspring.
    2. B The antibiotic changed the DNA of every bacterium in the population in the same way, making them all equally resistant.
    3. C Resistant bacteria already existed in the population due to random mutation before the antibiotic was applied; the antibiotic then killed the non-resistant bacteria, so resistant bacteria became more common in the surviving, reproducing population.
    4. D Resistant bacteria became more common only because they reproduce faster than non-resistant bacteria, regardless of whether the antibiotic was present.
15. 151 mark

    A pair of genetically identical twins is raised in different households. One twin takes up competitive swimming from a young age and develops noticeably larger shoulder muscles than the other twin. Which type of variation does this difference in muscle size best illustrate?

    1. A Genetic variation only, since any physical difference between individuals must be caused by a difference in their genes.
    2. B An equal combination of genetic and environmental variation, because muscle size in humans is always exactly 50 per cent determined by genes and 50 per cent by environment.
    3. C A new mutation that occurred in one twin's muscle cells after birth, changing that twin's genotype for muscle growth.
    4. D Environmental variation, since the twins have identical genes but differ in this characteristic because of a difference in their environment.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: B

   1. In eukaryotic cells the nucleus is the site of the genetic material (spec B4.1).
   2. Human red blood cells develop from nucleated precursor cells but expel their nucleus during maturation, leaving more space for haemoglobin.
   3. Because the nucleus and its DNA are lost, a mature red blood cell has no genetic material and cannot transcribe genes from a nuclear template.
   4. So the correct statement is B: the nucleus is lost and no genetic material remains.
   • Why not A: Confuses the small amount of mitochondrial DNA present outside the nucleus with the main genetic material, which the specification places in the nucleus.
   • Why not C: Assumes that losing a nucleus means a cell was never eukaryotic, when eukaryotic classification is a property of the organism, not of one specialised cell.
   • Why not D: Treats the missing nucleus as a resolution problem with the microscope rather than a genuine feature of a mature red blood cell.

2. Question 2Answer: D

   1. Genotype Tt has two different alleles of the same gene, so the plant is heterozygous, not homozygous (spec B4.2).
   2. Homozygous means both alleles are identical (TT or tt); heterozygous means they differ.
   3. Because T is dominant, its effect is expressed whenever at least one copy is present, so the phenotype of Tt is tall.
   4. So Tt is heterozygous with a tall phenotype, answer D.
   • Why not A: Mistakes carrying two different alleles for homozygous dominant; homozygous means both alleles are identical copies, which Tt is not.
   • Why not B: Correctly identifies the genotype as heterozygous but wrongly assumes the recessive allele is expressed whenever it is present, ignoring that the dominant allele always determines the phenotype in a heterozygote.
   • Why not C: Confuses having one recessive allele present with being homozygous recessive, which requires two copies of the recessive allele (tt).

3. Question 3Answer: A

   1. A TT parent crossed with tt can only produce Tt offspring, all of which are tall, so no short offspring at all are expected from that cross.
   2. A Tt parent crossed with tt produces Tt and tt offspring in a 1:1 ratio, so about half the offspring are expected to be short.
   3. The observed 8 tall to 9 short (17 offspring) is close to a 1:1 split and is inconsistent with a TT parent, which predicts zero short offspring.
   4. So the tall parent is most likely Tt, answer A.
   • Why not B: Assumes a TT parent could still occasionally give short offspring 'by chance', missing that T is fully dominant, so a TT x tt cross gives 100 percent tall offspring with no exceptions.
   • Why not C: Conflates carrying one hidden recessive allele (heterozygous, Tt) with carrying two recessive alleles (homozygous recessive, tt); a tt plant would itself be short, contradicting the stated tall phenotype of the parent.
   • Why not D: Denies that the ratios distinguish the genotypes, missing that the expected proportion of short offspring differs sharply between TT (0 percent) and Tt (about 50 percent), and the observed near 1:1 split matches only Tt.

4. Question 4Answer: B

   1. Since the child has cystic fibrosis (genotype ff), each parent must have supplied one f allele, so both unaffected parents must be carriers, Ff (spec B4.2, B4.3).
   2. A Punnett square for Ff x Ff gives genotypes FF, Ff, Ff and ff in equal proportions: 1 FF : 2 Ff : 1 ff.
   3. Only the ff genotype shows the condition, since F is dominant over f, and ff makes up 1 of these 4 equally likely outcomes.
   4. So the probability that the next child has cystic fibrosis is 1/4, answer B.
   • Why not A: Assumes phenotype directly reveals genotype, forgetting that an unaffected person can still be a heterozygous carrier of a recessive allele.
   • Why not C: Correctly works out that 3/4 of the offspring are unaffected but then reports that fraction instead of the fraction that is affected.
   • Why not D: Uses the 1:1 ratio that applies to a test cross of a carrier with an affected person (Ff x ff), not to a cross between two carriers (Ff x Ff).

5. Question 5Answer: A

   1. Spec B5.1 defines the genome as the full set of genetic material (DNA) of an organism.
   2. This DNA is organised into, and contained within, the organism's chromosomes.
   3. The genome is the DNA itself, not the proteins it encodes, and it is not restricted to a single chromosome.
   4. It also describes one organism rather than a whole population's alleles, so A is correct.
   • Why not B: Confuses the genetic material with the proteins it encodes, describing the proteome rather than the genome.
   • Why not C: Conflates the whole genome with a single chromosome, when the genome is the full set of DNA spread across all of an organism's chromosomes.
   • Why not D: Confuses an individual organism's genome with the 'gene pool' of alleles found across a whole population.

6. Question 6Answer: B

   1. In double-stranded DNA, adenine pairs with thymine and guanine pairs with cytosine, so the amount of A equals the amount of T, and the amount of G equals the amount of C (spec B5.2).
   2. If A is 30 per cent, then T must also be 30 per cent, so A and T together make up 60 per cent of all bases.
   3. The remaining 40 per cent of bases must be G and C combined, and since G equals C, each makes up half of that 40 per cent.
   4. So G is 20 per cent, answer B.
   • Why not A: Assumes guanine must equal adenine directly, extending the 'A equals T' pairing rule to guanine instead of pairing it correctly with cytosine.
   • Why not C: Correctly finds that guanine plus cytosine make up 40 per cent of the bases but forgets to halve this, since guanine and cytosine are present in equal amounts.
   • Why not D: Subtracts adenine's percentage directly from 100 per cent, ignoring that thymine also makes up 30 per cent, leaving only 40 per cent for guanine and cytosine combined.

7. Question 7Answer: B

   1. Spec B5.2 states that each nucleotide consists of a common sugar and phosphate group, plus one of four nitrogenous bases.
   2. A double helix (option A) describes the assembled double-stranded molecule made of many nucleotides on two strands, not one nucleotide.
   3. Removing the sugar (option C) or duplicating it (option D) leaves an incomplete or incorrect structure.
   4. So the correct description of a single nucleotide is B: a sugar, a phosphate group and one of four bases.
   • Why not A: Describes the assembled double helix, built from many nucleotides on two strands, rather than a single nucleotide monomer.
   • Why not C: Omits the sugar group entirely, leaving only two of the three parts a nucleotide must have.
   • Why not D: Mistakenly includes two sugar units in a single nucleotide instead of one.

8. Question 8Answer: C

   1. Spec B5.3 states that nucleotide bases in a gene are read as triplets, with each triplet of three bases coding for one amino acid.
   2. To find the number of amino acids coded for, divide the number of bases by 3.
   3. 300 divided by 3 is 100, so this section of the gene could code for a chain of 100 amino acids.
   4. The answer is C, 100 amino acids.
   • Why not A: Ignores that three bases are needed per amino acid, treating each single base as if it coded for one amino acid.
   • Why not B: Multiplies the base count by 3 instead of dividing by 3, inverting the triplet relationship between bases and amino acids.
   • Why not D: Divides by 2 rather than 3, confusing the two strands of double-stranded DNA with the three bases that make up one triplet.

9. Question 9Answer: D

   1. Spec B5.3 states that the sequence of amino acids in a polypeptide determines the protein's three-dimensional shape.
   2. A protein's shape in turn determines how it functions, for example how well an enzyme's active site fits its substrate.
   3. Changing the order of amino acids can therefore change the shape and disrupt or alter that function.
   4. So D correctly links sequence, shape and function; information flows from gene to amino acid sequence, not the reverse.
   • Why not A: Reverses the direction of genetic information flow; it is the gene's base sequence that determines the amino acid sequence, not the other way round.
   • Why not B: Assumes that only the number of amino acids matters, ignoring that their order is what specifies the resulting shape.
   • Why not C: Invents an irrelevant property (colour) instead of the shape-function link the specification describes.

10. Question 10Answer: B

    1. Spec B5.4 states that a mutation changes the sequence of nucleotides in the DNA.
    2. It also states that most mutations have no effect on the phenotype, some have a small effect, and occasionally a mutation determines the phenotype.
    3. This range of outcomes rules out claims that every mutation changes phenotype, and mutation is a different process from a change in chromosome number.
    4. Mutations can also arise in body cells, not only reproductive cells, so the correct statement is B.
    • Why not A: Assumes any DNA sequence change must alter the phenotype, missing that many changes fall in non-critical positions and have no visible effect.
    • Why not C: Wrongly restricts mutation to reproductive (germline) cells only, when mutations can also arise in body (somatic) cells and can still affect that organism.
    • Why not D: Confuses a gene mutation, a change within the DNA sequence of a gene, with a chromosome mutation, a change in chromosome number or structure.

11. Question 11Answer: C

    1. Spec B6.1 describes taking a copy of a gene from one organism's DNA and inserting it into another organism's DNA, naming restriction enzymes and ligases as the enzymes involved.
    2. Restriction enzymes recognise specific DNA sequences and cut the DNA there, both to remove the gene of interest and to open up the plasmid to receive it.
    3. DNA ligase then joins the cut ends of the gene and the plasmid together, sealing the new combination of DNA.
    4. So C correctly assigns cutting to restriction enzymes and joining to ligase.
    • Why not A: Swaps the roles of the two enzymes; restriction enzymes cut the DNA and ligase joins it, not the reverse.
    • Why not B: Assigns restriction enzymes the joining role, which belongs to ligase, and invents an unrelated uptake function for ligase.
    • Why not D: Assumes a single enzyme performs both the cutting and joining steps, missing that genetic engineering needs two distinct enzymes for these two distinct roles.

12. Question 12Answer: A

    1. Spec B6.2 states totipotent cells, found in very early embryos, have the potential to develop into a complete multicellular organism, the widest possible potential.
    2. Most embryonic stem cells are pluripotent, able to differentiate into any cell type of the body but not into a whole new organism on their own.
    3. Adult stem cells are multipotent, differentiating into only a limited number of related cell types.
    4. So the order from greatest to least potential is totipotent, then pluripotent, then multipotent, answer A.
    • Why not B: Inverts the entire ranking, placing the most restricted cell type, multipotent, at the top.
    • Why not C: Correctly places totipotent first but swaps the remaining two, forgetting that pluripotent cells can become any body cell type while multipotent cells are limited to a smaller, related range.
    • Why not D: Ranks pluripotent above totipotent, forgetting that only totipotent cells can form a complete organism, which is the widest potential of all.

13. Question 13Answer: B

    1. Spec B6.3 asks for the differences and similarities between natural selection and selective breeding.
    2. In selective breeding, humans deliberately choose which organisms with desirable characteristics are allowed to breed.
    3. In natural selection, no human chooses; organisms whose characteristics best suit the environment are more likely to survive and reproduce.
    4. So B correctly identifies human choice as the distinguishing feature of selective breeding, versus environmental pressure in natural selection.
    • Why not A: Fails to recognise the key difference: selective breeding is directed by human choice of desirable traits, while natural selection is driven by the environment, not that the two processes are identical.
    • Why not C: Swaps the two definitions, describing natural selection as human-directed and selective breeding as happening without human input.
    • Why not D: Invents an unsupported general rule about variation; selective breeding typically reduces genetic variation by repeatedly choosing similar parents, the opposite of what this option claims.

14. Question 14Answer: C

    1. Specs B7.1 and B7.2 describe extensive genetic variation existing within a population before selection acts; here, variation in antibiotic resistance arose from a random mutation.
    2. When the antibiotic is applied, bacteria without the resistance mutation are killed, while resistant bacteria survive because their phenotype suits this new environment.
    3. The surviving resistant bacteria reproduce, passing on the resistance allele, so the population's inherited characteristics change over time, which is evolution by natural selection.
    4. This matches option C; the antibiotic does not create resistance, and selection, not reproduction rate alone, explains why resistant bacteria became common.
    • Why not A: Assumes the antibiotic caused resistance to appear in response to exposure, an acquired characteristic, rather than recognising that resistance already existed as variation before selection acted.
    • Why not B: Assumes the antibiotic acts uniformly on every bacterium's DNA, missing that natural selection depends on existing variation between individuals and differential survival, not a shared change.
    • Why not D: Removes the antibiotic's selective role altogether, attributing the shift only to a general reproductive advantage rather than to survival under a specific selection pressure.

15. Question 15Answer: D

    1. Spec B7.2 states variation can be genetic/inherited or environmental, and both can affect a range of phenotypes.
    2. Genetically identical twins share the same genotype, so any difference between them cannot be due to genetic variation.
    3. The difference here, training as a competitive swimmer, is a difference in environment, which has produced a difference in phenotype (muscle size).
    4. So this is an example of environmental variation, answer D.
    • Why not A: Assumes all phenotypic variation must be genetic, overlooking that these twins share an identical genotype, so the difference cannot be explained by genetics.
    • Why not B: Invents a fixed universal ratio between genetic and environmental contributions, which the specification does not state and which is not fixed for every characteristic.
    • Why not C: Invents a mutation as an explanation instead of recognising a straightforward environmental cause, when no evidence of a mutation is given.