Edexcel International GCSE Biology 4BI1: Inheritance – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sub-topic 3(b), Inheritance (specification points 3.14–3.39), of the Pearson Edexcel International GCSE Biology (4BI1) specification, Issue 3. The qualification is untiered. Questions labelled (Paper 2 only) test the bold B-referenced points (3.16B–3.18B, 3.21B, 3.35B–3.37B); the rest can appear on either paper. They apply to the June and November series examined under Issue 3.

For whole-topic questions, including reproduction, see the Topic 3 practice questions; none is repeated here. Links to the notes, hub and checklist are under Next steps.

Questions

1. State what is meant by each term.

(a) allele [1] (b) heterozygous [1] (c) phenotype [1]

2. (Paper 2 only) One strand of part of a gene has the base sequence TAC CTG AAG CGT.

(a) Write the base sequence of the complementary DNA strand. [1] (b) The strand given above is used as the template in transcription. Write the mRNA sequence. [1] (c) State how many amino acids this section of mRNA codes for. [1] (d) Give the anticodon of the tRNA that pairs with the first codon. [1]

3. (Paper 2 only) Describe how the information in a gene is used to make a protein. [6]

4. In tomato plants, the allele for red fruit (R) is dominant to the allele for yellow fruit (r). A heterozygous red-fruited plant is crossed with a yellow-fruited plant.

(a) Draw a genetic diagram to show the possible genotypes and phenotypes of the offspring. [3] (b) State the probability that an offspring plant has yellow fruit. [1] (c) The cross produces 240 seeds. Calculate how many would be expected to grow into yellow-fruited plants. [1]

5. In fruit flies, the allele for normal wings (N) is dominant to the allele for vestigial wings (n). Describe how a breeding experiment could show whether a normal-winged fly is homozygous or heterozygous. [4]

6. A family pedigree shows a condition controlled by one gene.

• Generation I: I-1, an affected male, and I-2, an affected female, are the parents.
• Generation II: their children are II-1, an affected son; II-2, an unaffected daughter; II-3, an affected daughter.

(a) Explain the evidence in this pedigree that the allele causing the condition is dominant. [2] (b) Using D for the dominant allele, give the genotypes of I-1 and I-2. Explain your answer. [2] (c) State the probability that the next child of I-1 and I-2 is unaffected. [1] (d) II-2 has children with an unaffected man. State the probability that their child is affected. [1]

7. (Paper 2 only) Human ABO blood group is controlled by one gene with three alleles. Alleles IA and IB are codominant. Allele IO is recessive to both.

(a) Explain what is meant by codominance, using blood group AB as your example. [1] (b) A woman with genotype IA IO and a man with genotype IB IO have a child. Draw a genetic diagram to show the possible genotypes and blood groups of the child. [3] (c) State the probability that the child has blood group O. [1]

8. Sex determination in humans.

(a) State the sex chromosomes in a male. [1] (b) Explain why the father’s gamete determines the sex of a child. [2] (c) A couple have three daughters. They say their next child is “more likely to be a boy”. Explain why they are wrong. [2]

9. A mammal has a diploid number of 38.

(a) State the number of chromosomes in a skin cell produced by mitosis. [1] (b) State the number of chromosomes in one of its sperm cells. [1] (c) Explain why gametes must be haploid. [2] (d) A gardener takes cuttings from one plant. Explain why all the new plants are genetically identical. [2]

10. Differences in adult height (cm) within pairs of twins raised together:

Pair	1	2	3	4	5
Identical twins	1.8	2.4	1.2	2.0	1.6
Non-identical twins	4.6	5.8	3.2	6.1	4.3

The mean difference for identical twins is 1.8 cm.

(a) Calculate the mean difference for non-identical twins. [2] (b) Explain what these results suggest about the effect of genes on height. [2] (c) Suggest why identical twins still differ slightly in height. [1] (d) Explain why human height shows a wide range of values rather than a few distinct heights. [2]

11. (Paper 2 only) Mutation.

(a) Explain how a change in one base of a gene could change an organism’s phenotype. [3] (b) Suggest one reason why most mutations have no effect on the phenotype. [1] (c) Name one type of radiation and one chemical source that increase the rate of mutation. [2]

12. Most beetles of a bark-living species are pale; a few are dark. Over many years the bark in one area darkened and the proportion of dark beetles there rose.

(a) Use Darwin’s theory of natural selection to explain the increase in dark beetles. [5] (b) A hospital tested 300 bacterial samples each year. In year 1, 18 samples were resistant to an antibiotic. In year 6, 81 samples were resistant. Calculate the percentage of resistant samples in each year. [2] (c) Explain why this change could make infections difficult to control. [2]

Answers

1. (a) An alternative form of a gene. [1] (b) Having two different alleles of a gene. [1] (c) The characteristic that is expressed / the observable feature of an organism. [1] Examiner insight: One-mark definitions need the precise idea: “a type of gene” or “a different gene” for allele is not credited.

2. (a) ATG GAC TTC GCA [1] (b) AUG GAC UUC GCA [1] (c) 12 bases ÷ 3 = 4 amino acids [1] (d) UAC [1] Examiner insight: Any T in an mRNA or tRNA sequence loses the mark, even if every other base is correct.

3. Six from these marking points:

• In transcription, the DNA strands of the gene separate (in the nucleus). [1]
• mRNA is formed by complementary base pairing with one strand, with U in place of T. [1]
• mRNA leaves the nucleus and attaches to a ribosome. [1]
• In translation, the mRNA is read in codons of three bases. [1]
• tRNA molecules carry specific amino acids and have anticodons that pair with the codons. [1]
• Amino acids are joined in the order set by the codons, forming the protein chain. [1]

Examiner insight: Six separate points are needed; naming “transcription” and “translation” without saying what happens earns little.

4. (a) Parents’ genotypes Rr × rr [1]; gametes R, r and r [1]; offspring Rr (red) and rr (yellow), shown in a Punnett square with phenotypes labelled [1].

              r       r
   R         Rr      Rr
   r         rr      rr

(b) 1/2 (0.5 or 50%) [1] (c) 240 × 1/2 = 120 [1] Examiner insight: Allow error carried forward in (c) from a wrong probability in (b), but only if the working is shown.

5. Cross the normal-winged fly with a vestigial-winged fly (nn). [1] If any offspring have vestigial wings, the fly is heterozygous (Nn). [1] If all offspring have normal wings, the fly is probably homozygous (NN). [1] A large number of offspring is needed so the result is reliable. [1] Examiner insight: The cross must be with the homozygous recessive; another normal-winged fly earns no first mark.

6. (a) Both parents (I-1 and I-2) are affected [1] but their child II-2 is unaffected, so each parent carries a hidden unaffected allele; the condition allele must be dominant. [1] (b) Both are Dd. [1] II-2 is unaffected, so she is dd and received a d from each parent. [1] (c) Dd × Dd gives 1 DD : 2 Dd : 1 dd, so 1/4. [1] (d) dd × dd gives only dd, so 0. [1] Examiner insight: Name the individuals used as evidence; “it skips no generation” is not enough.

7. (a) In genotype IA IB both alleles are expressed in the phenotype, giving blood group AB. [1] (b) Gametes IA, IO and IB, IO [1]; four offspring genotypes IA IB, IA IO, IB IO, IO IO [1]; blood groups AB, A, B, O correctly matched [1].

              IB        IO
   IA       IA IB     IA IO
   IO       IB IO     IO IO

(c) 1/4 [1] Examiner insight: The gametes must be shown separately; a Punnett square with no gametes stated usually loses the gamete mark.

8. (a) XY [1] (b) All eggs carry an X chromosome. [1] Half of sperm carry X and half carry Y, so which sperm fertilises the egg decides the sex. [1] (c) The probability of a boy is 1/2 at each fertilisation. [1] Each fertilisation is independent, so earlier daughters do not change it. [1] Examiner insight: “50 : 50” alone does not explain independence; link the chance to each fertilisation.

9. (a) 38 [1] (b) 38 ÷ 2 = 19 [1] (c) At fertilisation two gametes fuse. [1] If each is haploid, the zygote is diploid, so the chromosome number stays the same each generation. [1] (d) Cuttings grow by mitosis (asexual reproduction). [1] Mitosis gives cells with identical sets of chromosomes, so every new plant has the parent’s genes. [1] Examiner insight: In (c), “so the number doesn’t double” needs to be linked to fusion of gametes at fertilisation to earn both marks.

10. (a) (4.6 + 5.8 + 3.2 + 6.1 + 4.3) ÷ 5 [1] = 24.0 ÷ 5 = 4.8 cm [1] (b) Identical twins have a smaller mean difference (1.8 cm) than non-identical twins (4.8 cm). [1] Identical twins have identical genes, so genes have a large effect on height. [1] (c) Environmental differences, such as diet. [1] (d) Height is controlled by several genes (polygenic inheritance). [1] Many alleles add their effects, plus environment, giving a range. [1] Examiner insight: Quote both means with units; “smaller for identical twins” alone usually earns one mark.

11. (a) The base sequence (a codon) changes. [1] A different amino acid may be placed in the protein. [1] The protein may have a different shape and not work, changing the phenotype. [1] (b) The changed codon may still code for the same amino acid / the protein may still work normally. [1] (c) Radiation: gamma rays, X-rays or ultraviolet rays. [1] Chemical: chemicals in tobacco. [1] Examiner insight: “It changes the gene” with no mention of amino acids misses the central mark.

12. (a) Beetles varied in colour because of different alleles. [1] On darker bark, pale beetles were easier for predators to see and more were eaten. [1] Dark beetles were more likely to survive and reproduce. [1] They passed on the alleles for dark colour to their offspring. [1] Over many generations, the proportion of dark beetles increased. [1] (b) Year 1: 18 ÷ 300 × 100 = 6% [1]; year 6: 81 ÷ 300 × 100 = 27% [1] (c) The antibiotic no longer kills the resistant bacteria. [1] The bacteria keep multiplying, so the infection continues or spreads. [1] Examiner insight: “Beetles turned dark because they needed to” scores nothing; variation must exist before selection.

Where marks are usually lost

• Gametes left out of genetic diagrams, or written as two alleles together.
• Answering a probability question with a ratio.
• Pedigree answers that do not name the individuals used as evidence.
• (Paper 2) T written in mRNA or anticodons; codon and anticodon swapped.
• Saying environment causes a mutation that the organism “needs”.

Next steps

• Review the Inheritance revision notes
• Re-read the Inheritance study guide
• Back to the Edexcel IGCSE Biology hub
• Tick off points on the printable checklist
• Browse all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Biology (4BI1) specification, Issue 3 (September 2024), Pearson Education Limited; first assessment June 2019. Topic 3, Reproduction and inheritance, sub-topic (b) Inheritance, points 3.14–3.39.