Pearson Edexcel International A Level Biology: Stem Cells, Gene Expression and Phenotype (YBI11) – 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 specification statements 3.17 to 3.21 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2: Topic 3, part of the International AS (IAS) content. Revise first with the study guide and revision notes. Course pages: Edexcel A Level Biology hub and printable checklist. The diagnostics show where to start.

Questions

1. State what is meant by the terms totipotent and pluripotent. [2]

2. Describe the structure of a blastocyst and identify the part that is the source of embryonic stem cells. [3]

3. A culture starts with 2000 pluripotent cells. They divide every 12 hours and none die. Calculate the number of cells after 60 hours. [2]

4. Discuss the use of embryonic stem cells compared with adult stem cells in medical therapies. [6]

5. A beta cell in the pancreas makes insulin. A muscle cell from the same person does not. Explain why. [4]

6. The table shows the CpG sites methylated at the promoter of one gene (out of 40 sites) and the relative mRNA from the gene.

Tissue	Sites methylated	Relative mRNA
Liver	6	100
Brain	20	41
Skin	34	9

(a) Calculate the percentage of sites methylated in skin. [1] (b) Describe the relationship shown. [2] (c) Explain how methylation could produce this relationship. [2]

7. A gene has four exons of 63, 48, 75 and 54 nucleotides, with the stop codon in exon 4. Exons 1 and 4 are always kept. Exons 2 and 3 are each kept or removed independently. (a) State the number of different mRNAs possible. [1] (b) Calculate the number of amino acids in the shortest and the longest polypeptide. [3] (c) Explain how one gene can give different polypeptides. [2]

8. Explain how histone acetylation increases the expression of a gene, and how a DNA methylation pattern can be kept by both daughter cells after mitosis. [5]

9. Genetically identical seedlings were grown with low or high nitrogen supply. Dry masses (g) after six weeks: low nitrogen 3.2, 3.6, 3.1, 3.5; high nitrogen 5.4, 5.9, 5.6, 5.5. (a) Calculate the mean mass at low nitrogen. [1] (b) Calculate the percentage increase in mean mass at high nitrogen, to 3 significant figures. [2] (c) Explain the difference between the groups and why the masses within one group still vary. [2]

10. A woman with blood group A and a man with blood group B have a child with group O. (a) State the genotypes of both parents. [2] (b) Calculate the probability that their next child has group AB. [2] (c) Explain why ABO inheritance is an example of multiple alleles. [1]

11. A plant trait is controlled by three unlinked genes, each with an additive allele (capital) and a non-additive allele (small). Two plants of genotype AaBbCc are crossed. (a) State the number of genetically different phenotype classes in the offspring. [1] (b) Calculate the probability of an offspring with all six capital alleles. [1] (c) Calculate the probability of an offspring with five or six capital alleles. [2] (d) Explain why the trait in a population shows continuous variation. [4]

12. Coat colour in rabbits is controlled by one gene with the dominance series C (full colour) > cch (chinchilla) > ch (Himalayan) > c (albino). A Himalayan rabbit has dark fur only on its ears, nose, feet and tail. (a) Two chinchilla rabbits of genotype cch ch are crossed. Give the probability that an offspring is Himalayan. [2] (b) The enzyme that makes dark pigment is active only at lower temperatures. Explain the pattern of dark fur. [2] (c) A Himalayan rabbit is kept in a warm room and its fur regrows. Predict and explain the change. [2]

Answers

1. Totipotent: can differentiate into any cell type, including extra-embryonic tissue such as the placenta, so can form a whole organism [1]. Pluripotent: can differentiate into any body cell type but not the extra-embryonic tissue [1]. Examiner insight: the placenta or extra-embryonic point is what separates the two terms, so a definition that only says “can become many cells” tends to gain little.

2. A hollow ball of cells with a fluid-filled cavity [1]. An outer layer, the trophoblast, which goes on to form part of the placenta [1]. An inner cell mass at one side of the cavity, which is the source of embryonic stem cells [1]. Examiner insight: name the layer and say what it becomes. Naming the inner cell mass but not stating it is the stem cell source leaves the question unanswered.

3. 60 ÷ 12 = 5 doublings [1]. 2000 × 2⁵ = 64 000 cells [1]. Examiner insight: show the number of doublings. Multiplying by 5 instead of 2⁵ is the common slip, and working lets the marker see where it went wrong.

4. Any six from: embryonic cells are pluripotent so can form a wide range of tissues [1]; they can be grown in large numbers in culture [1]; obtaining them destroys an embryo, which some people consider unacceptable [1]; adult cells do not need an embryo, so there are fewer ethical objections [1]; adult cells give a narrower range of cell types and are harder to culture [1]; a patient’s own adult cells reduce the risk of rejection, whereas embryonic cells may be rejected or form tumours [1]. Examiner insight: “discuss” needs both sources compared with reasons. A list of ethical views without a benefit or risk from the biology limits the credit available.

5. Both cells contain the same genes (all arise from one zygote by mitosis) [1]. In the beta cell the insulin gene is expressed, being transcribed [1]. The mRNA is translated to give insulin [1]. In the muscle cell the insulin gene is not transcribed, so no insulin mRNA or protein is made [1]. Examiner insight: write “expressed” or “transcribed”. Saying that muscle cells “do not have” the insulin gene is a factual error that costs the whole explanation.

6. (a) 34 ÷ 40 × 100 = 85% [1] (b) As the number of sites methylated increases, the relative mRNA decreases [1], from 100 at 6 sites to 9 at 34 sites [1]. (c) Methylated cytosine bases at the promoter reduce binding of transcription factors [1], so less transcription and less mRNA [1]. Examiner insight: quote paired values from the table in (b). Describing a trend with no figures is usually treated as incomplete.

7. (a) 2 × 2 = 4 mRNAs [1] (b) Shortest has exons 1 and 4: 63 + 54 = 117 nucleotides = 39 codons, minus the stop codon = 38 amino acids [1]. Longest has all exons: 63 + 48 + 75 + 54 = 240 nucleotides = 80 codons [1], giving 79 amino acids [1]. (c) Different combinations of exons are kept in the mature mRNA [1], so the amino acid sequence differs and a different polypeptide is made [1]. Examiner insight: remember the stop codon. Dividing by 3 alone gives 39 and 80, which are one too many in each case.

8. Acetyl groups are added to histone proteins [1]. This reduces the attraction between histones and DNA, so the chromatin loosens [1]. Transcription factors and RNA polymerase reach the promoter, so transcription rises [1]. After DNA replication the new strand is unmethylated while the parent strand is methylated [1]. Enzymes methylate the new strand to match the parent, so both daughter cells have the same pattern [1]. Examiner insight: link the structural change (loosened chromatin) to access for transcription. Stating only “the gene is switched on” is a result, not an explanation.

9. (a) (3.2 + 3.6 + 3.1 + 3.5) ÷ 4 = 3.35 g [1] (b) High mean = 22.4 ÷ 4 = 5.6 g [1]. (5.6 - 3.35) ÷ 3.35 × 100 = 67.2% [1] (c) The genotype is the same, so the difference is due to the environment (nitrogen supply, which gives more materials for growth) [1]. Variation within a group also arises from environment, such as small differences in light or water, because there is no genetic difference [1]. Examiner insight: give the final answer to the requested significant figures. Percentage change is a divide by the original (low-nitrogen) value, and dividing by the new value gives a different answer.

10. (a) Mother IA IO [1] and father IB IO [1] (b) Both parents are heterozygous. Gametes IA, IO and IB, IO give IA IB, IA IO, IB IO, IO IO [1]. Probability of AB = 1/4 [1]. (c) There are three alleles (IA, IB, IO) of one gene in the population, though each person has two [1]. Examiner insight: the group O child proves that each parent carries IO, so justify the genotypes from that fact. Writing IA IA for the mother ignores the evidence.

11. (a) 0 to 6 capital alleles gives 7 classes [1] (b) (1/2)⁶ = 1/64 [1] (c) Six capital alleles = 1/64 and five = 6/64 [1], so the total is 7/64 [1] (d) Many genes each have a small additive effect [1], so there are many genotypes and many phenotype classes [1]. The environment alters the phenotype within each class, so values overlap [1]. This gives a smooth, bell-shaped distribution rather than distinct categories [1]. Examiner insight: answers that mention only the number of genes, without the additive effect and the environment, usually miss part of the explanation of continuous variation.

12. (a) Gametes cch and ch from each parent [1]. Offspring: cch cch, cch ch, cch ch, ch ch, so Himalayan = 1/4 [1] (b) The extremities are cooler than the body core, so the enzyme is active there and makes dark pigment [1]. In the warmer body the enzyme is inactive, so the fur is pale [1]. (c) The new fur grows paler or lacks dark patches [1], because the enzyme is less active at the higher temperature [1]. Examiner insight: name the variable (temperature) and the enzyme activity in (b) and (c). “Because of the environment” alone does not explain the mechanism.

Where marks are usually lost

• Saying specialised cells contain different genes instead of expressing different genes.
• Describing a blastocyst as solid, which is the morula, or calling inner cell mass cells totipotent.
• Writing a one-sided answer to a “discuss” question about stem cells.
• Dividing exon length by 3 and forgetting the stop codon.
• Describing alternative splicing as removal of exons from the DNA.
• Claiming that methylation or histone modification changes the base sequence.
• Giving a trend from a table without quoting values.
• Calculating percentage change against the wrong value.
• Not stating both parental genotypes in a multiple alleles cross.

Next steps

• Return to the revision notes and the study guide.
• Check coverage on the Edexcel Biology checklist and the course hub.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology, Specification, Issue 2 (February 2021), Pearson Education Limited: Topic 3, statements 3.17 to 3.21.