Your answers will be automatically graded when you submit.
Question Navigation
This paper covers the whole of Topic 17. Like a real Cambridge paper it ranges across every sub-topic — 17.1 chromosomes, genes and proteins, 17.2 mitosis and stem cells, 17.3 meiosis, 17.4 monohybrid inheritance, and 17.5 codominance, ABO blood groups and sex linkage — and it mixes them inside single questions. All three Topic 17 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 — Four Sizes and One Protein
Total: 12 marks
Fig. 1.1 is not labelled with any names. Work from the size of each structure and from where it sits.
(a)[4]
Fig. 1.1 shows a zoomed-in view of part of a cell, with four structures lettered A to E. Name the structures A, B and C, and state what E is.
Model Answer — 1(a)
A — the cell [1]
B — the nucleus [1]
C — a chromosome [1]
E — a gene, which is a length of DNA that codes for a protein [1]
⚠ If you missed marks here: The step people slip on is C and E. A chromosome is the whole structure, made of DNA; a gene is a length of that DNA. If you wrote “DNA” for C you have named the material rather than the structure, and if you wrote “allele” for E you have named a version rather than the thing itself.
(b)[4]
F and G in Fig. 1.1 are at the same position on the two chromosomes of a homologous pair, and they control the same feature. State what F and G are, and explain how they differ from one another and from a gene.
Model Answer — 1(b)
F and G are alleles [1]
an allele is an alternative form of a gene [1]
there is one gene here, not two — F and G are two versions of it [1]
they are at the same position on each chromosome of a homologous pair, which is why an individual has two alleles of every gene [1]
⚠ If you missed marks here: The mark most often lost here is the third one. Writing “F and G are two genes” sounds harmless and destroys the definition: a gene is defined by its position and its product, so two things at the same position controlling the same feature are one gene in two forms.
(c)[4]
A single base in a gene coding for a digestive enzyme is changed. The cell still makes a protein of the normal length, but the enzyme no longer breaks down its substrate. Explain why.
Model Answer — 1(c)
the sequence of bases in the gene determines the sequence of amino acids in the protein [1]
a changed base changes one amino acid in that sequence [1]
a different sequence of amino acids gives the protein a different shape [1]
the substrate no longer fits the active site, so the reaction is not catalysed [1]
⚠ If you missed marks here: This is four separate marks for four separate links, and the usual answer — “the enzyme changed shape so it stopped working” — compresses them into one. Write the chain out in order: bases, amino acids, shape, fit. The fact that the protein is the normal length is a hint that nothing was cut short, so shape is the only remaining explanation.
Question 2 — One Field, Twelve Thousand Identical Plants
Total: 12 marks
(a)[3]
Define mitosis, and state what happens to the chromosomes immediately before and during the division.
Model Answer — 2(a)
nuclear division giving rise to genetically identical cells [1]
exact replication of the chromosomes occurs before mitosis [1]
during mitosis the copies separate, so the chromosome number is maintained in each daughter cell [1]
⚠ If you missed marks here: The word genetically is the mark. “Identical cells” on its own is regularly refused, because two cells can look identical without carrying the same genes. Notice also that naming any stage of mitosis earns nothing — the stages are not on this syllabus.
(b)[3]
A banana grower plants a field of 12 000 plants, all grown from suckers taken from a single parent plant. A new strain of fungal disease arrives and every plant in the field dies. Explain why every plant died rather than only some of them.
Model Answer — 2(b)
growing plants from suckers is asexual reproduction, which uses mitosis [1]
so all 12 000 plants are genetically identical to the parent and to each other, and there is no variation in the population [1]
if the parent plant had no resistance to the disease then no plant in the field has any, so all of them are equally vulnerable [1]
⚠ If you missed marks here: The mark scheme wants the mechanism, not just the outcome. “They are all clones” earns one mark at best; naming mitosis as the reason they are clones, and then linking the absence of variation to the absence of resistance, earns all three. An answer that blames the plants being close together explains a fast spread but not a total loss.
(c)[3]
Stem cells in the bone marrow of a mammal divide to produce both red blood cells and white blood cells. State what a stem cell is, and explain how two such different cells can be produced from one kind of cell.
Model Answer — 2(c)
a stem cell is an unspecialised cell that divides by mitosis to produce daughter cells that can become specialised [1]
because the division is mitosis, all the daughter cells contain the same genes as each other [1]
the cells differ because different genes are expressed in each — a cell only makes the proteins it needs [1]
⚠ If you missed marks here: The trap is to say that the specialised cells “lose the genes they do not need”. Mitosis copies the whole set, so nothing can be lost. Two other things to avoid: stem cells are not confined to embryos, and they divide by mitosis, never by meiosis.
(d)[3]
A different grower produces banana plants from seed instead. Suggest why this population would be less likely to be destroyed by the same disease, and state one disadvantage of growing plants from seed rather than from suckers.
Model Answer — 2(d)
seeds are produced by sexual reproduction, which involves meiosis and the fusion of two gamete nuclei [1]
so the plants are genetically different from one another, and some may have resistance to the disease and survive [1]
disadvantage: the crop would not be uniform — the desirable features of the parent are not reliably passed on, and the plants may ripen at different times [1]
⚠ If you missed marks here: The word to be careful with is “some may”. Variation does not guarantee that any plant survives; it means the population contains a range of responses, so survival becomes possible. Writing that the seed-grown plants “would be resistant” overstates it and is not credited.
Question 3 — Counting Chromosomes in One Animal
Total: 12 marks
The table gives the number of chromosomes in the nuclei of four cells taken from the same animal.
Cell
Where it came from
Chromosomes in the nucleus
P
lining of the small intestine
36
Q
a testis, after a division
18
R
a fertilised egg cell
36
S
the same animal, one day after cell R formed
36
(a)[4]
Name the type of division that produced cell Q, and explain fully how the numbers in the table support your answer.
Model Answer — 3(a)
meiosis [1]
cell P is a body cell, so 36 is the diploid number for this animal [1]
cell Q has 18, which is half of 36, so the chromosome number has been halved from diploid to haploid [1]
meiosis is the reduction division that produces gametes, and a testis is where male gametes are made [1]
⚠ If you missed marks here: Naming meiosis on its own scores one mark out of four. The evidence marks are for the arithmetic — establishing the diploid number from a body cell first, then showing that Q is half of it. The organ is supporting context, not the reason.
(b)[4]
Explain how cell R came to have 36 chromosomes, and explain what would happen to the chromosome number of this species over several generations if meiosis did not take place.
Model Answer — 3(b)
cell R is a zygote, formed at fertilisation [1]
the nuclei of two haploid gametes fused, 18 + 18 = 36, restoring the diploid number [1]
without meiosis the gametes would be diploid, so the zygote would have 72 chromosomes [1]
the number would double in every generation, so the chromosome number of the species could not stay constant [1]
⚠ If you missed marks here: The phrase Cambridge marks is fusion of the nuclei of two gametes. “The sperm joined the egg” is loose and is often only half credited, because fertilisation is defined by what happens to the nuclei. In the second half, giving the actual number 72 is worth more than saying “it would go up”.
(c)[4]
Cell S was produced from cell R by a different kind of division. Name that division and give three ways in which it differs from the division that produced cell Q.
Model Answer — 3(c)
mitosis [1]
mitosis produces two cells; meiosis produces four [1]
mitosis maintains the chromosome number; meiosis halves it [1]
⚠ If you missed marks here: A difference must state both sides. “Mitosis makes two cells” is half a difference and is normally not credited on its own; “mitosis makes two cells whereas meiosis makes four” is a whole one. Also avoid giving location as a difference — where a division happens is a consequence, not a difference in the process.
Question 4 — Wings, Ratios and an Unknown Genotype
Total: 12 marks
(a)[6]
In a species of fruit fly, normal wings (N) are dominant to vestigial wings (n). Use a full genetic diagram to predict the offspring of a cross between two heterozygous normal-winged flies. Set out every stage of your working.
a correctly drawn Punnett square with the gametes on the outside [1]
offspring genotypes: 1 NN : 2 Nn : 1 nn [1]
offspring phenotypes and ratio: 3 normal-winged : 1 vestigial-winged [1]
⚠ If you missed marks here: Six marks for six lines. If you wrote only “3 : 1” you have earned one of them. The two lines students leave out most often are the circled gametes and the offspring phenotypes in words — check both before you move on, because each is a whole mark for about five seconds of writing.
(b)[3]
The cross in (a) actually produced 148 normal-winged and 44 vestigial-winged offspring. Explain whether these results support the prediction.
Model Answer — 4(b)
148 ÷ 44 is about 3.4, so the results are approximately 3 : 1 [1]
this does support the prediction from the cross [1]
the numbers are not exact because fertilisation is random, so the actual numbers vary by chance around the expected ratio [1]
⚠ If you missed marks here: Two of these three marks are for words rather than numbers. Approximately is one, because a ratio is a probability and real data never come out exact; random fertilisation is the other. An answer that concludes the data disprove a 3 : 1 ratio has done no arithmetic and loses all three.
(c)[3]
A normal-winged fly of unknown genotype is available. Describe how you would find out whether it is homozygous, and state what result would prove that it is heterozygous.
Model Answer — 4(c)
carry out a test cross: cross it with a vestigial-winged fly, which must be homozygous recessive (nn) [1]
the recessive parent can only produce n gametes, so nothing the unknown parent contributes can be masked [1]
any vestigial-winged offspring proves the unknown parent is heterozygous, because that offspring must have received an n from it [1]
⚠ If you missed marks here: The middle mark is the one that separates candidates: it is not enough to say what to cross it with, you have to say why the recessive partner is the right choice. And notice the asymmetry — a vestigial offspring proves heterozygous, but no vestigial offspring never proves homozygous.
Question 5 — Reading a Family Tree
Total: 10 marks
Fig. 5.1 shows the inheritance of coat colour in a family of rabbits. Squares are males, circles are females, and a shaded symbol means the rabbit has a brown coat.
(a)[2]
State, with a reason, whether the allele for a brown coat is dominant or recessive.
Model Answer — 5(a)
recessive [1]
because I-1 and I-2 are both black and their daughter II-1 is brown — a dominant allele would have to be expressed in at least one parent [1]
⚠ If you missed marks here: The conclusion alone is worth one mark; the reason is the other, and it must name the individuals or at least describe the pattern. “Because it skips a generation” is too vague on its own — say that neither parent shows the feature and their child does.
(b)[4]
Give the genotypes of I-1, II-1, II-3 and II-2, using the symbols B and b.
Model Answer — 5(b)
I-1 is Bb — he is black but has a brown daughter, so he must carry b [1]
II-1 is bb — she shows the recessive phenotype [1]
II-3 is Bb — he is black but has a brown son, III-1, so he must carry b [1]
II-2 is BB or Bb — nothing in the pedigree distinguishes them [1]
⚠ If you missed marks here: The last one is the mark most often thrown away. II-2 is black with no affected children, so the diagram genuinely does not settle it, and “BB or Bb” is exactly what the mark scheme prints. Choosing one of them is marked wrong even in the cases where it happens to be right.
(c)[2]
Explain how the pedigree shows that the allele for a brown coat is not carried on the X chromosome.
Model Answer — 5(c)
a female showing a sex-linked recessive feature would have to receive the recessive allele from both parents, including her father, since she inherits his only X chromosome [1]
II-1 is a brown female and her father I-1 is black, so the allele cannot be on the X chromosome [1]
⚠ If you missed marks here: This is a two-sentence answer and most candidates never attempt it. Learn the move: an affected female with an unaffected father rules out X-linked recessive inheritance on its own. Scan any pedigree for that pairing before you write anything else.
(d)[2]
II-3 and II-4 have another offspring. State the probability that it has a brown coat, and explain your answer.
Model Answer — 5(d)
both II-3 and II-4 are Bb, so the cross is Bb × Bb [1]
one box in four is bb, so the probability is 1 in 4 (25%) [1]
⚠ If you missed marks here: The probability is unaffected by the offspring they already have — each fertilisation is independent, so having produced one brown offspring does not make the next less likely. The first mark is for establishing the genotypes; without it, the number is a guess.
Question 6 — When Both Alleles Show
Total: 12 marks
(a)[2]
Explain what is meant by codominance, using coat colour in cattle as an example.
Model Answer — 6(a)
both alleles in a heterozygous organism contribute to the phenotype [1]
a roan animal, CRCW, has red hairs and white hairs together — both alleles are expressed, rather than blending [1]
⚠ If you missed marks here: The mark scheme refuses “the colours mix” and “it is halfway between”. Nothing blends: the point of the cattle example is that you can see the two colours separately if you look closely enough at the coat.
(b)[5]
In cattle, CRCR animals are red, CWCW animals are white and CRCW animals are roan. Use a full genetic diagram to show the expected offspring of a cross between two roan cattle.
Model Answer — 6(b)
parental phenotypes: roan × roan [1]
parental genotypes: CRCW × CRCW [1]
gametes: CR and CW from each parent, circled [1]
Punnett square giving CRCR, CRCW, CRCW, CWCW [1]
phenotype ratio: 1 red : 2 roan : 1 white [1]
⚠ If you missed marks here: Write the superscripts small and clearly — CRCW is four alleles and means nothing. And notice the ratio: because the heterozygote has its own phenotype, the 1 : 2 : 1 pattern does not collapse into 3 : 1 as it would with ordinary dominance. Writing 3 : 1 here shows the codominance has not registered.
(c)[5]
In humans, blood group is controlled by three alleles: IA, IB and IO. A man of blood group A and a woman of blood group B have four children, of blood groups A, B, AB and O. Give the genotypes of both parents and explain how you deduced them, and explain why one child is group AB when neither parent is.
Model Answer — 6(c)
father IAIO and mother IBIO [1]
one child is group O, which can only be IOIO [1]
that child received an IO from each parent, so both parents must carry IO while showing groups A and B [1]
the group AB child received IA from the father and IB from the mother [1]
IA and IB are codominant, so both are expressed and the phenotype is AB — neither parent shows it because neither has both alleles [1]
⚠ If you missed marks here: Work backwards from the group O child, whose genotype is the only certain one in the family. The final mark needs the word codominant explicitly — describing the outcome without naming the relationship between the two alleles usually loses it.
Question 7 — One X, and No Second Chance
Total: 10 marks
(a)[3]
Red-green colour blindness is caused by a recessive allele carried on the X chromosome. Explain why it is more common in males than in females.
Model Answer — 7(a)
the allele is carried on the X chromosome, and the Y chromosome carries no allele for this gene [1]
a male has only one X, so a single recessive allele is expressed — there is no second X carrying a dominant allele to mask it [1]
a female has two X chromosomes, so she needs two copies of the recessive allele, XbXb, which is much less likely [1]
⚠ If you missed marks here: Two errors cost this question regularly. The first is putting the allele on the Y — if it were there, females could never be colour blind, and they can be. The second is stopping after the male half; the mark scheme wants the comparison, so the female side must be there too.
(b)[5]
A woman who is a carrier for red-green colour blindness has children with a man who has normal vision. Use a full genetic diagram to show the possible children. Use the symbols XB, Xb and Y.
Model Answer — 7(b)
parental phenotypes: carrier female × male with normal vision [1]
parental genotypes: XBXb × XBY [1]
gametes: XB and Xb from the mother, XB and Y from the father, circled [1]
Punnett square giving XBXB, XBXb, XBY, XbY [1]
phenotypes: a daughter with normal vision, a carrier daughter, a son with normal vision and a colour-blind son — 1 : 1 : 1 : 1 [1]
⚠ If you missed marks here: Two notation errors are fatal here. Writing Bb instead of XBXb hides which chromosome carries the allele, which is the whole point; and writing Yb invents an allele on the Y chromosome, whose absence is the reason the condition is commoner in males. The phenotype line must state the sex as well as the vision.
(c)[2]
Using your diagram, state the probability that a child of this couple is colour blind, and the probability that a son of this couple is colour blind.
Model Answer — 7(c)
a child: 1 in 4 (25%), because one of the four boxes is a colour-blind son [1]
a son: 1 in 2 (50%), because only two of the four boxes are sons and one of those two is colour blind [1]
⚠ If you missed marks here: This is the single most reliably lost mark in sex linkage: the same square gives two different answers depending on the denominator. Underline the word child, son or daughter in the stem before you count boxes, and say which boxes you counted.
Self-Assessment
Tick marks earned, then click Calculate Grade.
0
80
0%
A* : 56+
A : 48-55
B : 40-47
C : 32-39
D : 24-31
E : 16-23
U : <16
When finished, click Submit to see model answers.
Exam Submitted -- Marking Mode Active
Click "Show Model Answer" on each question to check your work.