Before you start
This review joins the five topics of Unit 5. Do the unit diagnostic first, then try these without looking at the answers.
Big ideas of the unit
- One copy, two divisions. DNA is copied once and the cell divides twice, giving four haploid cells. Meiosis I separates homologues; meiosis II separates sister chromatids (Topic 5.1).
- Every gamete is a new mixture. Independent assortment, crossing over and random fertilization make offspring unique. Nondisjunction gives gametes that are not haploid (Topic 5.2).
- Mendel’s laws are meiosis seen through alleles. Homologues separating in anaphase I is segregation; pairs lining up independently is independent assortment (Topic 5.3).
- Probability replaces big Punnett squares. Split a cross into one-gene crosses, multiply for “and”, add for “or”.
- Chi-square decides whether a difference is real, not what caused it (Topic 5.4).
- Where a gene sits changes the pattern. Genes on one chromosome travel together unless a crossover separates them; genes on the X show different results in sons and daughters; genes in mitochondria and chloroplasts follow the mother (Topic 5.4).
- How alleles act changes the ratio. Incomplete dominance and codominance give the heterozygote its own phenotype, so 1 : 2 : 1 replaces 3 : 1. One gene with several effects (pleiotropy) passes those traits on together.
- Phenotype = genotype acting in an environment. Temperature, light, diet or a signal can change gene expression or how well a gene product works (Topic 5.5). Hold genotype constant to see the environment’s effect.
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Counting through meiosis | chromosomes = centromeres; DNA ∝ chromatids; G1 value × 2 after S, ÷ 2 after each division | 5.1 cell and DNA questions |
| Chromosome combinations | 2ⁿ per gamete (n = pairs); 2ⁿ × 2ⁿ per offspring of one couple | 5.2 |
| Nondisjunction | meiosis I: all four gametes abnormal; meiosis II: two of four | 5.2, aneuploid offspring |
| Probability | independent: P(A and B) = P(A) × P(B); mutually exclusive: P(A or B) = P(A) + P(B) | 5.3 crosses and families |
| Chi-square | expected counts from the ratio; χ² = Σ (o − e)² ÷ e; df = categories − 1 | 5.3, 5.4 |
| Gene mapping | recombinants ÷ total × 100 = map units; at most about 50% | 5.4 linked genes |
| Sex linkage | sons get their X from the mother; fathers give X to every daughter | 5.4 crosses and pedigrees |
| Genotype vs environment | same genotype in two environments (clones, controlled cross) → any difference is environmental | 5.5 |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All organisms with letter-named genes, and all data sets, are fictional.
Question 1 (multiple choice · mixed)
A plant is heterozygous (Aa) for one gene. There is no crossing over between gene A and the centromere. At which stage of meiosis are allele A and allele a first pulled to opposite poles of the cell?
- (A) S phase, before meiosis begins
- (B) Prophase I
- (C) Anaphase I
- (D) Anaphase II
Answer and explanation
Answer: (C). A and a sit on the two homologues of one pair. With no crossover near the gene, both chromatids of one homologue carry A and both of the other carry a. Homologues separate in anaphase I: that is Mendel’s segregation.
- (A) copies each allele; it does not separate A from a.
- (B) is when homologues pair up and may cross over; they are still together.
- (D) separates sister chromatids, which carry the same allele here.
Question 2 (multiple choice · mixed)
In a hypothetical flower, colour alleles Cᴿ (red) and Cᵂ (white) show incomplete dominance: CᴿCᵂ plants are pink. An unlinked gene gives tall (T), dominant to dwarf (t). What fraction of the offspring of CᴿCᵂTt × CᴿCᵂTt are pink and dwarf?
- (A) 1/16
- (B) 1/8
- (C) 3/16
- (D) 3/8
Answer and explanation
Answer: (B). Colour: CᴿCᵂ × CᴿCᵂ gives 1/4 red, 1/2 pink, 1/4 white. Height: Tt × Tt gives 1/4 dwarf. The genes are unlinked, so multiply: 1/2 × 1/4 = 1/8.
- (A) uses 1/4 for pink, as for a homozygous class.
- (C) uses 3/4 for pink, as if the colour alleles showed complete dominance.
- (D) is pink and tall (1/2 × 3/4).
Question 3 (multiple choice · mixed)
In a hypothetical mammal, XXY individuals develop as males. An X-linked recessive allele Xᵈ gives pale eyes. A pale-eyed female (XᵈXᵈ) and a dark-eyed male (XᴰY) have a dark-eyed son with the chromosomes XᴰXᵈY. Which explanation fits?
- (A) Crossing over between the X and the Y in the father
- (B) Nondisjunction of the two X chromosomes in the mother at meiosis I
- (C) Nondisjunction of sister chromatids in the father at meiosis II
- (D) Nondisjunction of the X and Y in the father at meiosis I
Answer and explanation
Answer: (D). The mother has no Xᴰ, so the son’s Xᴰ came from the father, and so did his Y. One sperm carrying both X and Y means the father’s X and Y, which pair as homologues, went to the same pole in anaphase I. The egg was a normal Xᵈ.
- (A) does not change chromosome number, so it cannot add a chromosome.
- (B) would give an XᵈXᵈ egg; with a Y sperm the son would be XᵈXᵈY and pale-eyed.
- (C) separates sister chromatids, so it gives XᴰXᴰ or YY sperm, never one with both X and Y.
Question 4 (constructed response · mixed)
A fictional insect has 2n = 4. Gene D is on chromosome pair 1 and gene E is on chromosome pair 2. One insect has genotype DdEe, and a G1 cell in its ovary contains 2.4 pg of DNA.
(a) Give the number of chromosomes and the DNA content per cell at metaphase I, at the end of meiosis I and at the end of meiosis II. (b) List the gametes this insect can make, give the probability of each, and explain which event in meiosis makes all of them equally likely. (c) Two DdEe insects are crossed. Calculate the probability of an offspring with the dominant D phenotype and the recessive e phenotype. (d) In a related species, D and E are 8 map units apart on the same chromosome, and an insect inherited D with E from one parent and d with e from the other. Predict the percentage of each kind of gamete, and name the event that makes the rarer kinds.
Worked solution and suggested Marlbridge rubric
(a)
| Stage (per cell) | Chromosomes | Chromatids | DNA / pg |
|---|---|---|---|
| Metaphase I | 4 | 8 | 4.8 |
| End of meiosis I | 2 | 4 | 2.4 |
| End of meiosis II | 2 | 2 | 1.2 |
(b) DE, De, dE and de, each 1/4. At metaphase I each pair faces either way with equal chance, independently of the other pair.
(c) Dd × Dd: P(dominant D) = 3/4. Ee × Ee: P(ee) = 1/4. Product rule: 3/4 × 1/4 = 3/16.
(d) Recombinants are 8%, shared equally: De 4%, dE 4%. Parental types are 92%: DE 46%, de 46%. Recombinants come from crossing over between non-sister chromatids in prophase I, between D and E.
| Point | What earns it |
|---|---|
| 1 | All three rows of (a) correct, with DNA halving after each division |
| 1 | Four gametes at 1/4 each, explained by independent orientation of the pairs at metaphase I |
| 1 | 3/16 by the product rule |
| 1 | 46, 46, 4, 4% with the parental types as the large classes |
| 1 | Crossing over between non-sister chromatids in prophase I named as the source of recombinants |
Total: 5 points.
Question 5 (constructed response · mixed)
In a hypothetical beetle (XX females, XY males), black body (B) is dominant to tan (b) on an autosome. An eye-colour gene has a dominant allele for dark eyes and a recessive allele for pale eyes. A black, pale-eyed female is crossed with a black, dark-eyed male. Both are heterozygous for body colour. The 400 offspring are:
| Offspring | Black, dark-eyed females | Tan, dark-eyed females | Black, pale-eyed males | Tan, pale-eyed males |
|---|---|---|---|---|
| Number | 156 | 44 | 147 | 53 |
(a) Explain how the data show that the eye-colour gene is X-linked, and give the parents’ genotypes using Xᴰ and Xᵈ. (b) Use your genotypes and the product rule to give the expected number in each class. (c) Calculate χ², and state your conclusion. (d) Predict the eye colour of the sons and daughters if a pale-eyed male is crossed with a dark-eyed female from a true-breeding line.
Worked solution and suggested Marlbridge rubric
(a) Every daughter is dark-eyed and every son pale-eyed; an autosomal gene would not differ by sex. Sons get their X from the mother, so she is XᵈXᵈ. Daughters get the father’s X, so he is XᴰY. With body colour: BbXᵈXᵈ and BbXᴰY.
(b) P(female) = 1/2, all dark-eyed; P(male) = 1/2, all pale-eyed. P(black) = 3/4. So black dark females 1/2 × 3/4 = 3/8 → 150; tan dark females 1/8 → 50; black pale males 3/8 → 150; tan pale males 1/8 → 50.
(c) χ² = 36/150 + 36/50 + 9/150 + 9/50 = 0.24 + 0.72 + 0.06 + 0.18 = 1.20. Degrees of freedom = 3; critical value 7.81. Since 1.20 < 7.81, fail to reject the null hypothesis. The data are consistent with an autosomal body-colour gene and an X-linked eye-colour gene assorting independently.
(d) XᴰXᴰ × XᵈY: all offspring dark-eyed; daughters are XᴰXᵈ carriers. Reciprocal crosses differ, as expected for a sex-linked gene.
| Point | What earns it |
|---|---|
| 1 | Eye colour linked to sex of offspring, so the gene is on the X |
| 1 | Correct genotypes BbXᵈXᵈ and BbXᴰY |
| 1 | Expected 150, 50, 150, 50 |
| 1 | χ² = 1.20, df 3, fail to reject, with the critical value compared |
| 1 | All dark-eyed in (d), with daughters as carriers |
Total: 5 points.
Question 6 (constructed response · mixed)
In a hypothetical plant, purple flowers (P) are dominant to white (p). P codes for a pigment-pathway enzyme; p codes for an inactive version. A grower crosses two Pp plants and sows 400 seeds in a greenhouse at 20 °C and 400 at 32 °C:
| Greenhouse | Purple | White |
|---|---|---|
| 20 °C | 289 | 111 |
| 32 °C | 0 | 400 |
(a) Test the 20 °C results against a 3 : 1 ratio with chi-square. (b) Give the expected numbers of PP, Pp and pp plants in the 32 °C greenhouse, and suggest an explanation for the result there. (c) White plants from the 32 °C greenhouse are moved to 20 °C. Predict the fraction that will grow purple flowers, and explain. (d) Describe a test cross to find whether a plant is PP or Pp, and explain why its offspring must be grown at 20 °C.
Worked solution and suggested Marlbridge rubric
(a) Expected: 400 × 3/4 = 300 purple, 100 white. χ² = (289 − 300)² ÷ 300 + (111 − 100)² ÷ 100 = 0.403 + 1.210 = 1.61. Degrees of freedom = 1; critical value 3.84. Since 1.61 < 3.84, fail to reject: the data fit one gene with P dominant.
(b) Same cross, so the same genotypes: 100 PP, 200 Pp, 100 pp. Yet every plant is white. Likely explanation: the P enzyme is inactive at 32 °C (for example, it denatures), so no pigment forms in any genotype. The environment changed the phenotype, not the alleles.
(c) 3/4. Moving plants does not change genotypes. PP and Pp plants make working enzyme at 20 °C, so their new flowers are purple; pp plants stay white.
(d) Cross the plant with pp and grow the offspring at 20 °C. All purple → PP; about half white → Pp. At 32 °C every offspring would be white, hiding the genotypes.
| Point | What earns it |
|---|---|
| 1 | Expected 300 and 100, χ² = 1.61 |
| 1 | df 1, critical value 3.84, fail to reject |
| 1 | Genotypes 100 : 200 : 100 at 32 °C, with the genotype unchanged |
| 1 | Enzyme activity (or expression) depends on temperature, so no pigment at 32 °C |
| 1 | 3/4 in (c), because PP and Pp make working enzyme at 20 °C |
| 1 | Test cross with pp, expected results for PP and Pp, offspring grown at 20 °C |
Total: 6 points.
Question 7 (constructed response · mixed)
In a fictional mammal, a muscle weakness is caused by a mitochondrial DNA mutation. Cells can hold both normal and faulty mitochondria; animals with more than 50% faulty mitochondria in muscle show weakness.
- Family 1: an affected father and an unaffected mother have 6 offspring (3 daughters, 3 sons). None has any faulty mitochondria.
- Family 2: a mother with 40% faulty mitochondria and an unaffected father have 8 offspring. Percentage of faulty mitochondria in each offspring’s muscle: 15, 30, 42, 55, 70, 38, 22, 48.
(a) Calculate the mean and range for Family 2, and the percentage of the offspring that show weakness. (b) Explain why the offspring in Family 2 differ from each other. (c) Explain the result in Family 1. (d) A student claims the weakness is caused by a dominant allele on the X chromosome. Use Family 1 to evaluate the claim. (e) Explain why a Punnett square cannot predict the offspring in Family 2.
Worked solution and suggested Marlbridge rubric
(a) Mean = (15 + 30 + 42 + 55 + 70 + 38 + 22 + 48) ÷ 8 = 320 ÷ 8 = 40%. Range = 70 − 15 = 55 percentage points. Two offspring (55% and 70%) are above 50%: 2 ÷ 8 × 100 = 25% show weakness.
(b) Mitochondria are shared out at random when cells divide, including the divisions that make eggs. Different eggs get different proportions of faulty mitochondria: the mean matches the mother’s 40%, but individuals vary.
(c) The egg supplies the zygote’s mitochondria; sperm usually contribute none. So an affected father passes on no faulty mitochondria.
(d) Not supported. A father gives his X to every daughter, so with an X-linked dominant allele all 3 daughters in Family 1 would be affected. None is.
(e) Mendelian ratios need two alleles per gene, one from each parent, that segregate in meiosis. Mitochondrial genes come only from the mother, in many randomly shared copies, so there is no fixed ratio.
| Point | What earns it |
|---|---|
| 1 | Mean 40%, range 55 percentage points and 25% affected |
| 1 | Random sharing of mitochondria into eggs explains the variation |
| 1 | Mitochondria come from the egg, not the sperm |
| 1 | X-linked dominant rejected because the daughters of an affected father are unaffected |
| 1 | Mendel’s rules need paired nuclear alleles that segregate; organelle genes do not |
Total: 5 points.
How did you do?
Add up your points: 3 for multiple choice and 21 for constructed response. Compare only with your own next attempt; the total does not predict an exam score.
- Lost points on meiosis (Questions 1, 3, 4(a)–(b)): reread Topic 5.1 and the nondisjunction table in Topic 5.2.
- Lost points on probability or chi-square (Questions 2, 4(c), 5(b)–(c), 6(a)): rework the worked examples in Topic 5.3.
- Lost points on choosing an inheritance pattern (Questions 4(d), 5(a), 7): practise the clues for linkage, sex linkage and maternal inheritance in Topic 5.4.
- Lost points on Question 6(b)–(d): reread how the environment reaches the genes in Topic 5.5.
- Want a quick recheck? Retake the unit diagnostic.
Then tick off each topic checklist: