IB DP Biology: Dihybrid crosses, gene linkage and the chi-squared test (HL) (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover IB Diploma Programme Biology, sections D3.2.16 to D3.2.21 (dihybrid crosses, linkage, recombinants and the chi-squared test), from the IB Biology guide, first assessment 2025. All of this content is additional higher level, so every question is HL only. Species and data are invented. Learn the content from the study guide and the revision notes. Course hub: IB DP Biology. Checklist: printable checklist.

Critical values for χ² at p = 0.05: 1 degree of freedom 3.84; 2 degrees 5.99; 3 degrees 7.81; 4 degrees 9.49.

Questions

1. A cell of an insect has 2n = 10 chromosomes. Explain how meiosis I produces independent assortment of genes on different chromosomes, and state how many different combinations of maternal and paternal chromosomes are possible in the gametes, ignoring crossing over. [3]

2. A plant has the genotype GgRr. The genes for G and R are on different chromosomes.

(a) State the four types of gamete it makes and the proportion of each. [2] (b) The plant is crossed with another GgRr plant. State the number of different genotypes among the offspring. [1] (c) State the fraction of offspring that are GGRR. [1]

3. In an invented bird species, a crest (C) is dominant to a plain head (c), and green plumage (Y) is dominant to yellow plumage (y). The genes are on different autosomes. Two birds of genotype CcYy are crossed.

(a) Construct a Punnett grid and state the phenotypic ratio of the offspring. [3] (b) From 320 offspring, calculate the expected number with a plain head and yellow plumage. [2]

4. A CcYy bird is crossed with a ccyy bird. The genes are unlinked.

(a) State the expected phenotypic ratio. [1] (b) Explain how this ratio arises. [2] (c) A student says that a 1:1:1:1 result proves the two genes are on different chromosomes. Suggest why this is not certain. [1]

5. A student uses a database to record the loci of four human genes.

Gene	Polypeptide product	Locus
HBA1	alpha-globin	16p13.3
HBB	beta-globin	11p15.4
HBD	delta-globin	11p15.4
CFTR	membrane channel for chloride and bicarbonate ions	7q31.2

(a) Identify one pair of genes expected to assort independently and give the reason. [2] (b) HBB and HBD are close together on one chromosome. Predict how their alleles will be inherited and explain. [2]

6. In an invented plant, tall stem (T) is dominant to dwarf (t), and round pollen (R) is dominant to oval (r). The genes are on the same autosome. A plant with alleles TR on one chromosome and tr on the other was crossed with a plant homozygous recessive for both genes. The 600 offspring were: tall round 262, dwarf oval 258, tall oval 41, dwarf round 39.

(a) Show the genotype of the heterozygous parent using vertical lines for the chromosomes, and state its four possible gametes. [2] (b) Identify the recombinant offspring and calculate the recombinant frequency. [3] (c) Another heterozygote has the alleles Tr on one chromosome and tR on the other. State the two recombinant phenotypes in its test cross. [2]

7. (a) Explain how recombinant gametes arise from linked genes. [3] (b) Explain why the recombinant frequency in a test cross cannot be higher than 50%. [2]

8. Two plants heterozygous for two unlinked genes were crossed. Among 600 offspring the phenotypes were 330, 121, 106 and 43 in the order dominant-dominant, dominant-recessive, recessive-dominant, recessive-recessive. A chi-squared test is used to compare with 9:3:3:1.

(a) State the null hypothesis. [1] (b) Calculate the expected numbers. [1] (c) Calculate χ². [2] (d) State the conclusion at p = 0.05. [2]

9. A double heterozygote with alleles AB on one chromosome and ab on the other was test crossed. The 1000 offspring were: AB 392, ab 371, Ab 118, aB 119. The genes are autosomal.

(a) State the expected numbers if the genes assort independently. [1] (b) Calculate χ². [2] (c) State and explain the conclusion at p = 0.05. [2] (d) Calculate the recombinant frequency. [1]

10. (a) State what is meant by a result being statistically significant at the p = 0.05 level. [2] (b) Explain why the F2 generation of a cross is described as a sample. [1]

11. Two test crosses were carried out. In each, a double heterozygote with both dominant alleles on one chromosome was crossed with a double homozygous recessive. Cross 1 involves genes P and Q. Cross 2 involves genes R and S.

Cross	Parental class 1	Parental class 2	Other class 1	Other class 2
1	121	118	126	115
2	239	249	46	46

(a) Calculate χ² for each cross against 1:1:1:1. [4] (b) Draw a conclusion for each cross at p = 0.05. [2] (c) Calculate the recombinant frequency for cross 2. [1] (d) The 1:1:1:1 and 9:3:3:1 ratios are called Mendel’s second law. Discuss, using these data, whether it should be called a law. [2]

Answers

1. In metaphase I the five homologous pairs line up at the equator as bivalents [1]. Each pair faces the poles in either orientation, independent of the others, so alleles on different chromosomes are sorted independently [1]. Combinations = 2⁵ = 32 [1] [3] Examiner insight: Answers tend to gain credit for naming the stage (metaphase I) and for saying the orientation of each pair is independent; “chromosomes mix up” is too vague.

2. (a) GR, Gr, gR, gr [1], each ¼ [1] (b) 9 [1] (c) ¼ × ¼ = 1/16 [1] Examiner insight: For a gamete list, show every allele pairing once; state the proportion of each type as well as listing the types.

3. (a) Gametes from each parent: CY, Cy, cY, cy [1]. A 4 × 4 grid with 16 correct cells [1]. Ratio 9 crested green : 3 crested yellow : 3 plain green : 1 plain yellow [1] [3] (b) Plain and yellow is 1/16 [1]; 320 ÷ 16 = 20 [1] [2] Examiner insight: Label gamete types around the grid and say which phenotype each combination gives; a bare ratio without a visible grid or reasoning is less secure.

4. (a) 1:1:1:1 [1] (b) The ccyy bird makes only cy gametes [1]. The CcYy bird makes CY, Cy, cY, cy in equal numbers, so the four phenotypes appear equally [1] (c) Genes far apart on the same chromosome also give 1:1:1:1 because recombinant frequency reaches 50% [1] Examiner insight: Mention independent assortment or 50% recombination to explain why equal classes occur.

5. (a) HBA1 and HBB (or CFTR with any other) [1]; they are on different chromosomes (16 and 11), so homologous pairs orient independently in metaphase I [1] (b) They are likely to be inherited together [1]; the closer two loci are on a chromosome, the less likely a chiasma forms between them [1] Examiner insight: Quote the chromosome numbers from the table when reasoning.

6. (a) TR | tr [1]. Gametes TR, tr, Tr, tR [1] (b) Recombinants have a combination not on either parental chromosome: tall oval and dwarf round [1]. 41 + 39 = 80 [1]. 80/600 × 100 = 13.3% [1] (c) Parental gametes are now Tr and tR, so the recombinants are tall round (TR) and dwarf oval (tr) [2] Examiner insight: State the parental arrangement first. Do not assume the two largest classes are always parental.

7. (a) In prophase I, homologous chromosomes pair as bivalents [1]. Non-sister chromatids exchange segments at a chiasma between the two loci [1]. The chromatids now carry new allele combinations, which pass into gametes [1] [3] (b) Even if a chiasma forms between the loci in every meiosis, only two of the four chromatids are recombinant [1], so at most half the gametes, 50%, are recombinant [1] [2] Examiner insight: Say “non-sister chromatids” and “between the two loci”; crossing over elsewhere on the chromosome does not create a recombinant for these genes.

8. (a) There is no significant difference between the observed and expected numbers; any difference is due to chance [1] (b) 600 × 9/16, 3/16, 3/16, 1/16 = 337.5, 112.5, 112.5, 37.5 [1] (c) χ² = 56.25/337.5 + 72.25/112.5 + 42.25/112.5 + 30.25/37.5 [1] = 0.167 + 0.642 + 0.376 + 0.807 = 1.99 [1] (d) 3 degrees of freedom, critical value 7.81; 1.99 < 7.81 [1]. Do not reject the null hypothesis: the results fit 9:3:3:1, consistent with independent assortment [1] Examiner insight: Show the expected numbers and the degrees of freedom; a conclusion in words about the genes, not only “not significant”, completes the answer.

9. (a) 1000 ÷ 4 = 250 in each class [1] (b) χ² = (392 − 250)²/250 + (371 − 250)²/250 + (118 − 250)²/250 + (119 − 250)²/250 [1] = 80.66 + 58.56 + 69.70 + 68.64 = 277.6 [1] (c) 277.6 is much greater than the critical value 7.81 (3 degrees of freedom) [1]. Reject the null hypothesis: the difference is significant, so the genes are not assorting independently and are probably linked [1] (d) (118 + 119)/1000 × 100 = 23.7% [1] Examiner insight: Name the critical value you compare with, and avoid saying linkage is “proved”: the test shows chance alone is an unlikely explanation.

10. (a) If the null hypothesis were true, results at least this different from those expected would occur by chance less than 5% of the time [1]. So the difference is unlikely to be due to chance alone, and the null hypothesis is rejected [1] [2] (b) The F2 offspring are a limited group that represents all the offspring the cross could produce, the population [1] Examiner insight: Link p = 0.05 to the probability of chance; saying “95% certain the hypothesis is true” is a common misstatement.

11. (a) Cross 1: expected 480 ÷ 4 = 120. χ² = (1 + 4 + 36 + 25)/120 [1] = 0.55 [1]. Cross 2: expected 580 ÷ 4 = 145. χ² = (94² + 104² + 99² + 99²)/145 [1] = 39 254/145 = 270.7 [1] [4] (b) Cross 1: 0.55 < 7.81, so not significant; consistent with independent assortment [1]. Cross 2: 270.7 > 7.81, so significant; the genes are linked [1] [2] (c) (46 + 46)/580 × 100 = 15.9% [1] (d) Cross 1 follows the ratio, but cross 2 does not [1]. The ratios hold only for genes on different chromosomes or far apart, so there are exceptions and “law” should be applied with that condition [1] [2] Examiner insight: In a discuss question, use both data sets and state the condition under which the second law holds.

Where marks are usually lost

• Listing only two gametes for a dihybrid.
• Calling the largest classes recombinant when the parental arrangement is the opposite one.
• Giving recombinant frequency as a fraction of the recombinants instead of the total offspring.
• Using the wrong degrees of freedom, such as the number of classes.
• Concluding “the results are significant” without a comparison with a critical value.
• Stating the null hypothesis as the explanation, such as “the genes are linked”.
• Saying the null hypothesis is “accepted”, or that linkage is “proved”.

Next steps

• Revise with the revision notes and re-read the study guide.
• IB DP Biology hub and printable checklist.
• Find gaps with all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Sections D3.2.16 to D3.2.21 (additional higher level).