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Two genes at once: 9 : 3 : 3 : 1, the ways it breaks, and the chi-squared test questions

Dihybrid crosses and the 9 to 3 to 3 to 1 ratio, the dihybrid test cross, autosomal linkage and recombination frequency, epistasis with the 9:3:4, 12:3:1 and 9:7 ratios, and the chi-squared test with null hypothesis, degrees of freedom, critical values and conclusion.

6 original questions · 21 marks · the two genes at once: 9 : 3 : 3 : 1, the ways it breaks, and the chi-squared test notes · Inheritance and population genetics

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• 5 marks · Calculate

  Two rodents heterozygous at two loci are crossed and produce 160 offspring: 84 black, 36 brown and 40 white. Recessive epistasis predicts a 9 : 3 : 4 ratio. Calculate the value of chi-squared for these results, and state whether the difference between observed and expected numbers is significant at p = 0.05.

  Mark scheme

  1. M1 the parts of the ratio add to 16 and 160 ÷ 16 = 10, so the expected numbers are 90 black, 30 brown and 40 white, which add back to 160
  2. M1 each term is the difference squared divided by the expected value: 36 ÷ 90, 36 ÷ 30 and 0 ÷ 40
  3. A1 the terms are 0.40, 1.20 and 0, so chi-squared = 1.6
  4. B1 there are three phenotype classes, so the degrees of freedom are 3 − 1 = 2 and the critical value at p = 0.05 is 5.99
  5. A1 1.6 is below 5.99, so the difference is not significant, the null hypothesis is accepted and the deviation from 9 : 3 : 4 is attributed to chance

• 4 marks · Explain

  A pea plant of genotype RrYy is crossed with a plant of genotype rryy. Explain why four phenotypes are expected among the offspring in equal numbers, and explain what a large departure from those equal numbers would suggest.

  Mark scheme

  1. B1 a gamete receives one allele of each gene, and because the two genes are on different chromosomes the choice at one locus is independent of the choice at the other
  2. B1 this is independent assortment at metaphase I, so RrYy produces RY, Ry, rY and ry in equal numbers
  3. B1 the double recessive parent contributes ry to every offspring, so the offspring phenotypes read out directly the gametes the first parent made
  4. A1 a large excess of two classes and a shortage of the other two would suggest the genes are autosomally linked, so they cannot assort independently and only crossing over produces the scarce recombinant classes

• 4 marks · Suggest

  In a flowering plant, two enzymes act one after the other to convert a colourless precursor into a purple pigment. A cross between two plants heterozygous at both loci gives 9 purple : 7 white offspring. Suggest an explanation for this ratio.

  Mark scheme

  1. B1 this is complementary gene action, a form of epistasis: a dominant allele is needed at both loci for any pigment to be made
  2. B1 a plant of genotype A_bb makes the first enzyme but not the second, and a plant of genotype aaB_ makes the second but not the first, so in each the pathway is blocked and the flower stays white
  3. B1 those two classes, 3 and 3 in sixteenths, join the double recessive class of 1, which makes neither enzyme, giving 7 white in every 16
  4. A1 only the A_B_ class, 9 in every 16, makes both enzymes and completes the pathway to pigment, so the total is still in sixteenths and the classes have merged rather than changed in proportion

• 3 marks · Calculate

  A test cross between a doubly heterozygous fruit fly and a double recessive produces 300 offspring in four phenotype classes: 145, 15, 12 and 128. Calculate the recombination frequency for these two genes.

• 3 marks · Compare

  Compare the effect of autosomal linkage with the effect of epistasis on the offspring numbers obtained from a cross involving two genes.

• 2 marks · State

  State the phenotype ratio expected among the offspring of a cross between two organisms heterozygous at two unlinked loci, and state the ratio expected from a dihybrid test cross.

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