Questions on dihybrid inheritance, linkage, epistasis and the chi-squared test

Question 1

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.

Accepted Answer

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 M1 each term is the difference squared divided by the expected value: 36 ÷ 90, 36 ÷ 30 and 0 ÷ 40 A1 the terms are 0.40, 1.20 and 0, so chi-squared = 1.6 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 A1 1.6 is below 5.99, so the difference is not significant, the null hypothesis is not rejected and the deviation from 9 : 3 : 4 is attributed to chance

Question 2

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.

Accepted Answer

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 B1 this is independent assortment at metaphase I, so RrYy produces RY, Ry, rY and ry in equal numbers B1 the double recessive parent contributes ry to every offspring, so the offspring phenotypes read out directly the gametes the first parent made 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

Question 3

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.

Accepted Answer

B1 this is complementary gene action, a form of epistasis: a dominant allele is needed at both loci for any pigment to be made 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 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 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

Question 4

In a plant, allele A is epistatic to gene B: whenever a dominant A allele is present the flowers are white regardless of genotype at the B locus, and only aaB_ gives blue while aabb gives yellow. A cross between two double heterozygotes AaBb is carried out. Explain how many of the sixteen offspring in the standard dihybrid ratio would be expected to be white, and explain why this is called dominant epistasis.

Accepted Answer

B1 the standard dihybrid cross AaBb × AaBb gives the four genotype classes in the ratio 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb B1 the two classes carrying a dominant A allele, 9 A_B_ and 3 A_bb, both show white flowers because a dominant A allele masks the B locus entirely, so these merge into a white class of 9 + 3 = 12 B1 this leaves 3 aaB_ giving blue and 1 aabb giving yellow, so the overall ratio is 12 white : 3 blue : 1 yellow B1 it is called dominant epistasis because it is a single dominant allele, A, at the first locus, rather than the homozygous recessive genotype, that masks the expression of the second gene

Question 5

A test cross between a doubly heterozygous fruit fly and a double recessive fly produces 500 offspring: 195 and 205 in the two parental phenotype classes, and 55 and 45 in the two recombinant classes. Calculate the recombination frequency between the two genes. A second, independent cross of the same kind produces 800 offspring; calculate the number of recombinant offspring you would expect among them, assuming the same recombination frequency.

Accepted Answer

M1 recombinants = 55 + 45 = 100, out of a total of 500 offspring A1 recombination frequency = 100 ÷ 500 × 100 = 20% M1 for the second cross, apply the same percentage to the new total: 20% × 800 A1 = 160 recombinant offspring expected among the 800

Question 6

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. The recombination frequency is the number of recombinant offspring divided by the total number of offspring, multiplied by 100. Calculate the recombination frequency for these two genes.

Accepted Answer

M1 the genes are linked, so the two classes in large excess, 145 and 128, are the parental combinations and the recombinants are the two scarce classes: 15 + 12 = 27 M1 27 ÷ 300 × 100 A1 9 per cent

Question 7

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

Accepted Answer

B1 linkage arises because the two genes sit on the same chromosome and cannot assort independently, whereas epistasis arises because one gene affects the expression of the gene at the other locus, usually through a shared pathway B1 linkage leaves all the phenotype classes present but shifts the proportions heavily in favour of the parental combinations, whereas epistasis merges classes so fewer phenotypes appear B1 an epistatic ratio still adds to sixteen parts, such as 9 : 3 : 4 or 9 : 7, whereas a linked test cross departs from 1 : 1 : 1 : 1 by an amount that measures the distance between the loci

Question 8

Compare the 9 : 3 : 4 ratio produced by recessive epistasis with the 9 : 7 ratio produced by complementary gene action, referring to which genotypes are merged into the largest class in each case.

Accepted Answer

B1 in 9 : 3 : 4 the double recessive class, 1, merges with one of the single-recessive classes, 3, that shares the same masked phenotype, because the homozygous recessive genotype at the first locus masks the second gene in both classes B1 in 9 : 7, three classes merge instead of two: A_bb, aaB_ and aabb all lack a functioning pathway, because a dominant allele is needed at both loci for the product to be made, so 3 + 3 + 1 = 7 merge together and only the double dominant class, 9, shows the phenotype B1 the more classes that share the same underlying block in a shared pathway, the larger the merged class becomes and the fewer distinguishable phenotypes remain, which is why 9 : 7 shows only two phenotypes while 9 : 3 : 4 shows three

Question 9

A student concludes a chi-squared test on a dihybrid cross by writing 'the null hypothesis is proved correct'. Explain why this conclusion is not appropriate, and explain what should be written instead when a calculated χ² value is below the critical value.

Accepted Answer

B1 a χ² value below the critical value shows only that the data provide no evidence against the null hypothesis, not that the null hypothesis has been shown to be true, so 'proved correct' overstates what the test can show B1 the correct wording is that the null hypothesis is not rejected, because failing to find evidence against something is a different logical claim from confirming it B1 other explanations besides the null hypothesis being exactly true could also produce a low, non-significant χ² value, including a sample too small to detect a real but modest departure from the expected ratio

Question 10

A dihybrid cross produces offspring in four phenotype classes. Explain why the degrees of freedom used in a chi-squared test on this cross is 3 rather than 4.

Accepted Answer

B1 degrees of freedom is the number of categories minus one, so with four phenotype classes it is 4 − 1 = 3 B1 this is because once the total number of offspring and any three of the four class totals are known, the fourth class total is fixed automatically, so only three of the four values are genuinely free to vary B1 degrees of freedom is not related to the number of individual offspring in the cross, so a cross of 100 offspring and one of 1000 offspring in four classes both have 3 degrees of freedom

Question 11

A dihybrid test cross produces only 12 offspring in total, expected in a 1 : 1 : 1 : 1 ratio. Suggest why a chi-squared test would not be reliable on this data set, and suggest what the student should do instead.

Accepted Answer

B1 with only 12 offspring split across four classes the expected number in each class is 12 ÷ 4 = 3, well below the usual minimum of about 5 for a reliable chi-squared test B1 with such small expected values the test statistic does not follow the distribution the critical-value table assumes closely enough, so the conclusion drawn from comparing it with the table could be misleading B1 the student should repeat the cross to obtain a much larger sample of offspring before applying the test

Question 12

Describe how a dihybrid test cross is set out, and describe what result would be expected if the two genes assort independently.

Accepted Answer

B1 an individual heterozygous at both loci, such as RrYy, is crossed with a double recessive individual, rryy B1 because the double recessive parent can only contribute the recessive allele of each gene, the phenotypes of the offspring directly show which gametes the heterozygous parent produced B1 if the two genes assort independently, the heterozygous parent produces four kinds of gamete, RY, Ry, rY and ry, in equal numbers, so the offspring phenotypes are expected in a 1 : 1 : 1 : 1 ratio

Question 13

A dihybrid cross between two heterozygous mice gives approximately 9 : 3 : 3 : 1 in three separate trials, but a fourth trial with a different pair of genes gives approximately 13 : 3. Identify which phenomenon, linkage or epistasis, is more likely to explain the 13 : 3 result, and identify the pattern this ratio corresponds to.

Accepted Answer

B1 epistasis is more likely: the classes are still expressed in whole sixteenths, since 13 and 3 add to 16, which is the signature of epistasis merging categories, rather than linkage, which distorts the proportions among the parental and recombinant classes without keeping simple sixteenths B1 13 : 3 is 9 + 3 + 1 = 13 merged against 3 remaining, so three of the four dihybrid classes merge and only one class, 3 in sixteenths, remains phenotypically distinct B1 this is a form of dominant epistasis: every individual with a dominant allele at the first locus (9 + 3) shares a phenotype with the double recessive (1), and only the 3 that are homozygous recessive at the first locus but carry a dominant allele at the second differ

Question 14

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.

Accepted Answer

B1 9 : 3 : 3 : 1 from the cross between two double heterozygotes B1 1 : 1 : 1 : 1 from the dihybrid test cross

Question 15

In a test cross, the recombination frequency between two genes is the percentage of offspring that carry a new combination of the parents' alleles. Explain why the recombination frequency between two genes on different chromosomes is 50%, and explain why crossing over between two genes on the same chromosome can never make more than half of the gametes from a meiosis recombinant.

Accepted Answer

B1 genes on different chromosomes assort independently, so a test cross gives four classes in a 1 : 1 : 1 : 1 ratio, and the two recombinant classes together make up exactly half of the offspring B1 each crossover involves only two of the four chromatids of a bivalent, so even a crossover between the loci in every meiosis makes only half the gametes recombinant; multiple crossovers between distant loci average out to the same half