Dihybrid inheritance, linkage, epistasis and the chi-squared test

Two genes, four gametes, sixteen fertilisations

A dihybrid cross follows two genes at once. In Mendel's peas, R gives round seeds and is dominant to r for wrinkled, Y gives yellow cotyledons and is dominant to y for green; RrYy is heterozygous at both loci. The gametes are determined first.

A gamete carries one allele of each gene. Because the two genes are on different chromosomes, the allele received at one locus is independent of the allele received at the other, by independent assortment at metaphase I. RrYy therefore produces four kinds of gamete, RY, Ry, rY and ry, in equal numbers. Combining each allele of the first gene with each allele of the second lists all four.

Describe the drawing

A four-by-four Punnett square. The four gamete types R Y, R y, r Y and r y run along the top and down the side, and each of the sixteen cells holds the four-letter genotype the pair makes. The cells are shaded by phenotype: nine round and yellow, three round and green, three wrinkled and yellow, one wrinkled and green, with a colour key beneath naming each group and its count.

Four gametes from each parent give sixteen combinations, and sorting them by phenotype gives 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. The ratio 9 : 3 : 3 : 1 is the 3 : 1 of one gene multiplied by the 3 : 1 of the other, so the proportion of offspring that are round and green can be obtained directly as ¾ × ¼ = 3⁄16.

The dihybrid test cross is the more useful experiment. Cross RrYy with the double recessive rryy and the second parent contributes ry to every offspring, so the offspring phenotypes simply read out the gametes the first parent made. Four gamete types in equal numbers give 1 : 1 : 1 : 1, so any departure from that ratio is a departure in the proportions of the gametes themselves. That is why a test cross detects linkage.

Both ratios assume that the two genes assort independently, that neither affects the expression of the other, that fertilisation is random and that every genotype survives equally well; 9 : 3 : 3 : 1 also needs complete dominance at each locus. Observed numbers approach the ratios only in a large sample. Autosomal linkage and epistasis are the two departures treated below.

Autosomal linkage and recombination frequency

A human has around 20 000 genes and 23 chromosome pairs, so genes vastly outnumber chromosomes and most pairs of genes must share one. Two genes on the same autosome are autosomally linked, and they cannot assort independently, because they are on the same DNA molecule. The combination of alleles present on one chromosome is usually inherited together.

Crossing over is what stops linkage being absolute. If a chiasma forms between the two loci, the section carrying one of them is exchanged and a recombinant gamete results. The further apart the two genes sit on the chromosome, the more likely a chiasma is to fall between them, so the proportion of recombinants measures the distance between the loci.

Describe the drawing

Two bar charts side by side, each showing four offspring classes from a test cross totalling four hundred. On the left, for genes on different chromosomes, the four bars are equal at one hundred each. On the right, for linked genes, the two parental classes stand at one hundred and sixty-eight and one hundred and sixty-six while the two recombinant classes, drawn in coral, reach only thirty-four and thirty-two.

recombination frequency = (number of recombinant offspring ÷ total offspring) × 100%Calculated from a test cross, where the offspring phenotypes are a direct read-out of one parent's gametes.

In the figure's right-hand panel, 34 + 32 = 66 recombinants out of 400 offspring gives a recombination frequency of 16.5%. Values close to zero mean tightly linked genes; values approaching 50% are indistinguishable from independent assortment. The 50% ceiling arises because a crossover involves only two of the four chromatids of a bivalent: even if one fell between the loci in every meiosis, only half the products would be recombinant, and with the loci far apart, multiple crossovers average out to the same half. A value above 50% indicates that the parental classes have been counted as recombinant.

Fruit flies give the standard animal example, and Pearson Edexcel Biology B names it. In Drosophila, grey body (G) is dominant to black (g) and long wing (L) to vestigial (l), and the two genes sit on the same autosome. A GgLl female from a GGLL × ggll cross, test-crossed to a ggll male, gives mostly grey-long and black-vestigial offspring; grey-vestigial and black-long together make up only about 17%. The heterozygote used is a female, because male fruit flies do not cross over.

Identifying which classes are recombinant requires the parental chromosomes. In the figure the heterozygote came from EEFF × eeff, so it received EF on one chromosome and ef on the other, making EeFf and eeff the parental types. Had it come from EEff × eeFF the same four classes would appear with the labels swapped.

The two kinds of linkage are distinct. Sex linkage is one gene present on the X and not the Y; autosomal linkage is two genes on the same non-sex chromosome.

Autosomal linkage

The tendency of alleles of genes on the same autosome to be inherited together, unless crossing over separates them.

Recombination frequency

The percentage of offspring of a test cross that are recombinant. It rises with the distance between linked loci and does not exceed 50%.

Epistasis and modified dihybrid ratios

The second common cause of a departure from 9 : 3 : 3 : 1 is that the two genes do not act independently on the phenotype. Epistasis is one gene at one locus affecting the expression of a gene at a different locus. It usually happens because the two gene products act at different steps of the same metabolic pathway, and a block at an early step prevents any later step producing its effect.

Describe the drawing

A two-step pigment pathway drawn across the top: a colourless precursor, an arrow labelled with the enzyme made by gene A leading to brown pigment, and a second arrow labelled with the enzyme from gene B leading to black pigment, each arrow annotated with the homozygous recessive genotype that blocks it. Beneath, the four genotype classes of a dihybrid selfing are listed with the coat colour each gives and its share of sixteen: nine black, three brown, three white and one white.

Epistasis produces a modified dihybrid ratio: a set of numbers that add to 16 but are not 9 : 3 : 3 : 1, because two or more classes have merged. Three patterns are the ones commonly examined.

Read the merged classes to identify the pattern. In 9 : 3 : 4 the 4 is 3 + 1, the double recessive joined by the class sharing its epistatic genotype; in 12 : 3 : 1 the 12 is 9 + 3; in 9 : 7 the 7 is 3 + 3 + 1, so only the double dominant shows anything at all. A fourth pattern, 13 : 3, turns up occasionally. It is a form of dominant epistasis: a dominant allele at the first locus suppresses the second gene, and the double recessive lacks the dominant allele at the second locus altogether, so A_B_, A_bb and aabb look alike. The 13 is 9 + 3 + 1, and only aaB_, the 3, differs.

Epistasis and linkage are distinguished by what changes. Epistasis changes which phenotypes appear and leaves the totals in sixteenths; linkage leaves the phenotype categories unchanged and alters their proportions in favour of the parental combinations.

Epistasis

Interaction between genes at different loci, in which one gene affects or masks the expression of another.

The chi-squared test

Observed numbers do not match predicted ratios exactly, so a genetic experiment ends by asking whether the difference is consistent with sampling variation or whether the hypothesis should be rejected. The chi-squared test answers that question, and it is examined in closely similar terms by AQA, OCR A, Cambridge International and Edexcel B, and by Edexcel A as a maths skill. The formula and critical values are normally given in the paper, so the assessment is of method and conclusion rather than recall.

χ² = Σ (O − E)² ÷ EO is each observed count and E the expected count for the same category. Both must be actual numbers of individuals, never percentages, proportions or the ratio itself.

State the null hypothesis in full. It says that there is no significant difference between the observed results and those expected, here on a 9 : 3 : 3 : 1 ratio. The conclusion must refer back to it.

Calculate the expected values from the ratio and the total. With 160 offspring and a 9 : 3 : 3 : 1 prediction the parts add to 16, so one part is 10 and the expected numbers are 90, 30, 30 and 10. These should sum back to the total.

Describe the drawing

A five-column working table for a chi-squared test on a dihybrid cross. The rows are the four phenotype classes; the columns give the observed count, the expected count worked out from the nine to three to three to one ratio, the difference between them, and that difference squared over the expected value. The terms are totalled to the test statistic, and three lines beneath give the degrees of freedom, the critical value at the five per cent level, and the comparison between the two.

Degrees of freedom is the number of categories minus one, so here 4 − 1 = 3. It is not the number of individuals minus one. Once three class totals and the overall total are known, the fourth is fixed, so only three were free to vary.

Compare against the critical value at p = 0.05, the conventional threshold in biology: the value the statistic would exceed only 5% of the time if the null hypothesis were true.

Printed tables differ over one row. At three degrees of freedom the exact critical value is 7.8147, which rounds to 7.81; some school tables print 7.82, having rounded 7.815 upwards at an earlier step. Use the value supplied in your own examination materials. A χ² falling between 7.81 and 7.82 lies at the boundary, and the test does not separate the hypotheses at that value whichever table is used.

If χ² is below the critical value, the difference is not significant: you fail to reject the null hypothesis and attribute the deviation to chance. (Some mark schemes still accept the word 'accept', but the test has not proved the null true, only failed to find evidence against it, and 'fail to reject' says so.) If χ² is equal to or greater than the critical value, the difference is significant at the 5% level: reject the null hypothesis, because a difference this large would arise by chance less than 5 times in 100. Rejecting it does not say what is going on, only that chance alone is not enough: linkage, epistasis and differential survival are then the candidates.

Two limitations apply. Expected values below about 5 make the test unreliable, so a cross in which any class is expected to hold only three individuals does not meet the assumptions. And a non-significant result does not establish a hypothesis; it shows only that the data provide no evidence against it.