Biology › Inheritance and population genetics › Two genes at once: 9 : 3 : 3 : 1, the ways it breaks, and the chi-squared test

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

Follow two genes through a cross and the numbers get bigger but the logic does not change. What changes is what a departure from the expected ratio means — and deciding whether a departure is real is a job for a statistic.

Before this Monohybrid inheritance · Meiosis and the origin of variation

What you should be able to do

• Work out the gametes a dihybrid parent produces and use them to build a 16-cell Punnett square.
• Predict the 9 : 3 : 3 : 1 and 1 : 1 : 1 : 1 ratios and say what each assumes.
• Recognise autosomal linkage from a test cross result and calculate a recombination frequency.
• Identify epistasis from a modified dihybrid ratio and explain it through a pathway.
• Carry out a chi-squared test in full, from null hypothesis to a conclusion that names the probability.

Two genes, four gametes, sixteen boxes

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 first job is always the gametes.

A gamete gets one allele of each gene. Because the two genes are on different chromosomes, the choice at one locus is independent of the choice at the other — that is independent assortment at metaphase I, doing the work. So RrYy produces four kinds of gamete, RY, Ry, rY and ry, in equal numbers. Take each allele of the first gene with each allele of the second and you cannot miss one.

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. Notice that 9 : 3 : 3 : 1 is just the 3 : 1 of one gene multiplied by the 3 : 1 of the other, which is worth knowing because it means you can answer 'what proportion of offspring are round and green' as ¾ × ¼ = 3⁄16 without drawing anything.

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. Any departure from that is a departure in the gametes themselves, which is exactly what makes the test cross the tool for detecting linkage.

Both ratios assume the same four things: the two genes assort independently, fertilisation is random, every genotype survives equally well, and the sample is large. Each of those can fail, and the rest of this lesson is about what the failures look like.

Autosomal linkage: genes that travel together

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 physically attached. Whatever combination of alleles lies on a chromosome tends to leave in a gamete 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.

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, because a chiasma between the loci becomes as likely as not. A value above 50% means you have counted the parental classes as recombinant.

Deciding which classes are recombinant needs the parental chromosomes. In the figure the heterozygote came from RRYY × rryy, so it received RY on one chromosome and ry on the other, making RrYy and rryy the parental types. Had it come from RRyy × rrYY the same four classes would appear with the labels swapped.

Do not confuse the two kinds of linkage. Sex linkage is one gene sitting on the X and not the Y; autosomal linkage is two genes sharing a non-sex chromosome.

Epistasis: one gene overruling another

The other common reason a dihybrid ratio comes out wrong is that the two genes are not acting 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 blocking an early step makes everything downstream irrelevant.

The tell is 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 cover almost every question set.

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, where the 13 is 9 + 3 + 1.

Epistasis and linkage produce different symptoms and are diagnosed differently. Epistasis changes which phenotypes appear and leaves the totals in sixteenths; linkage leaves the phenotype categories alone and changes their proportions, usually dramatically, in favour of the parental combinations.

Chi-squared, set out the way it is marked

Observed numbers never match predicted ratios exactly, so every genetic experiment ends with the same question: is this difference the ordinary wobble of sampling, or is the hypothesis wrong? The chi-squared test answers it, and it is examined by AQA, OCR A and Cambridge International in almost identical terms. The formula and a table of critical values are provided on the paper or within the question in all three, so the marks are for 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.

Start with the null hypothesis, and write it down. It always says there is no significant difference between the observed results and those expected — here, on a 9 : 3 : 3 : 1 ratio. A conclusion that never mentions it is incomplete.

Get 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. Check they sum back to the total; if they do not, the arithmetic is already wrong.

Degrees of freedom is the number of categories minus one — here 4 − 1 = 3. It is not the number of individuals minus one, which is the commonest error on this topic and turns a correct statistic into a wrong conclusion. The reasoning is that 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.

One row of that table is worth a sentence, because printed tables disagree about it. At three degrees of freedom the exact critical value is 7.8147, so it rounds to 7.81 — but plenty of school tables print 7.82, having rounded 7.815 upwards at an earlier step. Use whichever your own booklet gives. The gap is far too small to change a verdict: if your χ² lands between 7.81 and 7.82 you are so close to the boundary that the honest conclusion is that the test cannot separate the hypotheses, whichever value you compare against.

If χ² is below the critical value, the difference is not significant: accept the null hypothesis and attribute the deviation to chance. 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 practical cautions. Expected values below about 5 make the test unreliable, so a cross producing only three individuals in a class is not worth testing. And a non-significant result never proves a hypothesis; it shows the data give no reason to abandon it.

In the exam

• Get the gametes right first. RrYy gives RY, Ry, rY and ry; writing RR or Yy as a gamete loses the question in its first line.
• 9 : 3 : 3 : 1 assumes the genes are on different chromosomes. If a question hands you numbers that are nowhere near it, linkage and epistasis are the two things to think about.
• Chi-squared uses raw counts. Converting to percentages first changes the answer and is a standard mark scheme rejection.
• Degrees of freedom is categories minus one. Not individuals, not individuals minus one.
• Write the conclusion in full: statistic against critical value, degrees of freedom, probability level, verdict on the null hypothesis. Each is separately credited.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

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