Biology by Bradford
Hardy–Weinberg and the 9:3:3:1 ratio look like two different topics. They are the same move twice: a model that predicts expected numbers, and a statistic that decides whether the real numbers agree.
p is the frequency of the dominant allele and q the recessive one. Between them they account for every allele at that locus, so p + q = 1.
Every zygote is two gametes drawn from that pool, so squaring gives the genotype frequencies: p² + 2pq + q² = 1. The 2 in front of pq is there because a heterozygote can be made two ways round.
Only the recessive phenotype tells you its genotype for certain. Count those, and q² is the only term you can measure directly: every other value comes from it.
Almost every exam question follows the same path: count the recessive phenotype, take the square root, subtract from one, then build whatever the question asked for.
Real populations break at least one of these · which is exactly why the model is useful as a comparison
Drag q, or type in how many individuals show the recessive phenotype and let the bench work backwards. Watch what happens to the carriers as the recessive allele becomes rare.
Set q to something small, 0.02, say, and compare the two ends of the bar. For a rare recessive condition the carriers outnumber the affected individuals by an enormous margin, and that single fact explains why recessive alleles are so hard to eliminate from a population.
Hardy–Weinberg predicts how many of each genotype a population should contain. A dihybrid cross predicts how many of each phenotype a set of offspring should contain. Neither prediction is ever met exactly, so both need the same question asking of them: is the gap too big to be chance?
Expected number = predicted proportion × total. For Hardy–Weinberg the proportions are p², 2pq and q². For two heterozygous parents they are 9/16, 3/16, 3/16 and 1/16. The arithmetic is identical; only the fractions change.
χ² = Σ (O − E)²/E, with degrees of freedom = classes − 1. For the four phenotype classes of a dihybrid cross that is 3, and the critical value at p = 0.05 is 7.815. Above it, reject the model.
Loaded with the snail data from the problem below. Change any observed count and everything recalculates.
320 offspring from two snails heterozygous for both genes. Expected values come straight from 9:3:3:1 × 320. Work out each (O−E)²/E on paper and check yourself against the last column.
Attempt each part on paper before opening the solution. Part b is where the two topics meet.
In a species of snail, dark shell (D) is dominant to light shell (d), and unbanded (U) is dominant to banded (u). The two genes are on different chromosomes.
(a) In a population of 400 snails, 64 have light shells. Calculate the frequency of each allele and the expected number of snails of each genotype.
(b) Two snails heterozygous for both genes are crossed. Of 320 offspring, 171 are dark unbanded, 63 dark banded, 71 light unbanded and 15 light banded. Use a chi-squared test to decide whether these results are consistent with the expected ratio.
Light-shelled snails must be dd, so they are the q² class. q² = 64 ÷ 400 = 0.16.
q = √0.16 = 0.4, and since p + q = 1, p = 1 − 0.4 = 0.6.
p² = 0.6² = 0.36 → 0.36 × 400 = 144 DD. 2pq = 2 × 0.6 × 0.4 = 0.48 → 192 Dd. q² = 0.16 → 64 dd. Check: 144 + 192 + 64 = 400. ✓
Note that 192 of the 336 dark snails are carriers: you could never have found that by counting.
Both parents are DdUu, so with unlinked genes the expected phenotype ratio is 9 : 3 : 3 : 1. Expected = ratio × total: 9/16 × 320 = 180, 3/16 × 320 = 60, 3/16 × 320 = 60, 1/16 × 320 = 20.
H₀: there is no significant difference between the observed and expected numbers. H₁: there is a significant difference.
χ² = 0.450 + 0.150 + 2.017 + 1.250 = 3.87. Four classes, so df = 4 − 1 = 3, and the critical value at p = 0.05 is 7.815. 3.87 is well below it, so p > 0.05.
The cross in Problem 1 is repeated with a different pair of snails. This time the 320 offspring are: 205 dark unbanded, 40 dark banded, 55 light unbanded, 20 light banded. Test these results against the expected ratio and explain what they suggest.
Same cross, same total, so the same prediction: 180 : 60 : 60 : 20. Only the observed numbers have changed.
χ² = 3.472 + 6.667 + 0.417 + 0.000 = 10.56, with df = 3. This exceeds 7.815, so p < 0.05 and the difference is significant.
Reject H₀. The two dominant traits appear together far more often than predicted and the mixed classes are depleted: the signature of gene linkage. A statistical result on its own is not a conclusion; you have to say what the deviation means.
In humans, the MN blood group is controlled by two codominant alleles, so all three genotypes are distinguishable: M (genotype MM), MN (genotype MN) and N (genotype NN). A sample of 500 people from one town gives 186 M, 228 MN and 86 N. Test whether this population is in Hardy–Weinberg equilibrium at this locus.
Because the alleles are codominant, every genotype is visible: there is no dominant phenotype hiding two genotypes. So you do not use √(q²) here; you count the alleles directly.
500 people carry 1000 alleles. M alleles: (2 × 186) + 228 = 600. N alleles: (2 × 86) + 228 = 400. So p(M) = 600/1000 = 0.6 and q(N) = 400/1000 = 0.4.
p²N = 0.6² × 500 = 180 MM. 2pqN = 2 × 0.6 × 0.4 × 500 = 240 MN. q²N = 0.4² × 500 = 80 NN. These are the expected values: and they came out of the population model, not out of a cross ratio.
χ² = 0.200 + 0.600 + 0.450 = 1.25.
Here df is not classes − 1. Because p and q were estimated from the sample itself, you lose one more: df = classes − alleles estimated − 1 = 3 − 1 − 1 = 1. The critical value at p = 0.05 with df = 1 is 3.841.
1.25 < 3.841, so p > 0.05: fail to reject H₀. A population that had, say, far more homozygotes than expected would push χ² above 3.841 and the same method would reject equilibrium. Try exactly that in the interactive below.
Two df, two topics. A cross tested against 9:3:3:1 uses df = 3, because the expected ratio is fixed in advance. A population tested against Hardy–Weinberg uses df = 1 for a two-allele locus, because you spent the data itself estimating p and q. Same statistic, different bookkeeping: and that difference is the clearest sign of which model you were really using.
Both halves of this lesson do the same thing: a model supplies E, and χ² decides whether O is close enough to keep believing the model
This is the type that ties the topic together: you cannot run the chi-squared test until you have run Hardy–Weinberg first, because the population model is the only thing that supplies the expected numbers. Enter the observed counts, count the alleles, and watch it chain through to the verdict.
The expected column is pure Hardy–Weinberg: p²N, 2pqN, q²N. Change a single observed count and the allele frequencies shift, which moves every expected value, which changes χ². The two models are not side by side here; one feeds the other.
Tap a concept to light up how it connects.
Both models exist to be compared against, not to be believed. Hardy–Weinberg describes a population where nothing is happening; 9:3:3:1 describes a cross where the genes are independent. The interesting biology is in the deviations.
Tap any concept to trace its connections · tap the background to reset
Drag each term into the gap it belongs in. Two terms are traps.
Single best answer, Paper 1 style. A calculator will help on two of them.