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Mendelian Probability and Allele Frequencies · Tutorial 19 of 989

Mendelian And Hardy–Weinberg Calculation Workshop

Practice connecting Mendelian transmission, sample allele frequencies, Hardy–Weinberg expectations, and uncertainty from finite samples.

Beginner 9 min read

What You'll Learn

  • Calculate offspring genotype probabilities from parental genotypes and transmitted gametes
  • Estimate allele frequencies from diploid genotype counts
  • Derive Hardy–Weinberg genotype probabilities from allele frequencies
  • Convert expected genotype frequencies into expected sample counts
  • Interpret a confidence interval for an allele frequency under a stated sampling model
  • Distinguish sample fluctuations from evidence of a population pattern

From Inheritance to Population Frequencies

A genetic calculation often begins with a family and ends with a population sample. For a family, the key question is which allele each parent can transmit and with what probability. For a population sample, the calculation usually begins with genotype counts, from which allele frequencies can be estimated. Under the Hardy–Weinberg model, those allele frequencies imply expected genotype frequencies. Keeping these steps separate prevents a family-level transmission probability from being confused with a population frequency.

This workshop uses one locus with two alleles, \(A\) and \(a\). The genotype \(AA\) has two copies of \(A\), \(Aa\) has one copy of each allele, and \(aa\) has two copies of \(a\). In a population, let \(p\) be the frequency of \(A\), and let \(q=1-p\) be the frequency of \(a\). A sample of \(n\) diploid individuals contains \(2n\) allele copies.

Definition: A genotype frequency is the proportion of individuals with a specified genotype. An allele frequency is the proportion of allele copies that are a specified allele. In a sample with counts \(n_{AA}\), \(n_{Aa}\), and \(n_{aa}\), the sample frequency of \(A\) is \(\hat{p}=(2n_{AA}+n_{Aa})/(2n)\); each homozygote contributes two copies of its allele, and each heterozygote contributes one.

Mendelian transmission describes how alleles pass from parents to offspring. A heterozygous parent \(Aa\) transmits \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\), assuming ordinary segregation and no transmission distortion. A homozygous \(AA\) parent transmits \(A\) with probability 1. These parental transmission probabilities determine offspring genotype probabilities; they do not, by themselves, describe genotype frequencies in an entire population.

Worked Example: Offspring Probabilities From Parental Genotypes

Worked Example: Offspring Probabilities From Parental Genotypes

Suppose two hypothetical parents are both \(Aa\). Assume each parent transmits either allele with probability \(1/2\), and the allele transmitted by one parent is independent of the allele transmitted by the other. Find the probabilities of the three offspring genotypes.

The first parent can transmit \(A\) or \(a\), each with probability \(1/2\). The second parent has the same transmission probabilities. The four equally likely combinations are \(A\) with \(A\), \(A\) with \(a\), \(a\) with \(A\), and \(a\) with \(a\). Therefore,

$$ P(AA)=\frac{1}{2}\frac{1}{2}=\frac{1}{4},\qquad P(Aa)=\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}=\frac{1}{2},\qquad P(aa)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}. $$

The two middle combinations both produce \(Aa\), which is why its probability is \(1/2\), not \(1/4\). The probabilities sum to \(1/4+1/2+1/4=1\), as they must. These are probabilities for an offspring of this particular parental pair; they are not estimates of population genotype frequencies.

Hardy–Weinberg Expectations

The Hardy–Weinberg model connects allele frequencies to genotype frequencies when alleles unite randomly in a population. In its basic two-allele form, the model assumes random union of gametes with allele frequencies \(p\) and \(q\). The familiar equilibrium interpretation also relies on conditions such as no selection, mutation, migration, or other systematic changes in allele frequencies between generations, and sufficiently large population size to make random sampling effects negligible for the model's population expectations.

Definition: Under Hardy–Weinberg equilibrium for two alleles with frequencies \(p\) and \(q=1-p\), the genotype probabilities are \(P(AA)=p^2\), \(P(Aa)=2pq\), and \(P(aa)=q^2\). In a sample of \(n\) individuals, the corresponding expected counts, when \(p\) and \(q\) are specified, are \(np^2\), \(2npq\), and \(nq^2\).

The probabilities follow directly from the possible allele pairs formed by random union of gametes. A gamete carries \(A\) with probability \(p\) and \(a\) with probability \(q\). The offspring is \(AA\) only if both gametes carry \(A\); it is \(aa\) only if both carry \(a\). Either of two gamete orders produces a heterozygote.

$$ P(AA)=p\cdot p=p^2,\qquad P(Aa)=p\cdot q+q\cdot p=2pq,\qquad P(aa)=q\cdot q=q^2. $$

The probabilities add to one because \(p^2+2pq+q^2=(p+q)^2=1\). Multiplying each probability by the sample size gives an expected count. These are model-based expectations, not a promise that a finite sample will contain exactly those counts.

Worked Example: Estimating Allele Frequencies and Calculating Expected Counts

In a hypothetical sample of \(n=200\) individuals, suppose the observed counts are 92 \(AA\), 92 \(Aa\), and 16 \(aa\). Estimate the allele frequencies, then calculate genotype counts expected under Hardy–Weinberg equilibrium using those estimated frequencies.

There are \(2(200)=400\) allele copies. The observed \(A\)-copy count is \(2(92)+92=276\), because each \(AA\) individual contributes two \(A\) copies and each \(Aa\) individual contributes one. Thus,

$$ \hat{p}=\frac{2(92)+92}{2(200)} =\frac{276}{400}=0.6900,\qquad \hat{q}=1-\hat{p}=0.3100. $$

The corresponding Hardy–Weinberg genotype probabilities, using these sample estimates, are

$$ \hat{p}^{\,2}=0.6900^2=0.4761,\qquad 2\hat{p}\hat{q}=2(0.6900)(0.3100)=0.4278,\qquad \hat{q}^{\,2}=0.3100^2=0.0961. $$

Multiplying each by 200 gives expected counts of \(200(0.4761)=95.22\) \(AA\), \(200(0.4278)=85.56\) \(Aa\), and \(200(0.0961)=19.22\) \(aa\). Their sum is \(95.22+85.56+19.22=200.00\). Expected counts need not be whole numbers: they are averages implied by the model, not literal fractional individuals. The observed and expected counts differ, but the difference alone does not establish a systematic departure from Hardy–Weinberg equilibrium; finite samples fluctuate.

Allele-Frequency Estimates Have Sampling Uncertainty

An allele frequency calculated from a sample is an estimate of a population quantity. If a sample contains \(2n\) independently sampled allele copies and the population frequency of \(A\) is \(p\), the count of \(A\) copies follows a binomial sampling model with \(2n\) trials and success probability \(p\). The sample proportion \(\hat{p}\) then has approximate standard error \(\sqrt{\hat{p}(1-\hat{p})/(2n)}\), using the estimate in place of the unknown population frequency.

When the sample is sufficiently large and the estimated frequency is not extremely close to zero or one, a normal-approximation 95% confidence interval is \(\hat{p}\pm1.96SE\). The interval describes uncertainty from sampling under the stated model. It does not account automatically for related individuals, population structure, genotyping errors, or other ways the sampling assumptions may fail.

Worked Example: A Confidence Interval for an Allele Frequency

Suppose a hypothetical random sample of 200 unrelated diploid individuals yields an estimated \(A\)-allele frequency of \(\hat{p}=0.7000\), based on 400 allele copies. Assume the allele copies are sampled independently under the binomial model. Calculate an approximate 95% confidence interval for the population frequency \(p\).

First calculate the estimated standard error:

$$ SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{2n}} =\sqrt{\frac{(0.7000)(0.3000)}{400}} =\sqrt{0.000525} \approx 0.02291. $$

The margin of error, using the unrounded standard error, is approximately \(1.96(0.0229129)=0.04491\). Therefore, the interval is

$$ 0.7000\pm 1.96(0.0229129) =0.7000\pm0.04491 \approx (0.6551,\ 0.7449). $$

Under the binomial sampling model and normal approximation, this procedure produces intervals that contain the population \(A\)-allele frequency in about 95% of repeated samples. For this sample, the calculated interval runs from approximately 0.6551 to 0.7449, or 65.51% to 74.49%. It is not correct to interpret the interval as a 95% probability statement about the fixed population frequency after the data have been observed.

A Short Derivation: Why the Sample Allele-Frequency Estimate Is Centered on \(p\)

Under the independent allele-sampling model, define \(X_i=1\) if sampled allele copy \(i\) is \(A\), and \(X_i=0\) otherwise, for \(i=1,\ldots,2n\). Each indicator has expected value \(E(X_i)=p\). The sample frequency is the average of these indicators, so its expected value is

$$ E(\hat{p})=E\left(\frac{1}{2n}\sum_{i=1}^{2n}X_i\right) =\frac{1}{2n}\sum_{i=1}^{2n}E(X_i) =\frac{1}{2n}(2np)=p. $$

Thus, under this sampling model, \(\hat{p}\) is unbiased: across repeated samples its average equals the population frequency. This does not mean every sample estimate equals \(p\); the standard error quantifies its typical sampling variability.

Putting the Calculations Together

A reliable sequence for an integrated problem is to identify what is being calculated at each stage. Parental genotypes determine transmission probabilities. Sample genotype counts determine allele-frequency estimates. A Hardy–Weinberg model converts specified or estimated allele frequencies into genotype probabilities and expected counts. A sampling model describes how much an estimate may vary from sample to sample.

1
Start with the inheritance or sample information.
For a family, list possible gametes and their probabilities. For a population sample, record the genotype counts and total sample size.
2
Calculate the relevant allele probabilities or estimate.
Use parental transmission probabilities for offspring, or count allele copies to calculate \(\hat{p}\) and \(\hat{q}\) from genotypes.
3
Apply the population model only when appropriate.
Under Hardy–Weinberg equilibrium, calculate \(p^2\), \(2pq\), and \(q^2\); multiply by \(n\) for expected counts.
4
Interpret the result at the correct level.
Distinguish an offspring probability from a population frequency, and distinguish an expected count from an observed count or a sample estimate.

Common Mistakes and Interpretation

The most common error is to combine quantities that answer different questions. The probability that an \(Aa\times Aa\) pair produces an \(aa\) offspring is \(1/4\). That does not imply that 25% of individuals in every population are \(aa\). A population prediction of \(q^2\) requires a population allele frequency \(q\) and the Hardy–Weinberg model.

  • Counting genotypes instead of allele copies. The frequency of \(A\) is not the fraction of individuals with at least one \(A\). Count two copies for each \(AA\), one for each \(Aa\), and none for \(aa\), then divide by \(2n\).
  • Using \(pq\) for the heterozygote frequency. The pair \(A,a\) can arise in either order, so the Hardy–Weinberg heterozygote probability is \(pq+qp=2pq\).
  • Treating expected counts as observed counts. A model expectation can be non-integer and need not match the sample. Differences can arise from sampling variation.
  • Overstating a confidence interval. A 95% confidence procedure has its repeated-sampling interpretation under its assumptions; it does not assign a 95% probability to the fixed parameter being in this particular interval.
Key takeaway: Keep transmission, frequency estimation, population expectations, and sampling uncertainty distinct. Each step uses different information and supports a different interpretation.

Check Your Understanding

Try each question by showing the calculation and stating what the result describes.

  1. For a hypothetical \(Aa\times aa\) cross, what is the probability that an offspring is \(Aa\), assuming ordinary Mendelian segregation?
  2. A sample contains 30 \(AA\), 40 \(Aa\), and 30 \(aa\) individuals. Calculate \(\hat{p}\) and \(\hat{q}\).
  3. If \(p=0.60\) under Hardy–Weinberg equilibrium, calculate the expected genotype frequencies and check that they sum to one.
  4. For a sample of 100 individuals with \(p=0.60\) used as the specified model frequency, what expected count of heterozygotes does Hardy–Weinberg equilibrium imply?
  5. Why does a difference between observed and Hardy–Weinberg expected counts not, by itself, show that a population departs systematically from the model?

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