Punnett squares: monohybrid and dihybrid crosses

A Punnett square is a bookkeeping grid. One parent's gametes go across the top, the other parent's gametes go down the side, and each box is one equally likely offspring. Nearly every error on a genetics exam comes from the step before the grid: writing the wrong gametes. Get those right and the rest is counting.

Everything below assumes complete dominance and genes on different chromosomes, which is what OpenStax Biology 2e (sections 12.2 and 12.3) uses for its examples.

Step 1: write the gametes

Mendel's law of segregation says each gamete carries one allele of each gene. So:

• Aa makes two kinds of gametes, A and a, in equal numbers.
• AA makes only A gametes. aa makes only a gametes.
• With two genes, the law of independent assortment says the alleles of one gene sort into gametes independently of the other. AaBb makes four kinds: AB, Ab, aB and ab, one quarter each.

A useful check: a parent that is heterozygous for n genes makes 2^n gamete types. One gene gives 2, two genes give 4, three give 8.

Example 1: a monohybrid cross, Aa x Aa

Put A and a across the top and A and a down the side:

• Genotype ratio: 1 AA : 2 Aa : 1 aa, or 1:2:1.
• Phenotype ratio: AA and Aa both show the dominant trait, so 3 dominant : 1 recessive, or 3:1. That is 75 percent and 25 percent.

This is Mendel's F2 result. His violet-by-white pea cross gave 705 violet and 224 white plants in the F2: 705 / 224 = 3.15, close to 3:1.

A common follow-up question: of the dominant-looking offspring, what fraction are heterozygous? There are 3 dominant boxes and 2 of them are Aa, so the answer is 2/3, not 1/2. The other 1/3 are AA.

Example 2: a test cross

You have a plant with the dominant phenotype and don't know if it is AA or Aa. Cross it with a homozygous recessive (aa):

• If it is AA, every offspring is Aa and shows the dominant trait.
• If it is Aa, you get 1 Aa : 1 aa, so half the offspring show the recessive trait.

One recessive offspring proves the parent is Aa. The reverse is weaker. If an Aa parent is test-crossed and three offspring are all dominant, that happens with probability (1/2)^3 = 1/8 = 0.125, so three dominant offspring can't prove the parent is AA.

Example 3: a dihybrid cross, AaBb x AaBb

Each parent makes four gamete types, so the grid is 4 x 4 = 16 boxes.

Count phenotypes, using A_ for "at least one A":

• A_B_ (both dominant): 9 of 16, 56.25 percent
• A_bb (dominant A, recessive b): 3 of 16, 18.75 percent
• aaB_ (recessive a, dominant B): 3 of 16, 18.75 percent
• aabb (both recessive): 1 of 16, 6.25 percent

That's the 9:3:3:1 ratio. There are nine different genotypes, in a 1:2:1:2:4:2:1:2:1 ratio, with AaBb the most common at 4 of 16.

Real data comes close to 9:3:3:1 but won't match it exactly. OpenStax gives an F2 of 2,706 : 930 : 888 : 300 (4,824 plants). The expected counts are 4,824 x 9/16 = 2,713.5, then 904.5, 904.5 and 301.5. Whether a gap like that is just chance is what a chi-square test answers.

Skip the grid: the product rule

The 9:3:3:1 ratio is two 3:1 ratios multiplied together. Because the genes assort independently, you can work each gene alone and multiply ("and" means multiply):

• P(A_) = 3/4 and P(B_) = 3/4, so P(A_B_) = 3/4 x 3/4 = 9/16.
• P(A_bb) = 3/4 x 1/4 = 3/16.
• P(AaBb) = P(Aa) x P(Bb) = 1/2 x 1/2 = 1/4, which matches the 4 of 16 boxes above.

For "or" questions, add the probabilities of the separate outcomes (the sum rule). The chance of showing exactly one recessive trait is 3/16 + 3/16 = 6/16 = 3/8.

This matters once there are more than two genes. A cross between two parents heterozygous for four genes needs 16 x 16 = 256 boxes, but the product rule gets P(aabbccdd) = (1/4)^4 = 1/256 in one line, and P(dominant for all four traits) = (3/4)^4 = 81/256.

Example 4: a mixed cross, AABb x aaBb

Split it into two monohybrid crosses:

• AA x aa: every offspring is Aa.
• Bb x Bb: 1 BB : 2 Bb : 1 bb.

Multiply: 1 AaBB : 2 AaBb : 1 Aabb, and a phenotype ratio of 3 A_B_ : 1 A_bb. A full 4 x 4 grid gives the same answer, just more slowly.

When 9:3:3:1 doesn't apply

The ratios above assume complete dominance and unlinked genes. Change either and the phenotype ratio changes:

• Incomplete dominance: red x white snapdragons give pink, and pink x pink gives 1 red : 2 pink : 1 white. The phenotype ratio matches the 1:2:1 genotype ratio.
• Epistasis: in OpenStax's mouse coat example, AaCc x AaCc gives 9 agouti : 3 solid (black) : 4 albino, because cc is albino no matter what the A gene says.
• Linkage: genes close together on one chromosome travel together, so gametes are mostly the parental types and the offspring ratio no longer comes out 9:3:3:1.

Practice

The Punnett square calculator builds the grid and both ratios for any one- or two-gene cross, which makes checking practice problems quick. These crosses sit in unit 12 of the course; what's in General Biology I shows where the rest of genetics comes in. Population-level genetics picks up in Hardy-Weinberg equilibrium explained, and how to memorize a list, a table, or a formula sheet helps with the ratios you need to know cold.

Encodr's free General Biology I course has a full chapter on Punnett squares and test crosses, with every ratio checked by script.