20 Punnett Square Problems to Practice On
A Punnett square is just a grid of every possible way two parents' alleles can combine. Once you can build that grid without thinking, every genetics problem turns into the same three moves: figure out the gametes, fill the grid, read off the ratio.
Twenty problems below, starting with a single gene and complete dominance, then working up through incomplete dominance, codominance, two genes at once, and sex-linked traits. Try each one yourself before opening the solution. That's where the pattern actually sticks.
Section 1: Monohybrid Crosses with Complete Dominance
Start with one gene, two alleles, and a clear dominant-recessive relationship. This is the grid mechanics you'll reuse in every section after this one.
Problem 1
In pea plants, tall () is dominant over short (). Two heterozygous plants () are crossed. What are the genotype and phenotype ratios of the offspring?
Step 1: Build the grid. Each parent produces and gametes.
Step 2: Read the ratios. Genotypes: . Since is dominant, and both look tall.
Problem 2
A homozygous tall pea plant () is crossed with a homozygous short plant (). What fraction of the offspring will be tall?
Step 1: Find the gametes. only produces gametes, only produces gametes.
Step 2: Fill the grid. Every offspring is .
Problem 3
A heterozygous tall plant () is crossed with a short plant (). What's the phenotype ratio of the offspring?
Step 1: Build the grid. gives ; gives .
Step 2: Read the ratio. Half the offspring are (tall), half are (short).
Problem 4
In pea plants, purple flowers () are dominant over white (). Two heterozygous plants () produce 160 offspring. How many are expected to have white flowers?
Step 1: Get the ratio. gives purple white, so white is of the offspring.
Step 2: Apply it to 160.
Section 2: Working Backward from Offspring Ratios
Now flip the question around. Instead of predicting offspring from known parents, you're given what showed up and have to figure out a genotype or a probability. Same grid, read in reverse.
Problem 5
In guinea pigs, black fur () is dominant over white (). A black guinea pig is crossed with a white guinea pig, and the offspring come out black to white. What is the genotype of the black parent?
Step 1: Test the possibilities. If the black parent were , every offspring would get a and be black, a black result, not .
Step 2: Try . gives , which is exactly .
Problem 6
Both parents are for a gene with complete dominance. What is the probability that a given offspring is homozygous, either or ?
Step 1: Get the genotype ratio. gives .
Step 2: Add the homozygous outcomes.
Problem 7
In fruit flies, red eyes () are dominant over sepia eyes (). Two red-eyed flies are crossed and produce 96 offspring, of which 24 have sepia eyes. What are the genotypes of the parents?
Step 1: Find the observed ratio.
Step 2: Match it to a cross. A recessive ratio only comes from two heterozygotes.
Problem 8
Both parents are carriers () for a recessive genetic disorder. What is the probability that a child is unaffected but still a carrier?
Step 1: Get the genotype ratio. gives .
Step 2: Pick out the carrier genotype. "Unaffected but a carrier" means specifically, not .
Section 3: Incomplete Dominance and Codominance
Not every trait follows a clean dominant-recessive rule. Sometimes the heterozygote blends both phenotypes, and sometimes it shows both at once. The grid mechanics don't change, but the phenotype you read off the genotype does.
Problem 9
In snapdragons, flower color shows incomplete dominance. Red () and white () plants are crossed. What phenotype do the offspring have?
Step 1: Build the grid. gives only , gives only .
Step 2: Read the phenotype. Every offspring is . Since neither allele is dominant, the heterozygote blends the two colors.
Problem 10
Two pink snapdragons () are crossed. What is the phenotype ratio of the offspring?
Step 1: Build the grid.
Step 2: Read the phenotypes. is red, is pink, is white.
Problem 11
In cattle, coat color is codominant: red () and white () alleles are both expressed in the heterozygote, producing roan. A roan bull () is crossed with a red cow (). What fraction of the offspring are expected to be roan?
Step 1: Build the grid. Bull gives ; cow gives only .
Step 2: Read the ratio. Half the offspring are (red), half are (roan).
Problem 12
In humans, ABO blood type alleles and are codominant, and both are dominant over . A parent with blood type AB () and a parent with blood type O () have a child. What blood types are possible, and in what ratio?
Step 1: Build the grid. AB parent gives ; O parent gives only .
Step 2: Read the phenotypes. is type A, is type B. Neither nor is possible.
Section 4: Dihybrid Crosses
Now track two genes at once. The grid grows to , but each gene still segregates on its own, independent assortment just means you can multiply the two single-gene ratios together instead of tracking every combination by hand.
Problem 13
In pea plants, round seeds () are dominant over wrinkled (), and yellow seed color () is dominant over green (). The genes assort independently. A double heterozygote is self-crossed (). What is the phenotype ratio of the offspring?
Step 1: Split into two single-gene crosses. gives round wrinkled. gives yellow green.
Step 2: Multiply the two ratios together.
Problem 14
From the same cross in Problem 13 (), 320 offspring are produced. How many are expected to have wrinkled, green seeds?
Step 1: Get the fraction. Wrinkled green is the class, of a ratio.
Step 2: Apply it to 320.
Problem 15
A double heterozygote is test-crossed with a double recessive (). What is the phenotype ratio of the offspring?
Step 1: Split into two single-gene crosses. gives round wrinkled. gives yellow green.
Step 2: Multiply the ratios.
Problem 16
Cross . What is the phenotype ratio of the offspring?
Step 1: Handle the gene first. gives only offspring, so every offspring is round. This gene doesn't split the ratio at all.
Step 2: Handle the gene. gives yellow green.
Step 3: Combine. Since every offspring is already round, the seed-color split is the whole story.
Section 5: Sex-Linked Traits and Combined Problems
X-linked genes behave differently because males only carry one X. These problems also stack a second skill on top, multiple alleles or a second gene, so you're combining two things you've already practiced.
Problem 17
In humans, red-green color blindness is X-linked recessive. A color-blind man () has children with a homozygous normal woman (). What are the possible genotypes of the daughters and sons?
Step 1: Find the gametes. Father gives or ; mother gives only .
Step 2: Build the grid.
Step 3: Read the results. Every daughter is (a carrier, but normal vision), every son is (normal vision).
Problem 18
A carrier woman () has children with a color-blind man (). What fraction of the daughters are expected to be color-blind, and what fraction of the sons?
Step 1: Find the gametes. Mother gives or ; father gives or .
Step 2: Build the grid.
Step 3: Split by sex. Daughters are or , a split. Sons are or , also .
Problem 19
For human ABO blood type, a heterozygous type A parent () and a heterozygous type B parent () have a child. What blood types are possible, and what fraction of each?
Step 1: Find the gametes. Type A parent gives or ; type B parent gives or .
Step 2: Build the grid.
Step 3: Read the phenotypes. Each of the four boxes is a different blood type: AB, A, B, O.
Problem 20
In fruit flies, red eye color is X-linked dominant () over white (), and long wings are autosomal dominant () over vestigial (). A female heterozygous for both traits () is crossed with a white-eyed, vestigial-winged male (). What fraction of the offspring are expected to be white-eyed sons with vestigial wings?
Step 1: Solve the eye-color cross on its own. gives four equally likely outcomes: , , , , each of the offspring. A white-eyed son is the class.
Step 2: Solve the wing cross on its own. gives long vestigial.
Step 3: Multiply, since the two genes assort independently.
A note on notation: superscript letters on the X, like or , mark which allele of an X-linked gene a chromosome carries. A chromosome never carries that gene, which is exactly why males only need one copy of an X-linked allele to show its phenotype.