Punnett square practice problems become easier when you name the event before calculating its probability. Are you finding a genotype, a visible trait, or a result among a smaller group of offspring? Those questions can use the same square and still have different answers.
These eight original problems use fictional plants so the assumptions stay clear. Work each one before reading its explanation. Unless a problem says otherwise, assume ordinary Mendelian segregation, equally likely gametes at each single locus, random fertilization, and no differences in survival between offspring genotypes. These are classroom models, not predictions about a person’s complex traits.
Build the square from gametes, not whole parent genotypes
At one locus, an Aa parent contributes A or a to a gamete. Put one parent’s possible gametes across the top and the other’s down the side. Each cell combines one contribution from each parent. Under equal gamete probabilities, each cell in a two-by-two square has probability one fourth.
For Aa crossed with Aa, the four cells are AA, Aa, aA, and aa. The two middle outcomes represent the same heterozygous genotype, normally written Aa. If A is completely dominant, AA and Aa share the dominant phenotype. See OpenStax’s explanation of segregation and independent assortment for the biological basis.
A square predicts probabilities under its assumptions. It doesn’t promise that four offspring will occupy its four cells once each. The University of Utah’s genetics teaching resource also explains why simple Mendelian examples shouldn’t be extended casually to complex traits.
1. One cross, two different ratios
Problem: In a fictional plant, P produces purple flowers and is completely dominant to p, which produces white flowers when homozygous. Cross Pp with Pp. Give both the genotype ratio and phenotype ratio.
Answer: The cells are PP, Pp, Pp, and pp. The genotype ratio is 1 PP : 2 Pp : 1 pp. The phenotype ratio is 3 purple : 1 white. For one offspring, the probability of being heterozygous is 2/4, or 1/2. The probability of purple flowers is 3/4.
If you wrote 3:1 for the genotype ratio, you combined distinct genotypes because they look alike. If you wrote 1:2:1 for the phenotype ratio, you treated the heterozygote as a third visible category despite the stated complete dominance. Label each ratio before filling in its numbers. That small step makes it much easier to see which outcomes belong together.
2. A testcross with a known heterozygote
Problem: Cross a purple Pp plant with a white pp plant. Find the chance of white offspring and the chance of PP offspring.
Answer: The first parent supplies P or p with probability 1/2 each. The second supplies only p. Half the offspring are expected to be Pp and half pp. White has probability 1/2. PP has probability zero because the pp parent cannot contribute P.
You can draw a two-by-two square with p written twice down one side. It will contain two Pp and two pp cells. You can also use a two-by-one layout, provided you attach the correct probabilities. The size of your drawing doesn’t create new kinds of gametes. Always check whether each proposed offspring allele could actually come from the parent assigned to that side.
3. Work backward without claiming more than the evidence shows
Problem: A purple plant of unknown genotype is crossed with pp. One offspring is white. Under the stated model, what does that show about the purple parent? Would ten purple offspring and no white offspring have proved the parent was PP?
Answer: A white offspring is pp, so the purple parent contributed p. Because that parent is purple, it also has P. It must therefore be Pp under this model.
Ten purple offspring alone wouldn’t prove PP. If the parent were Pp, each offspring would have probability 1/2 of being purple. Assuming independent offspring outcomes, the chance of ten purple offspring is (1/2)10 = 1/1024, approximately 0.098%. That’s unlikely, but not impossible. Don’t turn a finite sample into certainty. The conclusion from an observed recessive offspring differs from the conclusion drawn from not observing one.
4. Incomplete dominance changes the phenotype answer
Problem: At a different fictional flower-color locus, RR plants are red, RW plants are pink, and WW plants are white. Cross two pink plants. What proportion is expected to be pink?
Answer: Each RW parent makes R and W gametes. The cells are RR, RW, RW, and WW. Pink offspring have probability 2/4 = 1/2. The phenotype ratio is 1 red : 2 pink : 1 white, matching the three genotype categories in this example.
Don’t import the earlier purple-flower rule into this problem. The prompt explicitly defines the heterozygote’s phenotype as pink. Also, pink doesn’t mean the alleles have permanently blended into a new allele. An RW parent still transmits R or W in the model. Write the phenotype key above the square whenever a question changes the inheritance pattern.
5. Two loci: state independence before multiplying
Problem: Cross AaBb with Aabb. Assume the A and B loci assort independently. What is the probability of aaBb offspring? What is the probability of offspring showing both dominant phenotypes, assuming complete dominance at both loci?
Answer: Separate the crosses. At A, Aa crossed with Aa gives P(aa) = 1/4 and P(A_) = 3/4. The notation A_ means AA or Aa. At B, Bb crossed with bb gives P(Bb) = 1/2 and P(B_) = 1/2.
For aaBb, multiply 1/4 by 1/2 to get 1/8. For both dominant phenotypes, multiply 3/4 by 1/2 to get 3/8. These are different events, so write the event before the multiplication.
A larger square gives the same results: AaBb supplies AB, Ab, aB, and ab equally; Aabb supplies Ab and ab equally. There are eight equally likely combinations. One is aaBb, and three show both dominant phenotypes. If your square disagrees with the product method, check your gamete list first.
6. Two offspring: distinguish a particular order from either order
Problem: Return to Pp crossed with Pp. Assuming independent outcomes for two offspring, find the probability that both are white. Then find the probability that exactly one is white.
Answer: White has probability 1/4 for each offspring. Both white has probability (1/4)(1/4) = 1/16. Exactly one white can happen in two mutually exclusive orders: white then purple, or purple then white.
Each order has probability (1/4)(3/4) = 3/16. Add them to get 6/16 = 3/8. An answer of 3/16 describes only one specified order. An answer of 1/2 comes from adding two white probabilities without accounting for the required purple outcome and the possible overlap.
OpenStax’s probability section explains when multiplication and addition apply. In your working, connect independent events with multiplication and add mutually exclusive routes to the requested result. Don’t use an isolated word such as “and” or “or” as a substitute for identifying the events.
7. A conditional probability changes the denominator
Problem: From Pp crossed with Pp, choose an offspring known to have purple flowers. What is the probability that this offspring is heterozygous?
Answer: Restrict attention to the purple outcomes: PP, Pp, and Pp. Two of those three equally likely original cells are heterozygous, giving 2/3. The white pp outcome is excluded because the question already tells you the chosen offspring is purple.
The answer is not 1/2. That would be the probability of heterozygosity before knowing the phenotype. You can also calculate (1/2)/(3/4) = 2/3. Label the denominator “purple offspring” rather than “all offspring” to make the change visible. This is a useful check whenever a genetics question says “among,” “given,” or “known to be.”
8. When a familiar four-gamete list is unjustified
Problem: An organism is AaBb, and the two loci are linked. The prompt gives no arrangement of alleles on homologous chromosomes and no recombination information. Can you assume AB, Ab, aB, and ab each occur with probability 1/4?
Answer: No. The independent-assortment assumption used in problem five hasn’t been supplied and cannot simply be carried over. You need information about chromosome arrangement and recombination, or another explicit statement that determines the gamete frequencies.
For example, AB/ab and Ab/aB identify different parental arrangements. With complete linkage and no recombination, the first makes only AB and ab gametes, while the second makes only Ab and aB gametes, under ordinary segregation. The genotype label AaBb alone doesn’t tell you which arrangement applies. “Insufficient information” is a justified answer when you explain precisely what’s missing.
Make an answer sheet that explains your decisions
For each exercise, record the parent genotypes, allowed gametes, assumptions, requested event, and final probability. Include one sentence explaining why a tempting wrong answer fails. That makes your notes useful when the next quiz changes the letters or asks a conditional question.
If gamete formation is the confusing step, review mitosis and meiosis practice before drawing another large grid. Compare your course material with biology study documents, and use document browsing to find complementary explanations. Check the assumptions in a shared solution before copying its ratio.
Students should upload their own notes to help other students. An original worksheet with clear allele definitions and worked answers is useful to someone studying the same topic. Only share material you have permission to distribute, and keep restricted assessments and other students’ personal information out of the file.
Frequently asked questions
Does dominant mean more common?
No. Dominance describes the relationship between alleles and phenotype in a specified model. It doesn’t, by itself, tell you how frequent an allele is in a population.
Why can Aa appear twice in a square?
The A can come from either parent. Those are different gamete combinations that produce the same genotype. Count both cells when calculating that genotype’s probability.
Will four offspring always follow a 3:1 ratio?
No. The ratio describes model probabilities, not a required allocation for every small group. Actual small samples can differ, as the repeated-offspring calculations above illustrate.
When should I avoid a sixteen-cell square?
When separate single-locus calculations answer the question more clearly and independence is justified, multiply those probabilities. If the assignment asks for the square, draw it and use the shorter calculation to check your result.