20 Punnett Square Problems to Practice On (With Step-by-Step Solutions)

20 Punnett square practice problems covering monohybrid crosses, incomplete dominance, codominance, dihybrid crosses, and sex-linked inheritance, each with a full worked solution.

By Petrus Sheya

August 10, 2026 · 10 min read

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 (TT) is dominant over short (tt). Two heterozygous plants (Tt×TtTt \times Tt) are crossed. What are the genotype and phenotype ratios of the offspring?

Step 1: Build the grid. Each TtTt parent produces TT and tt gametes.

TTtt
TTTTTTTtTt
ttTtTttttt

Step 2: Read the ratios. Genotypes: 1 TT:2 Tt:1 tt1\ TT : 2\ Tt : 1\ tt. Since TT is dominant, TTTT and TtTt both look tall. Phenotype=3 tall:1 short\text{Phenotype} = 3 \text{ tall} : 1 \text{ short}

AnswerGenotype ratio 1 TT : 2 Tt : 1 tt; phenotype ratio 3 tall : 1 short

Problem 2

A homozygous tall pea plant (TTTT) is crossed with a homozygous short plant (tttt). What fraction of the offspring will be tall?

Step 1: Find the gametes. TTTT only produces TT gametes, tttt only produces tt gametes.

Step 2: Fill the grid. Every offspring is TtTt. 100% of offspring are Tt (tall)100\% \text{ of offspring are } Tt \text{ (tall)}

Answer100% tall

Problem 3

A heterozygous tall plant (TtTt) is crossed with a short plant (tttt). What's the phenotype ratio of the offspring?

Step 1: Build the grid. TtTt gives T,tT, t; tttt gives t,tt, t.

tttt
TTTtTtTtTt
tttttttttt

Step 2: Read the ratio. Half the offspring are TtTt (tall), half are tttt (short). 1 tall:1 short1 \text{ tall} : 1 \text{ short}

Answer1 tall : 1 short (50% each)

Problem 4

In pea plants, purple flowers (PP) are dominant over white (pp). Two heterozygous plants (Pp×PpPp \times Pp) produce 160 offspring. How many are expected to have white flowers?

Step 1: Get the ratio. Pp×PpPp \times Pp gives 33 purple :1: 1 white, so white is 14\frac{1}{4} of the offspring.

Step 2: Apply it to 160. 160×14=40160 \times \frac{1}{4} = 40

Answer40 white-flowered offspring

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 (BB) is dominant over white (bb). A black guinea pig is crossed with a white guinea pig, and the offspring come out 1:11:1 black to white. What is the genotype of the black parent?

Step 1: Test the possibilities. If the black parent were BBBB, every offspring would get a BB and be black, a 100%100\% black result, not 1:11:1.

Step 2: Try BbBb. Bb×bbBb \times bb gives Bb,Bb,bb,bbBb, Bb, bb, bb, which is exactly 1:11:1. Black parent=Bb\text{Black parent} = Bb

AnswerBb

Problem 6

Both parents are AaAa for a gene with complete dominance. What is the probability that a given offspring is homozygous, either AAAA or aaaa?

Step 1: Get the genotype ratio. Aa×AaAa \times Aa gives 1 AA:2 Aa:1 aa1\ AA : 2\ Aa : 1\ aa.

Step 2: Add the homozygous outcomes. 14+14=12\frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Answer50%

Problem 7

In fruit flies, red eyes (RR) are dominant over sepia eyes (rr). 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. 2496=14 sepia\frac{24}{96} = \frac{1}{4} \text{ sepia}

Step 2: Match it to a cross. A 14\frac{1}{4} recessive ratio only comes from two heterozygotes. Rr×RrRr \times Rr

AnswerBoth parents are Rr

Problem 8

Both parents are carriers (FfFf) for a recessive genetic disorder. What is the probability that a child is unaffected but still a carrier?

Step 1: Get the genotype ratio. Ff×FfFf \times Ff gives 1 FF:2 Ff:1 ff1\ FF : 2\ Ff : 1\ ff.

Step 2: Pick out the carrier genotype. "Unaffected but a carrier" means FfFf specifically, not FFFF. 24=12\frac{2}{4} = \frac{1}{2}

Answer50%

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 (CRCRC^RC^R) and white (CWCWC^WC^W) plants are crossed. What phenotype do the offspring have?

Step 1: Build the grid. CRCRC^RC^R gives only CRC^R, CWCWC^WC^W gives only CWC^W.

Step 2: Read the phenotype. Every offspring is CRCWC^RC^W. Since neither allele is dominant, the heterozygote blends the two colors. 100% pink100\% \text{ pink}

AnswerAll offspring are pink

Problem 10

Two pink snapdragons (CRCW×CRCWC^RC^W \times C^RC^W) are crossed. What is the phenotype ratio of the offspring?

Step 1: Build the grid.

CRC^RCWC^W
CRC^RCRCRC^RC^RCRCWC^RC^W
CWC^WCRCWC^RC^WCWCWC^WC^W

Step 2: Read the phenotypes. CRCRC^RC^R is red, CRCWC^RC^W is pink, CWCWC^WC^W is white. 1 red:2 pink:1 white1 \text{ red} : 2 \text{ pink} : 1 \text{ white}

Answer1 red : 2 pink : 1 white

Problem 11

In cattle, coat color is codominant: red (CRC^R) and white (CWC^W) alleles are both expressed in the heterozygote, producing roan. A roan bull (CRCWC^RC^W) is crossed with a red cow (CRCRC^RC^R). What fraction of the offspring are expected to be roan?

Step 1: Build the grid. Bull gives CR,CWC^R, C^W; cow gives only CRC^R.

CRC^RCRC^R
CRC^RCRCRC^RC^RCRCRC^RC^R
CWC^WCRCWC^RC^WCRCWC^RC^W

Step 2: Read the ratio. Half the offspring are CRCRC^RC^R (red), half are CRCWC^RC^W (roan). 24=12\frac{2}{4} = \frac{1}{2}

Answer50% roan

Problem 12

In humans, ABO blood type alleles IAI^A and IBI^B are codominant, and both are dominant over ii. A parent with blood type AB (IAIBI^AI^B) and a parent with blood type O (iiii) have a child. What blood types are possible, and in what ratio?

Step 1: Build the grid. AB parent gives IA,IBI^A, I^B; O parent gives only ii.

IAI^AIBI^B
iiIAiI^AiIBiI^Bi
iiIAiI^AiIBiI^Bi

Step 2: Read the phenotypes. IAiI^Ai is type A, IBiI^Bi is type B. Neither ABAB nor OO is possible. 1 type A:1 type B1 \text{ type A} : 1 \text{ type B}

AnswerType A or type B, each with 50% probability (never AB or O)

Section 4: Dihybrid Crosses

Now track two genes at once. The grid grows to 4×44 \times 4, 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 (RR) are dominant over wrinkled (rr), and yellow seed color (YY) is dominant over green (yy). The genes assort independently. A double heterozygote is self-crossed (RrYy×RrYyRrYy \times RrYy). What is the phenotype ratio of the offspring?

Step 1: Split into two single-gene crosses. Rr×RrRr \times Rr gives 33 round :1: 1 wrinkled. Yy×YyYy \times Yy gives 33 yellow :1: 1 green.

Step 2: Multiply the two ratios together. (3:1)×(3:1)=9:3:3:1(3:1) \times (3:1) = 9:3:3:1

Answer9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green

Problem 14

From the same cross in Problem 13 (RrYy×RrYyRrYy \times RrYy), 320 offspring are produced. How many are expected to have wrinkled, green seeds?

Step 1: Get the fraction. Wrinkled green is the rryyrryy class, 116\frac{1}{16} of a 9:3:3:19:3:3:1 ratio.

Step 2: Apply it to 320. 320×116=20320 \times \frac{1}{16} = 20

Answer20 wrinkled, green offspring

Problem 15

A double heterozygote is test-crossed with a double recessive (RrYy×rryyRrYy \times rryy). What is the phenotype ratio of the offspring?

Step 1: Split into two single-gene crosses. Rr×rrRr \times rr gives 11 round :1: 1 wrinkled. Yy×yyYy \times yy gives 11 yellow :1: 1 green.

Step 2: Multiply the ratios. (1:1)×(1:1)=1:1:1:1(1:1) \times (1:1) = 1:1:1:1

Answer1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green

Problem 16

Cross RRYy×rrYyRRYy \times rrYy. What is the phenotype ratio of the offspring?

Step 1: Handle the RR gene first. RR×rrRR \times rr gives only RrRr offspring, so every offspring is round. This gene doesn't split the ratio at all.

Step 2: Handle the YY gene. Yy×YyYy \times Yy gives 33 yellow :1: 1 green.

Step 3: Combine. Since every offspring is already round, the seed-color split is the whole story. 3 round yellow:1 round green3 \text{ round yellow} : 1 \text{ round green}

Answer3 round yellow : 1 round green (no wrinkled offspring at all)

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 (XcYX^cY) has children with a homozygous normal woman (XCXCX^CX^C). What are the possible genotypes of the daughters and sons?

Step 1: Find the gametes. Father gives XcX^c or YY; mother gives only XCX^C.

Step 2: Build the grid.

XCX^CXCX^C
XcX^cXCXcX^CX^cXCXcX^CX^c
YYXCYX^CYXCYX^CY

Step 3: Read the results. Every daughter is XCXcX^CX^c (a carrier, but normal vision), every son is XCYX^CY (normal vision).

AnswerAll daughters are carriers with normal vision; all sons have normal vision; no color-blind offspring

Problem 18

A carrier woman (XCXcX^CX^c) has children with a color-blind man (XcYX^cY). What fraction of the daughters are expected to be color-blind, and what fraction of the sons?

Step 1: Find the gametes. Mother gives XCX^C or XcX^c; father gives XcX^c or YY.

Step 2: Build the grid.

XCX^CXcX^c
XcX^cXCXcX^CX^cXcXcX^cX^c
YYXCYX^CYXcYX^cY

Step 3: Split by sex. Daughters are XCXcX^CX^c or XcXcX^cX^c, a 1:11:1 split. Sons are XCYX^CY or XcYX^cY, also 1:11:1. 12 of daughters and 12 of sons are color-blind\frac{1}{2} \text{ of daughters and } \frac{1}{2} \text{ of sons are color-blind}

Answer50% of daughters and 50% of sons are color-blind

Problem 19

For human ABO blood type, a heterozygous type A parent (IAiI^Ai) and a heterozygous type B parent (IBiI^Bi) have a child. What blood types are possible, and what fraction of each?

Step 1: Find the gametes. Type A parent gives IAI^A or ii; type B parent gives IBI^B or ii.

Step 2: Build the grid.

IBI^Bii
IAI^AIAIBI^AI^BIAiI^Ai
iiIBiI^Biiiii

Step 3: Read the phenotypes. Each of the four boxes is a different blood type: AB, A, B, O. 14 each\frac{1}{4} \text{ each}

AnswerType AB, A, B, or O, each with 25% probability

Problem 20

In fruit flies, red eye color is X-linked dominant (RR) over white (rr), and long wings are autosomal dominant (LL) over vestigial (ll). A female heterozygous for both traits (XRXrLlX^RX^rLl) is crossed with a white-eyed, vestigial-winged male (XrYllX^rY\,ll). 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. XRXr×XrYX^RX^r \times X^rY gives four equally likely outcomes: XRXrX^RX^r, XrXrX^rX^r, XRYX^RY, XrYX^rY, each 14\frac{1}{4} of the offspring. A white-eyed son is the XrYX^rY class. P(white-eyed son)=14P(\text{white-eyed son}) = \frac{1}{4}

Step 2: Solve the wing cross on its own. Ll×llLl \times ll gives 11 long :1: 1 vestigial. P(vestigial)=12P(\text{vestigial}) = \frac{1}{2}

Step 3: Multiply, since the two genes assort independently. 14×12=18\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

Answer1/8 (12.5%)

A note on notation: superscript letters on the X, like XRX^R or XcX^c, mark which allele of an X-linked gene a chromosome carries. A YY chromosome never carries that gene, which is exactly why males only need one copy of an X-linked allele to show its phenotype.