Punnett Square Practice- 16 Problems
Punnett Square Practice: 16 Problems with Answers
If you're struggling with Punnett squares, you're not alone. Most students find them confusing at first, but once you understand the mechanics, they're actually pretty simple. This guide gives you 16 practice problems ranging from basic monohybrid crosses to more challenging dihybrid combinations. Each problem includes the answer so you can check your work immediately.
Quick Refresher: What You Need to Know
Before diving into the problems, make sure you have these concepts down:
- Alleles are versions of a gene. Capital letters (B) represent dominant alleles; lowercase (b) represents recessive alleles.
- Genotype is the genetic makeup (BB, Bb, bb).
- Phenotype is the physical result (brown eyes, tall plant, etc.).
- A homozygous individual has two identical alleles (BB or bb).
- A heterozygous individual has two different alleles (Bb).
- Dominant alleles mask recessive alleles in heterozygotes.
How to Solve Punnett Square Problems
Follow these steps every time:
- Identify the genotypes of both parents.
- Write the possible gametes for each parent along the top and side of your square.
- Combine each gamete pair to fill in the four inner boxes.
- Count the genotypes and determine phenotypic ratios.
Example: Simple Monohybrid Cross
Problem: Cross a heterozygous tall plant (Tt) with a homozygous short plant (tt).
Step 1: Parent 1 = Tt, Parent 2 = tt
Step 2: Gametes from Tt = T or t. Gametes from tt = t and t.
| t | t | |
|---|---|---|
| T | Tt | Tt |
| t | tt | tt |
Results: 50% Tt (tall), 50% tt (short). Ratio is 1:1.
The 16 Practice Problems
Problems 1-6: Basic Monohybrid Crosses
Problem 1: In humans, brown eyes (B) are dominant over blue eyes (b). Cross a homozygous brown-eyed parent (BB) with a homozygous blue-eyed parent (bb).
Problem 2: In pea plants, tall stems (T) are dominant over short stems (t). Cross two heterozygous tall plants (Tt × Tt).
Problem 3: In dogs, floppy ears (F) are dominant over pointed ears (f). Cross a heterozygous dog (Ff) with a homozygous floppy-eared dog (FF).
Problem 4: In tomatoes, red fruit (R) is dominant over yellow fruit (r). Cross a homozygous red-fruited plant (RR) with a heterozygous red-fruited plant (Rr).
Problem 5: In rabbits, gray fur (G) is dominant over white fur (g). Cross two heterozygous gray rabbits (Gg × Gg).
Problem 6: In humans, attached earlobes (E) are dominant over free earlobes (e). Cross a homozygous attached earlobe parent (EE) with a free earlobe parent (ee).
Problems 7-10: Carrier and Test Crosses
Problem 7: A plant with purple flowers (P) is crossed with a plant with white flowers (p). All offspring are purple. What must be the genotype of the purple parent, and what will a cross of those offspring with the white parent produce?
Problem 8: A tall pea plant (unknown genotype) is crossed with a short plant (tt). Half the offspring are tall, half are short. What is the genotype of the unknown parent?
Problem 9: In mice, black fur (B) is dominant over brown fur (b). You have a black mouse but don't know its genotype. Design a test cross to determine this.
Problem 10: A farmer has a rooster with straight comb (S) but doesn't know if it's homozygous or heterozygous. Cross it with a rose-comb hen (ss). If 12 chicks hatch and 6 have straight combs and 6 have rose combs, what was the rooster's genotype?
Problems 11-14: Dihybrid Crosses
Problem 11: In peas, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are dominant over green (y). Cross two plants heterozygous for both traits (RrYy × RrYy). What fraction of offspring will be round AND yellow?
Problem 12: In humans, right-handedness (H) is dominant over left-handedness (h), and brown eyes (B) are dominant over blue eyes (b). A heterozygous right-handed, heterozygous brown-eyed person (HhBb) marries a left-handed, blue-eyed person (hhbb). What is the probability of having a left-handed, blue-eyed child?
Problem 13: In rabbits, black fur (B) is dominant over brown (b), and spotted (S) is dominant over solid (s). Cross a homozygous black, heterozygous spotted rabbit (BBSs) with a brown, solid rabbit (bbss). What are the phenotypes of the offspring?
Problem 14: A plant has genotype RrYy. What proportion of its gametes will be RY, Ry, rY, and ry?
Problems 15-16: Advanced Challenges
Problem 15: In snapdragons, flower color shows incomplete dominance: red (RR) crossed with white (WW) produces pink (RW). Cross two pink snapdragons (RW × RW). What are the genotypes and phenotypes of the offspring, and in what ratio?
Problem 16: Hemophilia in humans is X-linked recessive (Xh). A normal man (XY) marries a carrier woman (XH Xh). What are the possible genotypes of their children, and what is the probability of having an affected son?
Answers and Explanations
Answers to Problems 1-6
Problem 1: BB × bb
| b | b | |
|---|---|---|
| B | Bb | Bb |
| B | Bb | Bb |
All offspring are Bb (brown-eyed). 100% brown eyes, 0% blue eyes.
Problem 2: Tt × Tt
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
Results: 1 TT : 2 Tt : 1 tt. Phenotypically: 3 tall : 1 short.
Problem 3: Ff × FF
| F | F | |
|---|---|---|
| F | FF | FF |
| f | Ff | Ff |
Results: 1 FF : 1 Ff. All offspring have floppy ears (100%).
Problem 4: RR × Rr
| R | r | |
|---|---|---|
| R | RR | Rr |
| R | RR | Rr |
Results: 1 RR : 1 Rr. All offspring have red fruit (100%).
Problem 5: Gg × Gg
| G | g | |
|---|---|---|
| G | GG | Gg |
| g | Gg | gg |
Results: 1 GG : 2 Gg : 1 gg. Phenotypically: 3 gray : 1 white.
Problem 6: EE × ee
| e | e | |
|---|---|---|
| E | Ee | Ee |
| E | Ee | Ee |
All offspring are Ee (attached earlobes). 100% attached.
Answers to Problems 7-10
Problem 7: Since all offspring are purple, the purple parent must be homozygous PP (not Pp, because Pp × pp would produce some white offspring). Cross PP × pp gives all Pp offspring (all purple).
Crossing those offspring (Pp) with the white parent (pp):
| p | p | |
|---|---|---|
| P | Pp | Pp |
| p | pp | pp |
Results: 1 Pp : 1 pp. Half purple, half white.
Problem 8: The unknown parent must be Tt. Here's why: Tt × tt produces exactly 50% tall (Tt) and 50% short (tt). If the parent were TT, all offspring would be tall. If it were tt, all would be short.
Problem 9: Cross the black mouse with a brown mouse (bb). If the black mouse is BB, all offspring will be black (Bb). If it's Bb, you'll get roughly 50% black (Bb) and 50% brown (bb). This test cross immediately reveals the genotype.
Problem 10: The rooster must be Ss (heterozygous). Here's the cross:
| s | s | |
|---|---|---|
| S | Ss | Ss |
| s | ss | ss |
Since you got exactly 50% straight and 50% rose comb offspring, the rooster was heterozygous.
Answers to Problems 11-14
Problem 11: RrYy × RrYy. This is a 16-box dihybrid square. The fraction that is round AND yellow (R_ Y_) = 9/16. This is the classic 9:3:3:1 ratio.
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
Phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.
Problem 12: HhBb × hhbb
For each trait independently: Hh × hh gives 50% Hh (right-handed) and 50% hh (left-handed). Bb × bb gives 50% Bb (brown eyes) and 50% bb (blue eyes).
Combined probability of left-handed AND blue-eyed = ½ × ½ = 1/4 or 25%.
Problem 13: BBSs × bbss
| bs | bs | |
|---|---|---|
| BS | BbSs | BbSs |
| Bs | Bbss | Bbss |
Offspring phenotypes: 50% black spotted : 50% black solid.
Problem 14: RrYy produces four types of gametes in equal proportions: RY, Ry, rY, ry (each 1/4 or 25%). This follows the rule of independent assortment for genes on different chromosomes.
Answers to Problems 15-16
Problem 15: RW × RW (incomplete dominance)
| R | W | |
|---|---|---|
| R | RR | RW |
| W | RW | WW |
Results: 1 RR (red) : 2 RW (pink) : 1 WW (white). Ratio: 1 red : 2 pink : 1 white.
Problem 16: XH Xh × XY
| XH | Xh | |
|---|---|---|
| X | XH X | Xh X |
| Y | XH Y | Xh Y |
Offspring: XH X (normal female), Xh X (carrier female), XH Y (normal male), Xh Y (affected male).
Probability of an affected son = probability of being male (½) × probability of inheriting Xh from mother (½) = 1/4 or 25%.
Genotype/Phenotype Ratio Cheat Sheet
| Cross Type | Parents | Genotypic Ratio | Phenotypic Ratio |
|---|---|---|---|
| Monohybrid | Homozygous × Homozygous | All heterozygous | All dominant |
| Monohybrid | Heterozygous × Homozygous recessive | 1:1 | 1:1 |
| Monohybrid (complete dominance) | Heterozygous × Heterozygous | 1:2:1 | 3:1 |
| Dihybrid | Dihybrid × Dihybrid | 1:2:1:2:4:2:1:2:1 | 9:3:3:1 |
| Incomplete dominance | Heterozygous × Heterozygous | 1:2:1 | 1:2:1 (different phenotypes) |
| X-linked recessive | Carrier female × Normal male | Varies | 50% affected sons |
Common Mistakes to Avoid
- Forgetting that dominant alleles don't "dilute" in crosses. A heterozygous cross (Aa × Aa) still produces homozygous offspring.
- Mixing up genotype and phenotype ratios. Always state which one you're calculating.
- Writing the wrong gametes. Each gamete gets only one allele per gene. A Tt plant produces T or t, never Tt.
- Skipping the Punnett square entirely. Trying to calculate probabilities in your head is where most errors happen. Draw the square.
- Confusing heterozygous with homozygous. Heterozygous = different alleles. Homozygous = same alleles.
Final Tip
Work through each problem twice. First, solve it without looking at the answer. Then check your work. If you got it wrong, figure out where your reasoning broke down. That's how you actually learn this stuff. The square itself is just a tool—it doesn't care if you're "good" at biology. It just counts combinations.