Name _______________________________________
Due Monday after Break
How to set up and interpret Punnett Squares
PROBLEM 1. The Monohybrid cross
An example problem:
In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants
that are heterozygous for the seed shape trait, what
fraction of the offspring should have spherical
seeds?
The picture on the right above represents a monohybrid cross of F1-hybrid plants. Both parent
plants are heterozygous (Ss) for an allele that determines seed shape. Presence of the
dominant allele (S) in homozygous (SS) or heterozygous (Ss) plants results in spherical seeds.
Homozygous recessive (ss) plants have dented seeds.
To solve problem 1, you'll need to set up a Punnett square. This tutorial will walk you
through that process.
How do I set up a Punneett square to solve the problem?
1. Draw a 2 by 2 Punnett square.
Label each parent as “parent 1” or “parent two” (or if you know that parent 1
is the mom or the dad you can label it that way.
Identify both the genotype (the letters) and the phenotype (description) of the
parents if known.
In this example: S = round and s is dented.
S represents the dominant trait, so in a heterozygous individual the dominant
trait will be what you see (phenotype)
Both parents are heterozygous
P1
P2
Genotype: Ss Phenotype: round
Genotype: Ss Phenotype: round
2. Write the alleles for parent 1 on the left side of the Punnett
square.
Each gamete will have one of the two alleles of the parent. In
this particular cross, half of the gametes will have the dominant
Name _______________________________________
Due Monday after Break
(S) allele, and half will have the recessive (s) allele. We will use
blue and brown to keep track of the alleles of each parent.
3. Write the alleles from parent 2 above the Punnett square.
For this heterozygous parent (Ss), half of the gametes will have
the dominant (S) allele, and half will have the recessive (s) allele.
4. Fill the squares for parent 1.
Fill each square with the allele from Parent 1 that lines up with
the row.
5. Fill the squares for parent 2.
Fill each square with the allele from Parent 2 that lines up with
the column.
Interpreting the results of a Punnett square
We now have the information for predicting the outcome of the cross.
The genotypes in the four boxes of the Punnett square are each equally
likely to occur among the offspring of this cross. We may now tabulate
the results.
Probability of seeing the phenotypes and genotypes that resulted
from this monohybrid cross (Ss x Ss)
25%
homozygous
dominant
50%
heterozygous
25%
homozygous
recessive
Phenotypes that resulted from this monohybrid cross (Ss x Ss)
Remember….you are predicting the PROBABILITY that an event will occur…and NOT the certainty. It is
sort of an “average” of what will happen over a long period of time.
Name _______________________________________
Due Monday after Break
Practice problems: Monohybrid Cross
EXAMPLE
In pea plants, the allele for tallness is dominant (T) and shortness (t) is recessive. The first
problem is done for you as an illustration for how I would like you to set up the problems.
Parent 1 will be the male (if that information is known). Its alleles (letters) for the characteristic
(height in the firs example) will be shown on the left side of the punnett square. Parent 2 (the
female (if that is known) will have her alleles placed across the top of the Punnett square.
The combinations inside the four boxes (TT, Tt, tt) represent the possible combinations of the
alleles. To calculate the percentages you simply look at the chance of the combination in each
box being 25%. So TT appears in 1 of the 4 boxes. Its percentage is 25%, while Tt appears in 2
boxes, so it’s predicted percentage is 50%. We can also report thes as ratios.
1 TT: 2 Tt : 1 tt.
Because Tall is the DOMINANT phenotype, it will mask the short allele (not allow it to show)
when the two appear together. Because of this the predicted PHENOTYPE is 3 Tall (TT, Tt, Tt)
and 1 short Tt)
Do the remaining problems setting them up the same way as the first problem. Clearly label
every thing.
a. Create a Punnett square showing the possible outcomes for offspring of a homozygous
dominant (Tt) is cross with a homozygous recessive (Tt).
F (P2)
M( P1)
T
t
Predicted Phenotype ratio of the offspring: 3 Tall: 1 short
Tt
Predicted Genotype ratio of the offspring: _1 TT: 2 Tt : 1 tt
tt
Predicted Percentage phenotype 75% Tall and 25% Short
Name _______________________________________
1.
Due Monday after Break
In pea plants, the allele for tallness is dominant (T) and shortness (t) is recessive. A
heterozygous tall plant is crossed with another heterozygous tall plant. Create a Punnett square
to show possible outcomes, and show the RATIO of the phenotypes and the genotypes (be sure
to LABEL!!!
2. A heterozygous tall plant and a short plant are crossed with each other.
3. Yellow pea pods (Y) are dominant over green pods (y). A heterozygote is crossed with a
homozygous dominant plant.
Name _______________________________________
Due Monday after Break
4. A homozygous cream colored mouse (dd) is crossed with a heterozygous (Dd) dark mouse.
Create a Punnet square.
a. What are the odds (ratio or percentage) that this couple will have a cream colored
baby?
b. What are the odds of a dark mouse?
5. In sheep, white is due to a dominant gene (W), black is due to its recessive allele (w). A white
ewe mated to a white ram produces a black lamb.
a. Create a punnett square to show the possible offspring
b. How can two white sheep produce a black sheep?
c. Calulate the probable genotype and phenotype ratios
Name _______________________________________
Due Monday after Break
6. The polled (hornless) condition (P) is dominant over horned (p) in bulls . A polled bull is crossed
to 3 cows.
a. With cow A, which is horned, a polled cow is produced;
b. With cow B, also horned, a horned cow is produced;
c. with cow C, which is polled, a horned cow is produced. What are the genotypes of the
three calves and the four parents?
d. Create a Punnett square for each of the three scenarios above. Remember that in ALL cases the bull
MUST have the same genotype.
Name _______________________________________
Due Monday after Break
7. An albino is a person with no pigmentation in his skin, eyes or hair. Normal pigmentation is (A) is
dominant over albino (a). A normal pigmented man (whose mother was albino) marries a
homozygous normal woman. What would the expected appearances of the children be?
a. If the man in problem #7 married a normal woman (whose father was albino), what
percent of the children would be expected to be albino?
Name _______________________________________
Due Monday after Break
8. In chickens, yellow legs (Y) are dominant over white legs (y). A yellow legged rooster was
crossed with a white legged hen. Both kinds of offspring were produced. What are the
genotypes of the parents and the offspring? Make a punnet square too!
9. A yellow legged rooster was crossed with a white legged hen. All fifteen of the offspring were
yellow legged. What is the genotype of the rooster? Explain how you made your decision.
Could you be wrong? Why?
Name _______________________________________
Due Monday after Break
10. In a cross between two white mice, in several litters, 12 out of 50 offspring were black.
a. What color is dominant?
b. What are the genotypes of the parents?
c. What are the genotypes of the black offspring?
11. What would be the genotypes of both parents in problem #10 if all fifty offspring were white?
Explain the reasoning behind your answer.
12. If one fruit fly is heterozygous (red) and the other is homozygous recessive (white), what
percent of the offspring would be expected to have white eyes? Explain and show evidence on
which you base your decision (use probabilities.