homework

Genetics Packet ~ Punnett Square Practice KEY

Basics

1. The following pairs of letters represent alleles of different genotypes. Indicate which pairs

are Heterozygous and which are Homozygous. Also indicate whether the homozygous pairs

are Dominant or Recessive (*note heterozygous pairs don’t need either dominant nor

recessive labels.)

A. BB =

B. Bb =

C. Gg =

D. gg =

E. aa =

F. Ee =

2. In humans, brown eye color (B), is dominant over blue eye color (b). What are the

phenotypes of the following genotypes?

A. Bb =

B. BB =

C. bb=

​Monohybrid Crosses with Complete Dominance

3. A heterozygous smooth pea pod plant is crossed with a wrinkled pea pod plant. There are

two alleles for pea pod, smooth and wrinkled. Use R for seed texture. Predict the offspring from

this cross.

a. What is the genotype of the parents?

b. Set up a Punnett square with possible gametes.

c. Fill in the Punnett square for the resultant offspring.

d. What is the predicted genotypic ratio for the offspring?

e. What is the predicted phenotypic ratio for the offspring?

f. If this cross produced 50 seeds how many would you predict to have a wrinkled pod?

4. In humans, achondroplasia “dwarfism” (D) is dominant over normal (d).

A homozygous dominant (DD) person dies before the age of one.

A heterozygous (Dd) person is dwarfed. A homozygous recessive individual

is normal.

A heterozygous dwarf man marries a heterozygous dwarf woman…

a. What is the probability of having a normal child?

b. What is the probability that the next child will also be normal?

each child is a new shot at the same Punnett square!

c. What is the probability of having a child that is a dwarf?

d. What is the probability of having a child that dies at one from this disorder?

5. In humans, free earlobes (F) is dominant over attached earlobes (f).

If one parent is homozygous dominant for free earlobes, while the other

has attached earlobes, can they produce any children with attached

earlobes?

6. In humans widow’s peak (W) is dominant over straight hairline (w). A heterozygous man for

this trait marries a woman who is also heterozygous.

a. List possible genotypes of their offspring.

b. List the phenotypic ratio for their children.

Dihybrid Crosses

7. In humans there is a disease called Phenylketonuria (PKU), caused by a recessive allele

that doesn’t code for the enzyme that breaks down the amino acid phenylalanine. This

disease can result in mental retardation or death. Let “E” represent the normal enzyme. Also

in humans in a condition called galactose intolerance or galactosemia, which is also caused

by a recessive allele. Let “G” represent the normal allele for galactose digestion. In both

diseases, normal dominates over recessive.

a. Complete the Punnett Square for a cross between two adults were heterozygous for both

traits (EeGg):

What are the chances of having a child that is completely normal?

Has just PKU?

Has just galactosemia?

Has both diseases?

Incomplete Dominance

8. In humans’ straight hair (SS) and curly hair (CC) are incompletely dominant, that result in

hybrids who have wavy hair (SC). Cross a curly hair female with a wavy aired male.

a. Complete a Punnett square for this cross.

b. What are the chances of having a curly haired child?

c. What genotype(s) would you need to produce a curly haired child? __

Codominance

9. A black chicken (BB) is crossed with a speckled chicken (BW).

a. Show the Punnett square for the cross.

b. What is the predicted genotypic ratio for offspring?

c. What are the chances of having a white chick?

Codominance & Multiple Alleles

10. Suppose two newborn babies were accidentally mixed up in the hospital. In an effort to

determine the parents of each baby, the blood types of the babies and the parents were

determined.

Baby 1 had type O, Mrs. Brown had type B, Mrs. Smith had type B, Baby 2 had type A,

Mr. Brown had type AB, and Mr. Smith had type B.

a. Draw Punnett squares for each couple (you may need to do more than 1 square/ couple)

b. To which parents does baby #1 belong? Why? Hint you may want to refer to your Punnett

squares.

Sex-Linked Traits

11. Hemophilia is a sex-linked trait. A person with hemophilia is lacking certain proteins that are

necessary for normal blood clotting. Hemophilia is caused by a recessive allele so use “N” for

normal and “n” for hemophilia. Since hemophilia is sex- linked, remember a woman will have

two alleles (NN or Nn or nn) but a man will have only one allele (N or n). A woman who is

heterozygous (a carrier) for hemophilia marries a normal man:

a. What are the genotypes of the parents?

b. Make a Punnett square for the above cross.

c. What is the probability that a male offspring will have hemophilia?

d. What is the probability of having a hemophiliac female offspring? 0%