Download Practice Problems (Punnett Squares)

Practice Problems (Punnett Squares)
For each of the following crosses draw a punnett square and write the genotypic and phenotypic ratios.
Traits Dominant/Recessive: Seed coat shape Round (R) Wrinkled (r) Pod color Green (G) Yellow (g)
Height Tall (T) Short (t)
1. Rr X RR
2. Gg X GG
3. Tt X Tt
4. Rr X Rr
5. Cross two heterozygous green plants
6. Cross a heterozygous green plant with a homozygous green plant.
7. Cross a homozygous tall plant with a short plant
8. In mice the ability to run normally is a dominant trait. Mice with this trait are called running
mice (R), the recessive trait causes mice to run in circles only. Mice with this trait are called
waltzing mice (r) . Hair color is also inherited in mice. Black hair (B) is dominant over brown hair
(b). For each of the following I. Show the genotypes of the parents II. Show the possible
gametes produced by the parents III. Draw the punnett square IV. Give the phenotypic ratio of
the resultant offspring
a) Cross a heterozygous running, heterozygous black mouse with a homozygous running,
homozygous black mouse.
b) Cross a homozygous running homozygous black mouse with a heterozygous running,
brown mouse.
c) Cross a waltzing , brown mouse with a waltzing brown mouse
d) Cross a homozygous running, heterozygous black mouse with a waltzing brown
mouse.
e) Cross a heterozygous running, brown mouse with a heterozygous running,
homozygous black mouse.
f) Cross a heterozygous running, heterozygous black mouse with a heterozygous
running, heterozygous black mouse.
9. In Japanese four o'clock plants red (R) color is incompletely dominant over white (r) flowers,
and the heterozygous condition (Rr) results in plants with pink flowers. For each of the following
construct a punnett square and give the phenotypic ratios of the offspring.
a) a red plant and a white plant }
b) a red plant and a pink plant }
c) a white plant and a pink plant
d) two pink plants
10. In some cats the gene for tail length shows incomplete dominance. Cats with long tails and
cats with no tails are homozygous for their respective alleles. Cats with one long tail allele and
one no tail allele have short tails. For each of the following construct a punnett square and give
the phenotypic ratios of the offspring.
a) a long tail cat and a cat with no tail
b) a long tail cat and a short tail cat
c) a short tail cat and a cat with no tail
d) two short tail cats.
11. In some cattle the genes for brown hair and for white hair are co-dominant. Cattle with
alleles for both brown and white hair, have both brown and white hairs. This condition gives the
cattle a reddish color, and is referred to as Roan.
For each of the following construct a punnett square and give the phenotypic ratios of the
offspring.
a) a brown cow and a white bull
b) a brown cow and a roan bull
c) a white cow and a roan bull
d) a roan cow and a roan bull
12. Color-blindness is a sex-linked recessive trait in humans. The alleles for the hair color are
located on a pair of autosomes, and brown hair (B) is dominant to blond hair (b). A woman who
is homozygous for normal vision who has blond hair marries a man that is color blind and
heterozygous for brown hair. What is the probability that their first born daughter will have brown
hair and normal vision.
13. For each of the following crosses remember that hemophilia, muscular dystrophy, and color
blindness are recessive sex-linked traits in humans. In sex linked traits men will always express
the trait if they carry it on the X chromosome. Women can express the trait only if it is found on
both X chromosomes. Women have two normal phenotypes: homozygous normal and carrier.
Men have only one normal phenotype because they have only one X chromosome. Hemophilia
for example: Normal Male: XHY , Hemophiliac Male XhY, Normal Female: XHXH & XHXh (carrier),
Hemophiliac Female XhXh
A. A woman that is a carrier of hemophilia marries a hemophiliac man. What is the
probability that their first child will be a hemophiliac?
B. A hemophiliac woman has a mother who is phenotypically normal. What are the
genotypes of her mother and her father?
C. What is the probability that a normal vision woman who marries a man who is color blind,
will have a daughter who is color blind?
D. A phenotypically normal man, who has a brother with M.D., marries a homozygous
normal woman. What is the probability that any of their children will have M.D.
14. Blood types are controlled by three alleles A, B, O. A and B are co-dominants and O is
recessive to both A and B.
The following genotypes and resultant phenotypes are possible:
Type A blood can have these genotypes: LALA and LALO
Type B blood can have these genotypes: LBLB and LBLO
Type AB (universal recipient) can have this genotype: LALB
Type O (universal donor) can have this genotype: LOLO
A. A woman homozygous for blood type B marries a man that is heterozygous for blood type A.
State the possible phenotypic ratios of the offspring.
B. A man with blood type O marries a woman with blood type AB. State the possible phenotypic
ratios of the offspring.
C. A type B woman whose mother was type O marries a type O man. What are the possible
phenotypic ratios of their offspring?
D. A type A woman whose father was type B marries a type B man whose mother was was
type A. What are the possible phenotypes of their offspring?
E. What is the probability that a couple whose blood types are AB and O will have a type A
child?