Download Intro to mendelian genetics_problem set

Part A: Vocabulary
Match the definitions on the left with the terms on the right.
____ 1. genotypes made of the same alleles
____ 2. different forms of genes for a single trait
____ 3. gene that is always expressed
____ 4. gene that is expressed only in the homozygous state
____ 5. genotypes made of two different alleles
A. alleles
B. dominant
C. heterozygous
D. homozygous
E. recessive
Below each of the following words are choices. Circle the choices that are examples of each of
those words.
6. Dominant allele
DekLNnRS
7. Recessive allele
MndFGrkP
8. Homozygous dominant
AA Gg KK mm uu Rr TT
9. Homozygous recessive
ee Ff HH Oo qq Uu ww
10. Genotypes in which dominant gene must show
AA Dd EE ff Jj RR Ss
11. Genotypes in which recessive gene must show
aa Gg Ff KK rr Oo Tt
Part B: Monohybrid Cross Problems
12. In tomatoes, red fruit (R) is dominant over yellow fruit (r). A plant that is homozygous for
red fruit is crossed with a plant that has yellow fruit. What would be the genotypes and
phenotypes of the P1 and F1 generations?
13. If two of the F1 generation from the above cross were mated, what would be the genotype
and phenotype ratios of the F2?
14. Brown eyes in humans are dominant to blue eyes. A brown-eyed man, whose mother was
blue-eyed, marries a brown-eyed woman whose father had blue eyes. What is the probability that
this couple will have a blue-eyed child?
Part C: Multiple Allele and Codominance Problems
15. In humans straight hair (ss) and curly hair (cc) are codominant traits, that result in hybrids
who have wavy hair (sc). Cross a curly hair female with a wavy haired male. What are the
chances of having a curly haired child?
16. What is the probability that a couple whose blood types are AB and O will have a type A
child?
17. Suppose a newborn baby was accidentally mixed up in the hospital. In an effort to determine
the parents of the baby, the blood types of the baby and two sets of parents were determined.
Baby has type O
Mrs. Brown has type B, Mr. Brown had type AB
Mrs. Smith has type B, Mr. Smith had type B
To which parent does the baby belong?
Part D: Incomplete Dominance
18. 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
phenotypic and genotype 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.
19. from your textbook p.196 #3
Part E: Sex Linked Traits
20. Hemophilia is a sex linked recessive blood disorder which impairs the body’s ability to
control blood clotting. Genotypes in the disorder consist of: Normal Male XHY , Hemophiliac
Male XhY, Normal Female XHXH & XHXh (carrier), Hemophiliac Female XhXh
a. A women 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 women has a mother who is phenotypically normal. What are the genotypes of
her mother and her father?
21. What is the probability that a normal vision women who marries a man who is color blind ,
will have a daughter who is color blind?
22. A phenotypically normal man who has a brother with muscular dystrophy marries a
homozygous normal woman. What is the probability that any of their children will have
muscular dystrophy?
2. long tails (L), no tails (ll), short tails (Ll).
a) a long tail cat and a cat with no tail
LL X ll, gametes for LL (L), gametes for ll (l), resulting phenotype ratio: 100% short tails, resulting
genotype ratio: 100% Ll
b) a long tail cat and a short tail cat
LL X Ll, gametes for LL (L), gametes for Ll (L & l), resulting phenotype ratio: 50% long tails / 50% short
tails, resulting genotype ratio: 50% LL / 50% Ll
c) a short tail cat and a cat with no tail
Ll X ll, gametes for Ll (L & l), gametes for ll (l), resulting phenotype ratio: 50% short tails / 50% no tails,
resulting genotype ratio: 50% Ll / 50% ll
d) two short tail cats
Ll X Ll, gametes for Ll (L& l), gametes for Ll (L & l), resulting phenotype ratio: 25% Long Tailed, 50% short
tailed, and 25% no tailed, resulting genotype ratio: 25% LL / 50% Ll / 25% ll