Download Genetics Practice II

Genetics Practice II
1. Monohybrid Cross (1 Trait):
In humans, brown eyes (B) are dominant over blue (b). A brown-eyed man marries a
blue-eyed woman and they have three children, two of whom are brown-eyed and one of
whom is blue-eyed. Draw the Punnett square that illustrates this marriage. What is the
man’s genotype? What are the genotypes of the children?
2. Testcross:
In dogs, there is a hereditary deafness caused by a recessive gene, “d.” A kennel owner
has a male dog that she wants to use for breeding purposes if possible. The dog can
hear, so the owner knows his genotype is either DD or Dd. If the dog’s genotype is Dd,
the owner does not wish to use him for breeding so that the deafness gene will not be
passed on. This can be tested by breeding the dog to a deaf female (dd). Draw the
Punnett squares to illustrate these two possible crosses. In each case, what
percentage/how many of the offspring would be expected to be hearing? deaf? How
could you tell the genotype of this male dog? Also, using Punnett square(s), show how
two hearing dogs could produce deaf offspring.
3. Sex-Linked Genes:
In humans, the genes for colorblindness and hemophilia are both located on the X
chromosome with no corresponding gene on the Y. These are both recessive genes. If a
man and a woman, both with normal vision, marry and have a colorblind son, draw the
Punnett square that illustrates this. If the man dies and the woman remarries to a
colorblind man, draw a Punnett square showing the type(s) of children that could be
expected from this marriage. How many/what percentage of each could be expected?
4. Incomplete Dominance:
In the tropical Bayer’s Orchid O. Bayerididae, there are two alleles, red and blue.
Heterozygous individuals are purple. Set up a Punnett square to show a cross that will
exhibit all three phenotypes in the F1 generation.
5. Sex-Influenced Traits:
Baldness in humans is a dominant, sex-influenced trait. This gene is on the autosomes,
not the sex chromosomes. A man who is BB or Bb will be bald and will be normal only if
he is bb. A woman will only be bald if she is BB and normal if she is Bb or bb (it’s almost
like B is dominant in males and b is dominant in females). If two parents are
heterozygous for baldness, what are the chances of their children being bald? Use a
Punnett square to illustrate this. Note: because the sex of a person does make a
difference in how the gene is expressed, you need to set this up as a dihybrid cross to
account for the sex of the children.
Optional Challenges:
5. Dihybrid Cross (16 square Punnett Square):
Using mode of inheritance information from #1 and #3 what would be the genotypic and
phenotypic ratios of the offspring produced by a cross between a hemophiliac who is
heterozygous for brown eyes and a woman who is a carrier of the hemophilia allele and
is heterozygous for brown eyes?
6. Trihybrid Cross (64 square Punnett Square):
In Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is
dominant over smooth (r), and short hair (S) is dominant over long hair (s).
Assuming these genes are on separate chromosomes (they assort
independently), draw the Punnett square for a cross between two heterozygous
black, rough, short-haired Guinea pigs. What would the phenotype(s) of the
offspring be? If two of the F1 offspring were crossed, one homozygous dominant,
and the other homozygous recessive draw the Punnett square for this cross.
Hint: first make a list of the possible gametes, making sure each has exactly one
copy of each of the genes (one allele for each gene). What would the genotype
and phenotype ratios be for the F2 generation?