Genetics Practice Problems: KEY

Mendelian Genetics

1. Pattipan squash are either white or yellow. You start growing pattipans and find out that if you want to get white pattipans then at least one of the parents must be white. Which color is dominant? White alleles are dominant to yellow alleles. If yellow were dominant, then you should be able to get white squash from a cross of two yellow heterozygotes.

2. True-breeding tall red-flowered plants are crossed with dwarf white flowered plants. F1 generation consists of all tall pink-flowered plants. Assuming that height and flower color are each determined by a single gene locus, predict the results of an f1 cross of the TtRr plants. List the phenotypes and predicted ratios for the F2 generation Since folwer color shows incomplete dominance, there will be six phenotypic classes in the F2 instead of the normal four classes (ie note 9:3:3:1) You could find the answer using Punnett squares or by probabilities

Tall red T_RR = ¾ x ¼ 3/16Tall pink T_Rr = ¾ x ½ 3/8 or 6/16Tall white T_rr = ¾ x ¼ 3/16Dwarf red ttRR = ¼ x ¼ 1/16Dwarf pink ttRr = ¼ x ½ 1/8 or 2/16Dwarf white ttrr = ¼ x ¼ 1/16

3. Blood typing has often been used as evidence in paternity cases, when the blood type of the mother and child may indicate that a man alleged to be the father could not possibly have fathered the child. For the following mother and child combination, indicate which blood groups of potential fathers would be exonerated in court.

Blood group of mother

Blood group of child

Exonerated blood group

(s)AB A No groupsO B A or OA AB A or OO O ABB A B or O

4. In rabbits, CC = normal; Cc = deformed legs, cc = lethal BB = black coat color; Bb = brown coat color, bb = white coat colorWhat are the phenotypic ratios of offspring from a deformed-leg, brown rabbit and a deformed-leg, white rabbit? Parental cross CcBb x CcbbNote that cc offspring will die and no BB are possible. Therefore only 4 phenotypic classes are possible.Lethal (cc__) = ¼Normal brown (CCBb) = ¼ x ½ = 1/8Normal white (CCbb) ¼ x ½ = 1/8 1/8: 1/8: ¼: ¼ frequencyDeformed brown (CcBb) = ½ x ½ = ¼Deformed white (Ccbb) = ½ x ½ = ¼ 1:1:2:2 ratio

Or 1/6 normal brown: 1/6 normal white: 2/6 deformed brown: 2/6 deformed white

5. Fur color in rabbits is determined by a single gene locus for which there are four alleles. Four phenotypes are possible: black, Chinchilla (gray color caused by white hairs with black tips), Himalayan (white with black patches on extremities of rabbit), and white. The black allele (C) is dominant over all other alleles, Chinchilla (Cch) is dominant over Himalayan (Ch), and white © is recessive to all others.

a. A black rabbit is crossed with a Himalayan, and the F1 consists of a ratio of 2 black: 2 Chinchilla. What are the possible genotypes of the parents?

Parents are: black (C__) himalayan (Ch__), right away you can say that the black is (CCch) because half of the offspring are Chinchilla. Because Cch is dominant over both Ch and c the Himalayan parent could be either Ch Ch or Ch c. A test cross would be necessary to determine this genotype

b. A black rabbit is crossed with a Chinchilla rabbit. The F1 ratio was 2 black: 1 Chinchilla : 1 Himalayan. From these results determine which genotypes the parents can not be.

Although you can not directly determine the parental genotypes, you can eliminate some. The black parent could not be CC or C Cch , because both the C and Cch alleles are dominant to Ch, and Himilayan offspring were produced. The Chinchilla parent could not be CchCch for the same reason. One or both parents had to have Ch as their second allele; one, but not both, might have c as their second allele.

6. For the following crosses, determine the probability of obtaining the indicated genotype in an offspring

Cross Offspring ProbabilityAAbb x AaBb AAbb ½ Aa x ½ bb = ¼ AaBB x AaBb aaBB ¼ aa x ½ BB = 1/8

AABbcc x aabbCC AaBbCc 1 Aa x ½ Bb x 1 Cc =1/2 AaBbCc x AaBbcc aabbcc ¼ aa x1/4 bb x ½ cc = 1/32

7. A man who has type O blood has a child with a woman who has type A blood. The woman’s mother has AB blood, and her father, type O. What is the probability that the child has each of the following blood types? a. O 1/2 b. A 1/2 c. B 0 d. AB 0

Woman must be IAi man is ii8. Polydactyly (extra fingers and toes) is due to a dominant gene. A father is polydactyl, the mother

has the normal phenotype, and they have had one normal child. What is the genotype of the father? Of the mother? What is the probability that a second child will have the normal number of digits? Father’s geneotype must be Pp since polydactyly is dominant and he has had one normal child. Mother’s genotype is pp. The chance of the next child having normal digits is ½ or 50% because the mother can only donate a p allele and there is a 50% chance that the father will donate a p allele.

9. Achondroplasia is a common form of hereditary dwarfism that causes very short limbs, stubby hands, and an enlarged forehead. Above are four pedigrees depicting families with this specific type of dwarfism. What is the most likely mode of inheritance? Cite a reason for your answer.

Autosomal dominant – it affects both sexes, and occurs every generation10. Draw a pedigree to depict the following family:

One couple has a son and a daughter with normal skin pigmentation. Another couple has one son and two daughters with normal skin pigmentation. The daughter from the first couple has three children with the son of the second couple. Their son and one daughter have albinism; their other daughter has normal skin pigmentation.

11. Caleb has a double row of eyelashes, which he inherited from his mother as a dominant trait. His maternal grandfather is the only other relative to express the trait. Veronica, a woman with normal eyelashes, falls madly in love with Caleb, and they marry. Their first child, Polly has normal eyelashes. Now Veronica is pregnant again and hopes they will have a child who has double

eyelashes. What chance does a child of Veronica and Caleb have of inheriting double eyelashes 1/2 chance of that the child will have double eyelashes Draw a pedigree of this family.