Mendelian Genetics – Word problems with probability
1. Albinism is the absence of skin pigmentation and is a recessive trait found
in humans and other animals. In the human population about 1/20,000
individuals is an albino. Normal pigmentation (A) is dominant to albinism
(a). If an albino woman marries a homozygous normal man, what is the
likelihood that one of their children will display albinism?
- Parents’ Genotypes:
___aa__,
- Possible Gametes for each: __a___, ___a__
_______
- Punnet Square -draw and fill.
- Calculate probabilities.
The chances for having an albino child: 0%
___AA___
__A__A__,
2. A woman with normal pigmentation marries an albino man and their first
child is an albino. What are the genotypes of the couple?
Answer: The mother – Aa, The father - aa
3. A common squash in Texas is the Yellow crooked-neck squash. This fruit
is a source of vitamin A, B, and C. It also contains calcium and iron. Yellow
colored squash is recessive to white-colored squash. If a yellow male squash
is crossed with a heterozygous female white-squash, what are the predicted
genotypes and phenotypes of the offspring?
Parent squashes: Yellow male squash hh x Hh Female Heterozygote White
Accordingly: The phenotypic ratio: 1:1 for White:Yellow Squashes
The Genotypic ratio: 1:1:0 for Hh:hh:HH, respectively
4. Polydactylous cats have more than five toes. In fact, the author, Ernest
Hemingway is credited with establishing a large colony of about 50 feral
polydactylous cats in the Florida Keys. One of his cats, Princess six-toes
appeared in the New York Times. The polydactyl allele is dominant over the
allele for five toes and fingers. Predict the offspring of a mating between a
heterozygous polydactylous male cat and a female cat homozygous for five
toes and fingers.
Parent cats: heterozyg’ polydact’ male Pp x pp five toe female
Phenotypic ratio: 1:1 Polydact’ : five toe
Genotypic ratio: 1:1:0 Pp:pp:PP
5. In cattle the polled hornless condition (P) is dominant to the recessive
condition of horned (p). A heterozygous polled bull breaks out of his pen
and mates with the following three cows (cow #1) homozygous dominant
polled hornless (cow #2) horned, and (cow #3) heterozygous polled
hornless. What is the probability that all offspring will be horned? HINT:
The probability of three independent events occurring at the same time is the
product of the probability for each independent event.
For each cow, the chances for horned offspring:
Cow#1: Pp (hornless) x PP (hornless) – no chance for a hh horned.
Cow#2: Pp (hornless) x pp (horned) – 50% chance for a hh horned.
Cow#3: Pp (hornless) x Pp (hornless) – 25% chances for a hh horned.
The chances that all offspring will be horned is the product of all three cows, which
is 0x0.5x0.25 = 0%
The answers to the multiple choice questions that were on the back of this handout
were:
1c
2e
3d
4b
5c
6d