Ch. 1 Mendel's Principles of Heredity Practice Problems

An albino corn snake is crossed with a caramel colored corn snake. The offspring are all caramel colored. When these first generation progeny snakes are crossed among themselves, they produce 32 caramel colored snakes and 10 albino snakes. 

•  Which phenotype is controlled by the dominant allele? 

caramel (because it is the phenotype observed in the F1 hybrid)

An albino corn snake is crossed with a caramel colored corn snake. The offspring are all caramel colored. When these first generation progeny snakes are crossed among themselves, they produce 32 caramel colored snakes and 10 albino snakes. 

• What are the genotypes of the parents?

AA, aa (F2 ratio is approximately 3:1, therefore, F1 must be heterozygotes (Aa))

An albino corn snake is crossed with a caramel colored corn snake. The offspring are all caramel colored. When these first generation progeny snakes are crossed among themselves, they produce 32 caramel colored snakes and 10 albino snakes. 

• A caramel colored female snake of unknown genotype is involved in a test cross with an albino snake. This cross produces 10 caramel colored snakes and 11 albino snakes. What is the genotype of the caramel colored parent snake?

Aa (1:1 ratio of Aa (caramel) and aa (albino)

What is the probability of an Aa father and Aa mother having an aa child?

• In other words, what is the probability of a child inheriting a AND a?

1/4

What is the probability of an Aa father and Aa mother having a child with a dominant phenotype?

• Since a dominant phenotype can be seen in a heterozygous or homozygous genotype, we are asking what is the probability of a child having the genotype Aa, aA OR AA?

¾

• first, we can apply the product rule for each of these genotypes: AA, Aa, aa

• then, we can apply the sum rule: ¼ AA, ¼ Aa, ¼ aa

• ¼ + ¼ + ¼ = ¾

What is the possibility of getting Yy Rr progeny in F2?

• P: YY RR x yy rr

• F1: Yy Rr (all identical)

¼

• break down into two questions:

1. what is the probability of getting Yy?

2. what is the probability og getting Rr?

• use the punnett square to calculate each separately and then multiply the two answers

cross Aa Bb Cc Dd x Aa Bb Cc Dd

what proportion of progeny will be AA bb Cc Dd?

1/64

• Aa x Aa → ¼ AA

• Bb x Bb → ¼ bb

• Cc x Cc → ½ Cc

• Dd x Dd → ½ Dd

• ¼ x ¼ x ½ x ½ = 1/64

cross Aa Bb Cc Dd x Aa Bb Cc Dd

how many progeny will show the dominant traits for A, C, and D and the recessive trait for B?

27/256

• Aa x Aa → ¾ A-

• Bb x Bb → bb

• Cc x Cc → ¾ C-

• Dd x Dd → ¾ D-

• ¾ x ¼ x ¾ x ¾ = 27/256

In tomatoes, red fruit is dominant to yellow fruit and purple stems are dominant to green stems. The progeny from one mating consists of:

• 305 red fruit, purple stem plants

• 328 red fruit, green stem plants

• 110 yellow fruit, purple stem plants

• 97 yellow fruit, green stem plants

What were the genotypes of the parents in this cross?

RrPp x Rrpp

For the following cross: AA Bb Cc Dd x Aa Bb cc DD

How many different types of gametes are possible for the parent on the left?

8

For the following cross: AA Bb Cc Dd x Aa Bb cc DD

What is the probability of having a child that phenotypically resembles the parent on the left?

3/8

Huntington’s Disease is a rare fatal, degenerative neurological disease in which individuals start to show symptoms in their 40s. It is caused by a dominant allele. Joe, a man in his 20s, just learned his father has the disease.

What is the probability that Joe will develop the disease?

½

Huntington’s Disease is a rare fatal, degenerative neurological disease in which individuals start to show symptoms in their 40s. It is caused by a dominant allele. Joe, a man in his 20s, just learned his father has the disease.

What is the probability that Joe’s first child will also develop the disease?

¼

Tay-Sachs disease is a recessive early-onset neurodegenerative disease. The disease is rare in a population overall, but is found at a relatively high frequency in Ashkenazi Jews from eastern Europe. A woman whose maternal uncle had the disease is trying to determine the probability that she and her husband could have an affected child. Her father does not come from a high-risk population, but her husband’s sister died of the disease at a young age.

What is the probability of their first child being affected by the disease?

1/18

• three steps:

  • calculate probability of a person being a carrier

  • calculate probability of passing it to their offspring if they are a carrier

  • use product rule to multiply the above probability together

• see ppt