Genetics Bio 206 Section BC

HW#1

1. Testcross- When individual expressing the dominant phenotype (Y-) is crossed with an

individual which expresses the recessive phenotype (yy).

2. Backcross- the crossing of a heterozygote and crossing it with one of its parents or one

genetically similar to one of its parents.

3. F1- the first generation and the progeny of the P generation.

4. F2- second generation the progeny of the F1 generation.

5. Dominant-  Gene that produces the same phenotype in the organism whether or not its

allele identical

6. Recessive- the trait that is not expressed in a heterozygote, rather remains hidden through

generations

7. Co-dominant- genes of equal dominance and are both expressed in the phenotype of the

individual

8. Partially Dominant- the situation in which both alleles of a heterozygote influence the

phenotype

Q5) Consider the cross between pure lines (i.e. homozygotes)

AABB x aabb

Write the genotypes of the F1 and the F2 generations

The F1 generations will all be AaBb

The F2 generation will consist of:

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb Aabb

6) A rare recessive trait in a pedigree is indicated by which pattern of inheritance?

B. horizontal

7) Sickle cell anemia is a recessive trait in humans. In a cross between a father who has

sickle cell anemia and a mother who is heterozygous for the gene, what is the probability that

their first child will have the normal phenotype?

-50 percent. (1/2)

What is the probability that their second child will have the normal phenotype?

-50 percent. (1/2)

What is the probability that both children will have the normal phenotype?

(1/2)(1/2)= 1/4

8) After a cross between two mice, the F1 offspring all had the same phenotype.

The F2 consisted of 91 short tails and 29 normal tails.

Identify the phenotypes and genotypes of the two parent mice.

Answer:

The F1 shows only one phenotype; the F2 shows two phenotypes. This implies that the

parents must have had different alleles, otherwise the F2 would exhibit only one phenotype.

Given that the two parents have different genotypes and that the F1 shows a single phenotype

one of the alleles must be dominant to the other. The dominant allele must be the one that

determines the phenotype that is found in excess in the F2, i.e. short tail. We can call (S) the

allele for short tail and (s) the allele for short tail.

The phenotypic ratio of the F2 is 91/120=0.76 or 76% and 29/120=0.24 or 24%.

These proportion are very close to 3:1. This is the ratio that is obtained in the F2 from

an F1 of the type Ss x Ss

( work out the cross Ss x Ss).

If the F1 is all Ss the parents must be SS and ss, i.e one short tail and one normal tail.

9) (from text book)

2. Breeders failed to uncover the principle that traits are governed by discrete units f

inheritance because people believed in the idea where one parent contributes to most of the

progenies characteristics, and also the idea of blended inheritance.

4.

(a) The normal-colored snake phenotype is controlled by the dominant allele.

(b) The female snake is Aa and the male snake is aa.

6.

(a)The piebald spotting trait is dominant because both parents show this condition. By having

child with this skin condition and another child with normal skin it that at least one of the

parents also carries the allele for the normal skin pigmentation.

(b) Both the parents must be a heterozygote genotype. If parents were not both heterozygote,

then the recessive (normal) pigmentation would not occur.

8. Since all the F1 progeny were open, the dominant trait is open flowers, so closed trait can

only be passed by a homozygote of the recessive allele.

32. (a) The probability that Joe will also develop Huntington disease is 50%.

(b) The probability that Joe’s first child will develop Huntington disease is 25%.