Question

1. Let's say an individual is of the genotype AABbCcDdEe. a) How many different gamete types could he/she produce for could he/she produce for these traits ? another of the same genotype, what proportion of their b) If this individual were to mate with another of the same genotype, offspring would we expect to show all 5 of the dominant phenotypes expect to show all 5 of the dominant phenotypes? How about all 5 of the recessive phenotypes ? c) Again, assuming this individual mates with another of the same phenotype, what is the probability that a randomly selected offspring will have the genotype of AAbbccDdEE? 2. In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A_B_ = gray A _bb = yellow aa B _ = black aa bb = cream A third gene pair on a separate autosome determines whether or produced. The CC and Cc genotypes allow color according to and B alleles. However, the cc genotype results in albino rats alleles present. Determine the F-1 phenotypic ratio of the follo rate autosome determines whether or not any color will be types allow color according to the expression of the A bino rats regardless of the A and B otypic ratio of the following crosses. a. AAbbCC x aaBBcc b. AaBbCc x AABbcc с. АаВВСc x AaBBCс

Answer 1

Answer:

1).

a).

AA produces only one type of gametes = 1 (A)

BB produces two types of gametes = 2 (B & b)

Cc = 2

Dd = 2

Ee = 2

Number of different gamete types could he/she produce for these traits = 1*2*2*2*2 = 16

b).

AABbCcDdEe x AABbCcDdEe ---Parents

AA xAA = AA (1)

Bb x Bb = BB (1/4), Bb (1/2) & bb (1/4)

Cc x Cc = CC (1/4), Cc (1/2) & cc (1/4)

Dd x Dd = DD (1/4), Dd (1/2) & dd (1/4)

Ee x Ee = EE (1/4), Ee (1/2) & ee (1/4)

B_ = BB + Bb = 3/4

Proportion of their offspring would be 5 dominant phenotypes = AAB_C_D_E_ = 1*3/4*3/4*3/4*3/4 = 81/256

Proportion of their offspring would be 5 recessive phenotypes = aabbccddee = 0*1/4*1/4*1/4*1/4 = 0

c).

AAbbccDdEE = 1*1/4*1/4*1/2*1/4 = 1/128

2).

a).

AAbbCC (yellow) x aaBBcc (albino)---Parents

AaBbCc (gray)-----F1

All progeny are with gray color.

b).

AaBbCc (gray) x AABbcc (albino) –Parents

Aa x AA = A_ (1)

Bb x Bb = B_(3/4) & bb (1/4)

Cc x cc = Cc (1/2) & cc (1/2)

Gray = A_B_C_ = 1 * ¾ * ½ = 3/8

Yellow = A_bb C_= 1 * ¼ *1/2= 1/8

Black = aaB_C_ = 0*3/4*1/2 = 0

Cream = aabbC_ = 0*1/4*1/2 = 0

All colors = 3/8+1/8 = 4/8

Albino = 1- 4/8 = ½

c).

AaBBCc x AaBBCc ---Parents

Aa x Aa = A_ (3/4) & aa (1/4)

BB x BB = BB (1)

Cc Cc = C_ (3/4) & cc(1/4)

Gray = A_B_C_ = ¾ * 1 * ¾ = 9/16

Yellow = A_bb C_= 3/4 * 0 *3/4= 0

Black = aaB_C_ = 1/4*1*3/4 = 3/16

Cream = aabbC_ = ¼*0*3/4 = 0

All colors = 9/16+3/16 = 12/16

Albino = 1- 12/16 = 4/16 = 1/4