Question

1. Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:
Flower position: Axial (A) is dominant to Terminal (a)
Stem length: Tall (T) is dominant to dwarf (t)
Seed shape: Round (R) is dominant to wrinkled (r)

If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows?

a. homozygous for the three dominant traitsb. homozygous for the three recessive traitsc. heterozygous for all three charactersd. homozygous for axial and tall, heterozygous for seed shape

Answer 1

Answer:

a. homozygous for the three dominant traits AATTDD= 1/64

b. homozygous for the three recessive traits: aattdd = 1/64

c. heterozygous for all three characters AaTtDd  =   proportion of

  Aa = 2/4, Tt = 2/4, Dd = 2/4. thus we have 2/4 x 2/4 x 2/4 =8/64 = 1/8.

d. homozygous for axial and tall, heterozygous for seed shape.

 AATTDd -  homozygous axial AA-1/4  homozygous tall TT -1/4     heterozygous for seedshape Dd - 2/4

AATTDd 1/4 x 1/4 x 2/4 = 2/64 = 1/32

Explanation:

If a plant that is heterozygous for all three characters is allowed to self-fertilize:

      AaTtRr x self

    ATR, ATr, AtR, Atr

    aTR, atR, aTr, atr

a. homozygous for the three dominant traits AATTDD= 1/64

b. homozygous for the three recessive traits: aattdd = 1/64

c. heterozygous for all three characters AaTtDd  =   proportion of

  Aa = 2/4, Tt = 2/4, Dd = 2/4. thus we have 2/4 x 2/4 x 2/4 =8/64 = 1/8.

d. homozygous for axial and tall, heterozygous for seed shape.

 AATTDd -  homozygous axial AA-1/4  homozygous tall TT -1/4     heterozygous for seedshape Dd - 2/4

AATTDd 1/4 x 1/4 x 2/4 = 2/64 = 1/32

Answer 2

Answer:

a. 1/64 AATTDD

b. 1/64 aattdd

c. 8/64=1/8 AaTtDd

d. 2/64 = 1/32 AATTDd or aattDd or AAttDd or aaTTDd

Explanation:

Available data:

• Flower position: Axial (A) is dominant to Terminal (a)

• Stem length: Tall (T) is dominant to dwarf (t)

• Seed shape: Round (R) is dominant to wrinkled (r)

Cross:

Parental) AaTtDd   x   AaTtDd

Knowing that these four genes assort independently, then we can infer that the F1 will have the next genotypic proportions for each gene:

• AA= 1/4

       Aa =2/4

       aa = 1/4

• TT = 1/4

        Tt = 2/4

         tt = 1/4

• DD = 1/4

        Dd = 2/4

        dd = 1/4

This is the same as developing a Punnett square for each gene separately.

Knowing this, to get each of the mentioned genotypes for each trait, we must multiply their respective genotypic proportions, like this:

a. homozygous for the three dominant traits, AATTDD

   AA1/4 x TT 1/4 x DD 1/4 = AATTDD 1/64

b. homozygous for the three recessive traits, aattdd

   aa 1/4 x tt 1/4 x dd 1/4 = aattdd 1/64

c. heterozygous for all three characters, AaTtDd

   Aa 2/4 x Tt 2/4 x Dd 2/4 = AaTtDd 8/16=1/8

d. homozygous for axial and tall, heterozygous for seed shape. There are four possibilities, all of them with the same result:

AATTDd --> 1/4  x 1/4 x 2/4 = 2/64=1/32

aattDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

AAttDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

aaTTDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32