1. Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

 

Character                        Dominant                                         Recessive

Flower position              Axial (A)                                        Terminal (a)

Stem length                    Tall (L)                                           Dwarf (l)

Seed shape                      Round (R)                                      Wrinkled (r)

 

 

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.)

 

                  i.    homozygous for the three dominant traits

                  ii.    homozygous for the three recessive traits

                  iii.   heterozygous at all three traits

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

 

2. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.

 

3. The genotype of F₁ individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F₂ offspring would have the following genotypes?

 

                  i.    aabbccdd

                  ii.    AaBbCcDd

                 iii.    AABBCCDD

                 iv.    AaBBccDd

                  v.   AaBBCCdd

 

4.What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs)?

 

                  i.    AABBCC x aabbcc --- AaBbCc

                  ii.    AABbCc x AaBbCc --- AAbbCC

                 iii.    AaBbCc x AaBbCc --- AaBbCc

                 iv.    aaBbCC x AABbcc --- AaBbCc

 

 

 

 

 

 

5.         B is a dominant allele coding for black fur on rabbits and b is a recessive allele coding for white fur on rabbits. Fill in the following blanks with the correct cross of the following:

(1) BB x bb, (2) Bb x Bb, (3) bb x bb, (4) Bb x bb

 

                  i.    All (100%) of the offspring are white: __________

                  ii.    One quarter (25%) of the offspring are white: __________

                  iii.   All (100%) of the offspring are black: ___________

                  iv.   Three-quarters (75%) of the offspring are black: __________

                  v.   One-half (50%) of the offspring are white: __________

 

6.         In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F₁ generation?

 

 

ANSWERS

 

1. The cross is the same for all the questions. The plant is heterozygous at all three loci (genes) and is self-fertilizes, therefore

                                                AaLlRr  X AaLlRr

 

i. homozygous dominant at all three traits = AALLRR

probability = ¼ X ¼ X ¼ or 1/64

 

ii. homozygous recessive at all three traits = aallrr

probability = ¼ X ¼ X ¼ or  1/64

 

iii heterozygous at all three traits = AaLlRr

probability = ½ X ½ X ½ = 1/8

 

iv. homozygous axial and tall, heterozygous for seed shape = AALLRr

probability = ¼ X ¼ X ½ = 1/32

 

 

2. The best explanation is that albinism is recessive. In the first cross, the albino guinea pig (aa) was likely mated to a black guinea pig that was homozygous (AA). In the second cross, the black guinea pig had to have been heterozygous (Aa).

How does this question relate to our discussion about test crosses?

 

 

 

 

 

 

3.  The cross is a tetrahybrid cross AaBbCcDd X AaBbCcDd

 

i.          aabbccdd         probability =  ¼ X ¼ X ¼  X ¼ or 1/256

ii.          AaBbCcDd      probability =   ½ X ½ X ½ X ½ or 1/16

iii.         AABBCCDD   probability =   ¼ X ¼ X ¼ X ¼ or 1/256

iv.         AaBBccDd      probability =   ½ X ¼ X ¼ X ½ or 1/64

v.         AaBBCCdd      probability =  ½ X ¼ X ¼ X ¼ or 1/128

 

 

4. What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs)?

 

i.          AABBCC x aabbcc --- AaBbCc    probability = 1 X 1 X 1 X 1 or 100%

ii.          AABbCc x AaBbCc --- AAbbCC  probability =  ½ X ¼ X ¼  or 1/32

iii.         AaBbCc x AaBbCc --- AaBbCc   probability =   ½ X ½  X ½ or 1/8

iv.         aaBbCC x AABbcc --- AaBbCc   probability =  1 X ½  X 1 or  1/2

 

 

5.         (1) BB x bb, (2) Bb x Bb, (3) bb x bb, (4) Bb x bb

 

                  i.    All (100%) of the offspring are white: ___3_______

                  ii.    One quarter (25%) of the offspring are white: _____2_____

                  iii.   All (100%) of the offspring are black: _____1______

                  iv.   Three-quarters (75%) of the offspring are black: _____2_____

                  v.   One-half (50%) of the offspring are white: ____4______

 

6.  IiPp X IiPp

 

This is an example of epistasis. Do the math on the control gene first.

 

Ii X Ii 

 

¾ of the offspring will inherit at least one I allele so will have white kernels.

 

Of the remaining ¼ that will possess the ii alleles and are able to express colour. ¾ will have purple kernels (P-) and ¼ will have red kernels (pp).

So we predict from this cross that the ratio of white:purple:red will be 12:3:1

Remember ¾ = 12/16

¼ X ¾ = 3/16

and ¼ X ¼ = 1/16