Answers to Homework Assignment #2

ANSWERS: BI311 Homework Assignment #2

Chapter 2

2.2:   2/3. Because the seed is round it cannot be ww. Ww seeds give 2 bands in the gel. Of the round seeds, 1/3 are WW and 2/3 are Ww.

2.4.    Note that you assume ‘rare’ alleles to exist in heterozygous state.

a) parents must be Aa x aa so the risk of having an Aa child is 1/2

b) parents must be Aa x Aa so the risk of having an aa child is 1/4

c) parents must be Aa x aa so the risk of having an aa child is 1/2

 

2.8.

a) Individual II-2 is aa

b) Both individuals I-1 and I-2 must be Aa

c) Individuals II-1 and II-3 are either AA or Aa

d) Because II-3 is phenotypically normal, the odds that they are heterozygous carrier is 2/3

 

2.16:   Due to Mendelian segregation, the expected numbers are: a) 0; b) 100; c) 200; d) 100; e) 0; f) 0

 

2.21.

a) Recessive allele because the affected individual had unaffected parents. Also, the parents of the affected individual are related.

b) The double line indicates a mating between relatives.

c) III-1 and III-2 are first cousins.

d) I-1 and I-2 could be (AA) or (Aa). At least one individual must be (Aa) or both could be (Aa). Not possible for both to be (AA). II-1 and II-4 are most likely (AA) because they a allele is a rare trait. Individuals that marry into a family are assumed to be wildtype unless there is evidence to the contrary.   III-1 and III-2 are both (Aa) because they have an affected offspring (IV-4).

 

Chapter 3

3.1. After the centromeres have split, each former sister chromatid is counted as a chromosome in its own right, so at anaphase there are 92 chromosomes.

 

Additional Questions not in your textbook:

#1. On the basis of Mendel’s observations, predict the results from the following crosses with peas:

a) true-breeding tall peas x true-breeding dwarf peas

all tall

b) the offspring of (a) self-fertilized

¾ tall, ¼ dwarf

c) offspring of (a) crossed with the original tall parent

all tall

d) the offspring of (a) crossed with original dwarf parent

½ tall, ½ dwarf

 

#2. In pigeons, a dominant allele C causes a checkered pattern in feathers, the recessive allele c produces a plain pattern. Feather color is controlled by an independently assorting gene; where the dominant B allele causes red feathers and the recessive allele b produces brown feathers. You cross a true-breeding checkered red pigeon with a true-breeding plain brown pigeon.

a) What is the phenotype of the F1 offspring?

checkered and red feathers

b) If these F1 offspring are crossed with each other, what phenotypes will appear in the F2 generation, and in what proportions?

9 checkered red : 3 checkered brown : 3 plain red : 1 plain brown

 

 

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