Ch 13 17 Please disregard the problems that need a Figure to

Ch. 13 17>.

Please disregard the problems that need a Figure to solve, the picture is not showing up. Thank you so much for your help.

Solution

Question 17

Figure 3.6 shows the results of a cross between a tall pea plant and a short pea plant.
a. What phenotypes and proportions will be produced if a tall F1 progeny is backcrossed to the short parent?

The tall F1 progeny are all heterozygous (Tt). The short parent is homozygous for the short allele (tt).

F1 tall pea plant × parental short pea plant
         (Tt) x       (tt)

T

t

t

Tt

tt

t

Tt

tt

½ tall pea plants (Tt)
½ short pea plants (tt)

C. 1/2 tall and 1/2 short

===================

Question 18

The tall F1 progeny are all heterozygous (Tt). The tall parent is heterozygous (Tt) as well.

F1 tall pea plant × parental tall pea plant
(Tt) (Tt)

T

t

T

TT

Tt

t

Tt

tt

answer
¾ tall pea plants (TT and Tt)
¼ short pea plants (tt)

==================================================================================

Question 19

Alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A).

Sally has normal metabolism, but her brother has alkaptonuria. Sally\'s father has alkaptonuria, and her mother has normal metabolism. What is Sally\'s genotype?

Answer = Sally -----Aa      

======================================================

Question 20

Answer : option E

Sally\'s father, who has alkaptonuria, must be aa. Her brother, who also has alkaptonuria, must be aa as well. Because both parents must have contributed one a allele to her brother, Sally\'s mother, who is phenotypically normal, must be heterozygous (Aa).

Sally, who is normal, received the A allele from her mother but must have received an a allele from her father.The genotypes of the individuals are:

Sally (Aa),

Sally\'s mother (Aa),

Sally\'s father (aa),

And

Sally\'s brother (aa).