Module 6 Mendelian Genetics
Explore the chapters to learn the unit 1 content.
Lesson 3.6.5
3.6.5 page 4
Self-Check
Â
Try out these questions to see if you have mastered the use of probabilities.
-
The ability to taste the chemical PTC is determined by a single gene in humans, with the ability to taste PTC indicated by the dominant allele T, and the inability to taste PTC by the recessive allele t. Suppose two heterozygous tasters (Tt) have a large family.Â
-
Predict the proportion of their children who will be tasters and non-tasters. Use a Punnett square to illustrate how you made this prediction.
-
What is the likelihood that their first child will be a taster? What is the likelihood that their fourth child will be a taster?
-
What is the likelihood that the first three children of this couple will be non-tasters?
-
-
A husband and wife are both heterozygous for a recessive gene, c, for albinism. They were informed that the twins they are expecting are dizygotic, a boy and a girl.
-
Draw a Punnett square of this cross.
-
What are the chances that one child will be albino?
-
What are the chances that both children will be normal?
-
What are the chances that both babies will have the same phenotype for skin pigmentation?
-
Self-Check Answers
Â
-
Parents:         Tt                    x         Tt
Gametes:        [T], [t]              x         [T], [t]
|
 |
T |
t |
|
T |
TT |
Tt |
|
t |
tT |
tt |
¼ TT + ½ Tt = ¾ tasters
¼ tt = ¼ non-tasters
-
1st child-- 3/4
4th child-- 3/4 (each child is independent)
These are not linked. -
Here they are linked probabilities, so we use the multiplication rule.
1/4 for each child; 1/4 x 1/4 x 1/4 = 1/64 that all three will be non-tasters.
|
 |
C |
c |
|
C |
CC |
Cc |
|
c |
Cc |
cc |
Â
-
1/4 or 25% or 0.25
-
3/4 x 3/4 = 9/16; this is the multiplication rule.
-
First: 1/4 x 1/4 = 1/16 for two albinos
Next: 3/4 x 3/4 = 9/16 for two normal
Then: 9/16 + 1/16 = 10/16 or 5/8 for either!