Page 46 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

38 Chapter 2 Mendel’s Principles of Heredity

II. In tomatoes, red fruit is dominant to yellow fruit, Answer
and purple stems are dominant to green stems. The This problem requires an understanding of dominance/
progeny from one mating consisted of 305 red fruit, recessiveness and probability. First diagram the pedigree,
purple stem plants; 328 red fruit, green stem plants; and then assign as many genotypes as possible using the
110 yellow fruit, purple stem plants; and 97 yellow following allele designations:
fruit, green stem plants. What were the genotypes
of the parents in this cross?
T = normal allele; t = Tay-Sachs allele
Tt Tt
Answer I 1 2
This problem requires an understanding of independent as- TT Tt Tt
sortment in a dihybrid cross as well as the ratios predicted II tt
1
from monohybrid crosses. A ected 2 3 4 5
Designate the alleles: uncle
R = red, r = yellow III 1 2 3 tt
A ected
P = purple stems, p = green stems ? sister
In genetics problems, the ratios of offspring can indicate
the genotype of parents. You will usually need to total the The genotypes of the two affected individuals, the woman’s
number of progeny and approximate the ratio of offspring uncle (II-1) and the husband’s sister (III-3), are tt. Because
in each of the different classes. For this problem, in which the uncle was affected, both of his parents must have been
the inheritance of two traits is given, consider each trait heterozygous.
independently. For red fruit, there are 305 + 328 = 633 Similarly, as the husband’s sister (III-3) is affected,
red-fruited plants out of a total of 840 plants. This value both of her parents (II-4 and II-5) must be heterozy-
(633/840) is close to 3/4. About 1/4 of the plants have yel- gotes. Finally, because individual II-3 is not from a
low fruit (110 + 97 = 207/840). From Mendel’s work, high-risk population, the most likely assumption is that
you know that a 3:1 phenotypic ratio results from crosses he is TT.
between plants that are hybrid (heterozygous) for one You next need to determine the chance that a child of
gene. Therefore, the genotype for fruit color of each par- III-1 and III-2 (that is, individual IV-1) would have Tay-
ent must have been Rr. Sachs (tt). For that to be possible, both III-1 and III-2 must
For stem color, 305 + 110 or 415/840 plants had pur-
ple stems. About half had purple stems, and the other half be Tt given that neither is tt. For III-1 to be Tt, II-2 must be
Tt. Calculating the chance that II-2 is Tt is a bit tricky. At
(328 + 97) had green stems. A 1:1 phenotypic ratio occurs first, it appears that the chance is 1/2 that the daughter of
when a heterozygote is mated to a homozygous recessive two heterozygous (Tt) parents would be Tt: the expected
(as in a testcross). The parents’ genotypes must have been progeny ratio is 1 TT : 2 Tt : 1 tt. However, in this case you
Pp and pp for stem color. have additional information to consider: II-2 is unaffected
The complete genotype of the parent plants in this
cross was Rr Pp × Rr pp. and thus the genotype tt is ruled out. That leaves 1 TT :
2 Tt, or a 2/3 chance that II-2 is Tt. If so, the chance that
II-2 would transmit the t allele to III-1 is 1/2. Thus, the
III. Tay-Sachs is a recessive lethal disease in which there probability that III-1 is Tt is 2/3 × 1/2 = 1/3. This fact
is neurological deterioration early in life. This disease implies that II-2 could be either TT (probability = 1/3) or
is rare in the population overall but is found at rela- Tt (probability = 2/3). If II-2 is Tt, the chance that she
tively high frequency in Ashkenazi Jews from Eastern would transmit the t allele to III-1 is 1/2. Thus, the proba-
Europe. A woman whose maternal uncle had the dis- bility that III-1 is Tt is 2/3 × 1/2 = 1/3.
ease is trying to determine the probability that she What is the chance that III-2 is Tt? Both of his parents
and her husband could have an affected child. Her fa- are heterozygous, and he is unaffected. Thus, using similar
ther does not come from a high-risk population. Her logic, the likelihood that III-2 is Tt is 2/3.
husband’s sister died of the disease at an early age. The probability that both III-1 and III-2 are Tt is 1/3 ×
a. Draw the pedigree of the individuals described. 2/3 = 2/9. The chance that the child of two Tt parents would
Include the genotypes where possible. be tt is 1/4. Thus, the overall likelihood that IV-1, the child
b. Determine the probability that the couple’s first of III-1 and III-2, would have Tay-Sachs is 2/9 × 1/4 =
child will be affected. 1/18.