1. Probability has no memory- there is a 100% probability that the previous events
have occurred and they have led to all girls. Regardless of what the relatives say,
the chance for the next birth remains approximately 0.5 for either a boy or a girl, and
all previous children do not figure into the probability. Ans. (e) 1/2.
2.
parents
progeny
type of cross
pin 1 x pin 2
44 pin
recessive x recessive
thrum 3 x thrum 3
23 thrum
dominant x dominant
thrum 3 x pin 1
32 thrum
dominant x recessive
thrum 4 x pin 2
20 thrum 17 pin
heterozygote x recessive
3.
The last cross indicates that the thrum 4 plant must be a hybrid, since it gives
approximately a 1:1 ratio of dominant to recessive phenotypes. The genotypes of the 4
plants must therefore be: hh, hh, HH, Hh. Ans: (b).
3. Plant thrum 4 must be a heterozygote since it produces recessive phenotype progeny
in a testcross. Thrum 4 x Thrum 4 would therefore be a monohybrid cross, and should
produce a 3 thrum to 1 pin ratio. Ans: (d).
AABbCcddEe x AaBbCCDdEe
4. P (A_B_C_ddE_) = 1 x 3/4 x 1 x 1/2 x 3/4 = 9/32 Ans: (a)
5. P [(A_B_C_ddE_) or (A_B_CCDdEe)] = 9/32 + (1 x 3/4 x 1 x 1/2 x 3/4) =
9/32 + 9/32 = 18/32 = 9/16 Ans: (e) – none of the above.
6. Probability of neither parent = 1 - [probability of either parent) = 1 - (9/16) = 7/16 Ans:
(c)
7. Ans: (e) None of the above- please see ‘fast forward’ box on pages 21-22 in your
textbook.
8. There are only 20 different amino acids that are encoded into proteins, which can
differ in the number of residues in the polypeptide chain containing these amino acids.
The number of genes in E. coli is approximately 4,500- the human genome contains
between 20,000-30,000 genes. Gamete cells have undergone meiotic divisions- their
chromosomes are no longer paired (i.e. diploid) and contain a haploid genome (23
chromosomes in humans. See figure 1.12 in textbook, and accompanying text.
The wife had a maternal uncle with TSD and it is known that the uncle’s parents were
related. The husband had a male first cousin on his father’s side who died from TSD,
and it is known that this first cousin’s parents were related. There are no other known
cases and no other known instances of consanguinity. Assume that the trait is very rare
in the general population.
Pedigree problem…first draw the pedigree:
or
Aa
Aa
aa
Aa
r
AA
AA AA
⅔Aa
½ Aa
½ Aa
½ Aa
Aa
Aa
aa
¼ aa
9. Since individual II-1 is affected, then the parents must be heterozygotes. This means
individual II-2 has a 2/3 change of inheriting a recessive allele. If the trait is rare,
individual II-3 is presumed to be normal. There is then a ½ chance individual will pass
on the TSD gene in her gametes. The odds that individual III-1 will therefore be a
heterozygote that has the disease gene is: 2/3 x ½ = 1/3. Ans: (b).
10. Since individual II-VI is an obligate heterozygote, he must have received a disease
gene from a parent. They are not related, so presumably one of the parents was a
heterozygote for the rare TSD allele; that parent had a ½ chance of passing on the
disease gene to II-5, the husband’s father. He had a ½ chance of passing the rare allele
to III-2. The probability the husband will therefore be a heterozygote that has the
disease gene is ½ x ½ = ¼. The odds that two heterozygotes will pass the disease
gene on two their progeny is ¼. Therefore, the combined probability is:
P( III-1 is heterozygote) x P (III-2 is heterozygote) x P (aa child) =
1/3 x ¼ x1/4 = 1/48. Ans: (d).