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1. The web site takes you to Online Mendelian Inheritance in Man, #219700, which
discusses the gene encoding CFTR, the gene defective in cystic fibrosis. Ans. (b)
2. The web site gives the gene locus at ‘7q31.2’. The first number here designates the
chromosome, chromosome 7. The other designations indicate where the gene is located
on chromosome 7, and we will discuss this convention later in the course. They refer to a
specific position on the longer arm of the acrocentric chromosome 7. Ans. (c)
3. One hypothesis is that heterozygotes may have a selective advantage against the
effects of bacterial endotoxins that can cause secretory diarrhea. As discussed in the text:
Baxter et al. (1988) presented actual observations indicating that intestine in CF
homozygotes fails to exhibit a secretory response on exposure to bacterial toxins that
would normally induce a secretory diarrhea. In a study of CFTR homozygous mutant
mice, transgenic mice that expressed no CFTR protein did not secrete fluid in response to
cholera toxin. Heterozygotes expressed 50% of the normal amount of CFTR protein in
the intestinal epithelium and secreted 50% of the normal fluid and chloride ion in
response to cholera toxin. The findings suggested that CF heterozygotes may possess a
selective advantage of resistance to cholera. Ans. (a)
AaBbccDdEe x aaBbCcDDEe
4. P(AaBbccDdEe) = ½ x ½ x ½ x ½ x ½ = 1/32; Ans. (b)
5. (A_B_ccD_E_) = ½ x ¾ x ½ x 1 x ¾ = 9/64; none of the above Ans. (e)
6. For eye color, the progeny show ¾ red and ¼ sepia, therefore parents must be
heterozygous for mutant allele (se+/se x se+/se). For wing length, the progeny show a 1:1
ratio of mutant to wild type, therefore one parent must be heterozygous and one
homozygous recessive (vg+/vg x vg/vg). se+/se vg+/vg x se+/se vg/vg Ans. (d)
7. At the end of meiosis I, there has been a reductional division, and the two homologs
have separated. Each homolog, however, has replicated prior to the beginning of meiosis,
and consists of two sister chromatids. There are 23 chromosomes per cell, or 46
chromatids. Ans. 46 (b).
8. For the answer to this question, you need to be able to read the codon table in Chapter
1, page 6. Both UCA and AGC code for serine. Ans. (d)
9. You need to draw the pedigree first: Brad’s brother is affected, and Janet’s maternal
grandfather had a brother and sister who died of the disease:
Aa
Aa
2/3
Aa
AA
1/2
Aa
Aa
Aa
AA
1/2
Aa
2/3
Aa
1/4
Aa
2/3 x ½ x ½ x 2/3 x ¼ = 1/36 ans: (c)
10. Additional information: Janet’s aunt just had a child with Pompe’s disease…But we
already know that Janet’s mother is an obligate heterozygote from the birth of her sister.
There would not be any change in the odds on the basis of this information. Ans: 1/36
(c). It would be odd that so many people marrying into the family were
heterozygotes…on closer inspection was there evidence of consanguinity?
Aa
Aa
2/3
Aa
Aa
Aa
1/2
Aa
1/2
Aa
2/3
Aa
1/4
Aa
?