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ch04 - Chapter 4 4.1 What blood types could be observed in...

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Chapter 4 4.1 What blood types could be observed in children born to a woman who has blood type M and a man who has blood type MN? ANS:  M and MN. FEEDBACK:  4.1 DIFFICULTY: easy 4.2 In rabbits, coloration of the fur depends on alleles of the gene  c.  From information given in the chapter, what phenotypes and proportions would be expected from the following   crosses:   (a)   ; c c cc + + ´   (b)   ; c c c c + + ´   (c)   ; h ch c c c c + + ´   (d)   ; ch cc cc ´   (e) ; h c c c c + + ´  (f)  ? h c c cc ´   ANS:  (a) all wild-type; (b) 3/4 wild-type, 1/4 albino; (c) 3/4 wild-type, 1/4 chinchilla; (d) 1/2 chinchilla, 1/2 albino; (e) 3/4 wild-type, 1/4 himalayan; (f) 1/2 himalayan, 1/2 albino. FEEDBACK:  4.1 DIFFICULTY: medium 4.3    In mice, a series of five alleles determines fur color. In order of dominance, these alleles are:   , Y A   yellow fur but homozygous lethal;   , L A   agouti with light belly;   , A + agouti (wild-type);  , t a  black and tan; and  a,  black. For each of the following crosses, give the coat color of the parents and the phenotypic ratios expected among the progeny: (a)  ; Y L Y L A A A A ´  (b)  ; Y L t A a A a ´  (c)  ; t Y a a A a ´  (d)  ; L t L L A a A A ´  (e)  ; L L Y A A A A + ´ (f)  ; t t A a a a + ´  (g)  ; t a a aa ´  (h)  ; Y L t A A A a + ´  and (i)  . Y L Y A a A A + ´ ANS:  
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Parents Offspring (a) yellow   yellow 2 yellow: 1 light belly (b) yellow   light belly 2 yellow: 1 light belly: 1 black and tan (c) black and tan   yellow 2 yellow: 1 black and tan: 1 black (d) light belly   light belly all light belly (e) light belly   yellow 1 yellow: 1 light belly (f) agouti   black and tan 1 agouti: 1 black and tan (g) black and tan   black 1 black and tan: 1 black (h) yellow   agouti 1 yellow: 1 light belly (i) yellow   yellow 2 yellow: 1 light belly FEEDBACK:  4.1 DIFFICULTY: hard 4.4 In several plants, such as tobacco, primrose, and red clover, combinations of alleles in eggs and pollen have been found to influence the reproductive compatibility of the plants. Homozygous combinations, such as  1 1 , S S  do not develop because  1 S  pollen is not effective on  1 S -- stigmas. However,  1 S  pollen is effective on  2 3 S S  stigmas. What progeny might be expected from the following crosses (seed parent written first): (a) 1 2 2 3 ; S S S S ´  (b)  1 2 3 4 ; S S S S ´  (c)  4 5 4 5 ; S S S S ´  and (d)  3 4 5 6 ? S S S S ´   ANS:  (a) S 1  S 2 , S 1  S 3 , S 2  S 3 ; (b) S 1  S 3 , S 1  S 4 , S 2  S 3 , S 2  S 4 ; (c) S 4  S 5 ; (d) S 3  S 5 , S 3  S 6 , S 4  S 5 , S 4  S 6 .
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