Chapter 3 Genetics
49.) A corn geneticist has three pure lines of genotypes a /a ; B /B ; C /C , A /A ; b /b ; C /C , and A /A ; B /B ; c /c . All the phenotypes determined by a , b , and c will increase the market value of the corn, so naturally he wants to combine them all in one pure line of genotype a /a ; b /b ; c /c . a. Outline an effective crossing program that can be used to obtain the a /a ; b /b ; c /c pure line. b. At each stage, state exactly which phenotypes will be selected and give their expected frequencies. c. Is there more than one way to obtain the desired genotype? Which is the best way?
Begin with any two of the three lines and cross them. If, for example, you began with a/a ; B/B ; C/C A/A ; b/b ; C/C, the progeny would all be A/a ;B/b ; C/C. Crossing two of these would yield: 9 A/- ; B/- ;C/C 3 a/a ; B/- ; C/C 3 A/- ; b/b ; C/C *1 a/a ; b/b ; C/C* The *a/a ; b/b ; C/C* genotype has two of the genes in a homozygous recessive state and occurs in 1/16 of the offspring. If that were crossed with *A/A ; B/B ;c/c, the progeny would all be A/a ; B/b ; C/c*. *Cross A/a ; B/b ; C/c x A/a ; B/b ; C/c* Crossing two of them (or "selfing") would lead to a 27:9:9:9:3:3:3:1 ratio, and the plant occurring in 1/64 of the progeny would be the desired a/a ; b/b ; c/c. There are several different routes to obtaining a/a ; b/b ; c/c, but the one outlined above requires only four crosses.
21. Suppose you discover two interesting rare cytological abnormalities in the karyotype of a human male. (A karyotype is the total visible chromosome complement.) There is an extra piece (satellite) on one of the chromosomes of pair 4, and there is an abnormal pattern of staining on one of the chromosomes of pair 7. *With the assumption that all the gametes of this male are equally viable, what proportion of his children will have the same karyotype that he has?*
His children will have to inherit the satellite-containing 4 (probability= 1/2), the abnormality staining 7 (probability= 1/2), and the Y chromosome (probability= 1/2). To inherit all three, the probability is *(1/2) (1/2) (1/2)= 1/8.*
18.) When a cell of genotype A/a; B/b: C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells?
Mitosis produces daughter cells having the same genotypes at that of the original cell: A/a; B/b.
13.) Normal mitosis takes place in a diploid cell of genotype A/a; B/b. Which of the following genotypes might represent possible daughter cells? a. A;B b. a;b c. A;b d. a;B e. A/A; B/B f. A/a; B/b g. a/a; b/b
The genotype of the daughter cells will be identical with that of the original cell: (f) A/a; B/b.
26.) Assuming independent assortment of all genes, develop formulas that show the number of phenotypic classes and the number of genotypic classes from selfing a plant heterozygous for n gene pairs.
With the assumption of independent assortment and simple dominant-recessive relations of all genes *genotypic: 3n* *phenotypic: 2n* Hh x Hh would give a 1:2:1 genotypic ratio
29.) In tomatoes, two alleles of one gene determine the character difference of purple (P) versus green (G) stems, and two alleles of a *separate, independent gene* determine the character difference of "cut" (C) versus "potato" (Po) leaves. The results for five matings of tomato-plantphenotypes are as follows:progeny Mating P pheno P,C P, Po G,C G,Po1 P,C x G, C 321 101 310 1072 P,C x P, Po 219 207 64 713 P,C x G,C 722 231 0 04 P,C x G,Po 404 0 87 05 P,Po x G,C 70 91 86 77A) Determine which alleles are dominant. B) What are the most probably genotypes for the parents in each cross?
a. Cross 2 indicates that *purple (G) is dominant to green (g),* and cross 1 indicates *cut (P) is dominant to potato (p).* 1:1 ratio means hetero to recessive (Hh x hh) 3:1 ratio means hetero x hetero (Hh x Hh) if there is absence of one trait then (HH x hh) recessive trait is hidden Cross 1: Gg Pp x gg Pp - There are 3 cut: 1 potato and 1 purple : 1 green Cross 2: Gg Pp x Gg pp - There are 3 purple: 1 green and 1 cut: 1 potato Cross 3: GG Pp x gg Pp - There are *no green* and there are 3 cut: 1 potato Cross 4: Gg PP x gg pp - There are no potatoes, and there are 1 purple: 1 green Cross 5: Gg pp x gg Pp - There are 1 cut: 1 potato adn there are 1 purple : 1 green
A man is brachydactylous (very short fingers; *rare autosomal dominant*), and his wife is not. Both can taste the chemical phenylthiocarbamide (*autosomal dominant*; common allele), but their mothers could not. a. Give the genotypes of the couple. If the genes assort independently and the couple has four children, what is the probability of b. all of them being brachydactylous? c. none being brachydactylous? d. all of them being tasters? e. all of them being nontasters? f. all of them being brachydactylous tasters? g. none being brachydactylous tasters? h. at least one being a brachydactylous taster?
a. Let B=brachydactylous, b=normal, T=taster and t=nontaster. - The genotypes of the couple are *MALE= B/b; T/t* and *FEMALE= b/b; T/t* b. B/b x bb 1:1 ratio For all four children to be brachydactylous, the chance is (1/2)^4 = 1/16. For none of the four children to be brachydactylous, the chance is (1/2)^4 = 1/16. c. For all to be tasters, the chance is (3/4)^4 = 81/256. d. For all to be nontasters, the chance is (1/4)^4 = 1/256. f. For all to be brachydactylous tasters, the chance is (1/2 x 3/4)4 = 81/4096. *g*. Not being a brachydactylous taster is the same 1- (the chance of being a brachydactylous taster) or 1 - (1/2 x 3/4) = 5/8. The chance that all four children are not brachydactylous tasters is (5/8)4 = 625/4096. h. The chance that at least one is a brachydactylous taster is 1 - (the chance of none being a brachydactylous taster) or 1 - (5/8)4.