Wi09 Homework 3 key

Wi09 Homework 3 key - Genetics Homework #3 key 1. The...

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Unformatted text preview: Genetics Homework #3 key 1. The flower colors of plants in a particular population may be blue, purple, turquoise, light‐blue, or white. A series of crosses between different members of the population produced the following results: Cross Parents Progeny 1 2 3 4 5 6 7 8 9 10 11 12 13 purple x blue purple x purple blue x blue purple x turquoise purple x purple purple x blue purple x blue turquoise x turquoise purple x blue light‐blue x light‐blue turquoise x white white x white purple x white all purple 76 purple, 25 turquoise 86 blue, 29 turquoise 49 purple, 52 turquoise 69 purple, 22 blue 50 purple, 51 blue 54 purple, 26 blue, 25 turquoise all turquoise 49 purple, 25 blue, 23 light‐blue 60 light‐blue, 29 turquoise, 31 white all light‐blue all white all purple How many genes and alleles are involved in the inheritance of flower color? Indicate all possible genotypes for the following phenotypes: purple; blue; turquoise; light‐blue; white. Start by trying to see if a single gene with multiple alleles can satisfy the genes. Fill in the table with potential genotypes: Cross Parents Progeny 1 CPCP x CBCB CPCB 2 CPCT x CPCT 1 CPCP : 2 CPCT : 1 CTCT 3 CBCT x CBCT 1 CBCB : 2 CBCT : 1 CTCT PCT x CTCT 4 C 1 CPCT : 1 CTCT 5 CPCB x CPCB 1 CPCP : 2 CPCB : 1 CBCB 6 CPCB x CBCB 1 CPCB : 1 CBCB PCT x CBCT 7 C 1 CPCB : 1 CPCT : 1 CBCT : 1 CTCT 8 CTCT x CTCT CTCT 9 CPCT x CBCW 1 CPCB : 1 CPCW : 1 CBCT : 1 CTCW TCW x CTCW 10 C 1 CTCT : 2 CTCW : 1 CWCW 11 CTCT x CWCW CTCW 12 CWCW x CWCW CWCW PCP x CWCW 13 C CPCW (You were probably tempted to try a light­blue allele, CL, but this would not work for cross 10! Knowing this would then make you change cross 9.) In this allelic series, purple (CP) is dominant over all other alleles, since any cross involving purple always give at least some purple progeny. The blue allele (CB) is next in the dominance hierarchy, followed by turquoise (CT) and then white (CW), which is recessive to all the other alleles but is incompletely dominant with turquoise to form light­blue. Possible genotypes: Purple (CPCP, CPCB, CPCT, CPCW); Blue (CBCB, CBCT, CBCW); Turquoise (CTCT); Light­blue (CTCW); White (CWCW). 2. Consider the following hypothetical scheme of determination of coat color in a mammal. Gene A controls the conversion of a white pigment P into a gray pigment Q; the dominant allele A produces the enzyme necessary for this conversion, and the recessive allele a produces an enzyme without biochemical activity. Gene B controls the conversion of the gray pigment Q into a black pigment R; the dominant allele B produces the active enzyme for this conversion, and the recessive allele b produces an enzyme without activity. The dominant allele C of a third gene produces a polypeptide that completely inhibits the activity of the enzyme produced by gene A; that is, it prevents the reaction P Q. Allele c of this gene produces a defective polypeptide that does not inhibit the reaction P Q. Genes A, B and C assort independently, and no other genes are involved. In the F2 of the cross AAbbCC x aaBBcc, what is the expected phenotypic segregation ratio? Parents: AAbbCC x aaBBcc F1: AaBbCc F2: 8 different phenotypic classes in a 27:9:9:9:3:3:3:1 ratio. You could draw out a Punnett square, but this will be VERY cumbersome. It’s better to use the laws of probability to calculate the fraction of each possible phenotypic class. For the aabbcc class, each parent had to produce an “abc” gamete, which has a ½*½*½, or 1/8 probability. 1/8 * 1/8 = 1/64. For the other classes, you’ll also need to add in the various combinations that would produce the given genotypes. Fraction Genotype Color 27/64 A­B­C­ white 9/64 aaB­C­ white 9/64 A­bbC­ white 9/64 A­B­cc black 3/64 aabbC­ white 3/64 aaB­cc white 3/64 A­bbcc gray 1/64 aabbcc white 52/64 will be white, 9/64 will be black, and 3/64 will be gray. ...
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