Homework 6 Key

Homework 6 Key - BICD 100 Spring 2008 30 pts total 1. a)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: BICD 100 Spring 2008 30 pts total 1. a) Page 1 of 4 HOMEWORK SET #6 -- KEY S. Brody, ext. 42619 4125 Muir Biology (14 pts total) (2 pts) Let b = bristles missing, n = nicked wings, the transmission is typical sex-linkage, thus P1 generation: (female) F1 bn bn bn Y and X++ Y ++ bn (male) b) (2 pts) The 2:1 sex ratio in the F2 suggests that the F1 female received from her father an X chromosome bearing a recessive lethal. (5 pts) The order of b, n and l (lethal) may be one of three possibilities: 1 2 3 c) + b n ________ l + + b + n _______ + l + b n + _______ + + l Double recombinants would be: 1 2 3 Genotype Phenotype Observed ++n +nl nicked lethal 9 lb+ lethal 0 bln lethal 0 +++ normal 7 b++ bristles 1 0 Since the rare class (i.e., the product of double exchange) is missing bristles, normal wing, then order 3 is indicated. BICD 100 Spring 2008 d) (5 pts) Page 2 of 4 S. Brody, ext. 42619 4125 Muir Biology b-n = 1 + 9 = 20 m.u. 50 n-1 = 1 + 7 = 16 m.u. 50 double single b 20 n 16 l ___________________ 0.0 20.0 36.0 2. (4 pts total) a) Designate the chromosomes as AA'aa'. The types of gametes then would be: Aa; A'a'; AA'; aa'; Aa'; A'a. This assumes no crossing over between A (or a) and the centromere and equal disjunction of all 4 chromosomes. The sum of the gametes would then be 5A_:1aa, which would give phenotype ratio of 5A:1a since it is a test cross. (4 pts) 3. a) Let XT be an X chromosome with the mutant allele. Xt has the normal allele. (12 pts total) (1 pt) I-1 is XTY II-1 is XTXt (1.5 pts) III-1 is XTY -- but see (c) below III-2 is XTXt III-5 is XtXt (3.5 pts) Some or all of the germ cells of III-1 must carry a translocation between the X and an autosome such that the X no longer carries T (such an X is now X0) but the autosome does (AT versus A normal). (2 pts) IV-4 is XtX0, AA IV-5 is XtY, ATA e) (4 pts) V-9 is XtXt, AA V-10 is XtY, ATA V-11 is XtY, AA V-12 is XtXt, ATA b) c) d) ...
View Full Document

This note was uploaded on 06/07/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

Ask a homework question - tutors are online