{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 5 Key

Homework 5 Key - BICD 100 Spring 2008 30 pts total 1 a(10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
BICD 100 Page 1 of 4 S. Brody, ext. 42619 Spring 2008 4125 Muir Biology HOMEWORK SET #5 — KEY 30 pts total 1. ( 10 pts total ) a) ( 4 pts ) b) Origin of extra chromosome: ( 4 pts ) Assume that similar events happen with equal frequency, i.e. MII female leads to equal numbers of bb and aa gametes. Then, looking at Down's offspring, some genotypes are unique to a certain type of nondisjunction. These numbers can then be used (by subtracting from the total) to assign numbers to those genotypes which are not unique to a particular nondisjunction event. genotype total MI female MII MI male MII aab 2 2 aac 2 2 abb 16 11 5 abc 36 11 25 acc 5 5 bbb 7 2 5 bbc 27 2 25 bcc 5 5 = 22 = 8 = 50 = 20 c) Down's from all female nondisjunction: ( 1 pt ) 22 + 8 = 30/100 30/100 × 2 × 10 -3 = 6 × 10 -4
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
BICD 100 Page 2 of 4 S. Brody, ext. 42619 Spring 2008 4125 Muir Biology d) Down's from all MII nondisjunction ( 1 pt ) 8 + 20 = 28/100 28/100 × 2 × 10 -3 = 5.6 × 10 -4 2. (
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 4

Homework 5 Key - BICD 100 Spring 2008 30 pts total 1 a(10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online