Chapter 8

Chapter 8 - Chapter 1. 2. a5, b10, c8, a4, d12, e6, b6, f2,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 8 solutions: 1. a5, b10, c8, d12, e6, f2, g9, h14, i3, j13, k1, l7, m15, n11, o4, p16 2. a4, b6, c1, d2, e3, f5 4 - a. Comparing the mutant to the wild-type sequences one can determine insertions and deletion sites, corresponding to + and mutations respectively. b. Lys Ser Pro Ser Leu Asn Ala wild-type 5’ AAA AGT CCA TCA CTT AAT GCC 3’ (-) (+) mutant 5’ AAA GTC CAT CAC TTA ATG GCC 3’ Lys Val His His Leu Met Ala The five amino acids in between the – and the + mutations are different from wild-type c. The substitutions of amino acids between the – and + mutations in the mutant must not alter the structure of the protein significantly enough to alter protein function 9. DNA sequences are written with the 5’ to 3’ strand on the top and the 3’ to 5’ strand on the bottom. If the protein coding sequence for gene F is read from left (N terminus) to right (C terminus), then the top strand is the RNA-like strand which is read 5’ to 3’ and the template strand must be the bottom strand of DNA. The template for gene G is the opposite since the coding
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/07/2010 for the course BIO 105 taught by Professor Sullivan during the Summer '08 term at UCSC.

Page1 / 2

Chapter 8 - Chapter 1. 2. a5, b10, c8, a4, d12, e6, b6, f2,...

This preview shows document pages 1 - 2. Sign up to view the full document.

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