Genetic Analysis: An Integrated Approach (2nd Edition)
Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 9, Problem 34P
Summary Introduction

To analyze:

Table-C contains DNA-sequence information compiled by Marilyn Kozak (1987). The data consists of the percentage of A, C, G and T at each position among the 12 nucleotides preceding the start codon in 699 genes from various vertebrate species and at the first nucleotide after the start codon.(The start codon occupies positions +1 to +3, and the first nucleotide immediately after the start codon occupies +4).Use the data to conclude the consensus order for the 13 nucleotides (-12 to -1 and +4) surrounding the start codon in vertebrate genes.

Genetic Analysis: An Integrated Approach (2nd Edition), Chapter 9, Problem 34P , additional homework tip  1

Introduction:

Consensus sequences are a short stretch of DNA sequences that are commonly located nucleotides found at a specific location of DNA and RNA. These sequences are used for inter or intramolecular interactions. These sequences are similar in structure and function in all organisms. The consensus sequence is a calculated order that states the most frequent nucleotide at that position in the alignment.

Genetic Analysis: An Integrated Approach (2nd Edition), Chapter 9, Problem 34P , additional homework tip  2

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A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation?   The cell membranes are more resistant than normal.   An isotonic solution had been added instead of water.   A solution of 0.1 M NaCl had been added instead of water.   Not enough water had been added to the red blood cell pellet.   The man had sickle-cell anaemia.
A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation?   The cell membranes are more resistant than normal.   An isotonic solution had been added instead of water.   A solution of 0.1 M NaCl had been added instead of water.   Not enough water had been added to the red blood cell pellet.   The man had sickle-cell anaemia.
With reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobin

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Genetic Analysis: An Integrated Approach (2nd Edition)

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