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 21, Problem 11P
Summary Introduction

To analyze:

A statistical project has been conducted in which, total of 20 men and 20 women participated. The height and weight of each individual are given in the table.

Subject Men Women
Height(in.) Weight(lb) Height(in.) Weight(lb)
1 65 136 60 95
2 66 146 61 103
3 67 141 62 110
4 67 148 62 109
5 68 147 62 118
6 68 166 63 137
7 69 165 63 152
8 69 173 64 134
9 69 159 64 127
10 70 188 64 166
11 70 183 65 129
12 70 179 65 130
13 70 190 66 148
14 71 169 66 152
15 71 186 67 155
16 71 190 67 149
17 72 206 68 157
18 72 210 68 138
19 73 238 69 162
20 74 267 70 169
  1. The question asked to draw one histogram for the height of the individuals and one more histogram for the weight of the individuals. Note: Plot weights in 10-pound intervals (i.e., 9099 lbs, 100109 lbs, 110119 lbs, etc.).

  2. The question asked to estimate the mean, variance, and standard deviation for height and weight in men and women.

  3. The question asked to compare the numerical values with the graphical distribution of heights and weights drawn in the histograms and the question also asked to describe whether graphical impression matches the numerical values.

Introduction:

Histogram is the graphical representation of numerical data. With the help of a histogram, complex data can be analyzed easily. In statistics, mean is the average of data, to obtain the Mean, sum of total observations is divided by the total number of observations. Variance is the middling of the squared differences. Standard deviation is square root of variance.

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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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