Principles of Biology by Brooker et al.
Answers to Quantitative Analysis’s Crunching the Numbers
Chapter 2
To begin, convert 5.5 mM to M units. 1 mM is equivalent to 0.001 M. Thus, 5.5 mM is equal to 0.0055
M. Remember that “M” stands for moles/liter, and that the molecular mass of glucose is 180 grams. To
find the number of grams/liter blood, perform the following calculation: (0.0055 moles/liter) x (180
grams/mole) = 0.99 grams/liter, or approximately 1g/L.
Chapter 4
The volume of the spherical cell is 38,792 m3. Its surface area is 5,542 m2.
The volume of the cylindrical cell is also 38,792 m3. However, its surface area is much larger,
25,918 m2.
Chapter 6
The KMapp would be 0.65 mM.
Chapter 8
For Keisha, over 50% of her receptors would be bound by insulin. For Ryan, it would be much
less than 50%. Ryan needs to take insulin injections so that his receptors are activated by insulin,
enabling his muscle (and fat cells) to properly metabolize glucose.
Chapter 11
For humans: 22,000 x 0.70 x 5 = 77,000. These are the genes that are alternatively spliced. The
other 30% are not, which equals 6600. If we add the two together, we get:
77,000 + 6600 = 83,600
For Drosophila: 14,000 x 0.07 x 3 = 2940. These are the genes that are alternatively spliced.
The other 93% are not, which equals 13,920. If we add the two together, we get:
2940 + 13,920 = 15,960.
Chapter 12
The average mutation rate is 60 divided by 2 million, which equals 3 x 10-5.
If we conduct a (two-tailed) t test, the probability value, p, equals 0.00013. Because this value is
much less than 0.01 (<1%), we reject the null hypothesis and accept the hypothesis that the
control and mutagen-exposed samples are different from each other. These results suggest that
mutagen X is a mutagen.
Chapter 13
The answer is 223, which equals 8,388,608. Crossing over would make this number even larger.
Chapter 16
EcoRI recognizes a 6-bp sequence. Because there are 4 types of bases, this sequence will occur
on average 46, or every 4096 bp. If we divide 3 billion by 4096, we obtain an estimate of about
732,422 fragments of DNA.
Chapter 17
If we plug these values into the equation P = e- m, we get a value of 0.0083, or 0.83% of the
genome. With a genome size of 4.6 Mb, this amounts to 38,180 bp.
Chapter 18
We first need to calculate the number of atoms of U-235 and Pb-207. The masses of U-235 and
Pb-207 are approximately 235 g/mole and 207 g/mole, respectively. For U-235, the number of
atoms in the sample is 0.11/235 x 6.022 x 1023 = 2.82 x 1020, which is N. For Pb-207, the number
of atoms in the sample is 0.035/207 x 6.022 x 1023 = 1.02 x 1020. We assume that all of the Pb207 is the decay product of U-235. Therefore, the number of original atoms of U-237 would be
2.82 x 1020 + 1.02 x 1020 = 3.84 x 1020, which is N0. We can plug these values into our equation
and solve for t. Note: the half-life (T1/2) for U-235 is 710 million years. The answer is that the
fossil is approximately 316 million years old.
Chapter 19
The value of 0.01 is a genotype frequency, such as q2. If we take the square root of this value,
q = 0.1. Assuming that p and q are the only two alleles, q = 1 – p, so p = 0.9. The frequency of
heterozygotes equals 2pq, which equals 2(0.1)(0.9) = 0.18, or 18%.
Chapter 21
The tree at the top would be the least likely because it contains the highest number of genetic
changes compared to the other two trees.
Chapter 25
2,500,000
Chapter 29
(-0.15 MPa) + (+0.15 MPa) = 0 MPa
Reference to examples of water potential calculations shown in Figure 29.16 reveals that a water
potential of 0 indicates that the cell is turgid.
Chapter 32:
The equilibrium potential is 60 mV x log10 [6.5/145] = -81 mV. This is different from the resting
membrane potential of -70 mV, and therefore K+ is not in equilibrium. Under these conditions, K+ would
diffuse out of the cell, because the concentration force is greater than the electrical force. In other words,
the tendency for K+ to diffuse down its concentration gradient out of the cell is greater than the electrical
attraction tending to draw K+ into the cell.
Chapter 36:
Heart mass varies roughly in proportion to body mass, such that an animal with twice the body
mass of another will have a heart that is roughly twice the mass, too. However, heart rate varies
inversely with body mass, and the relationship is not linear. In other words, smaller animals not
only have greater heart rates than do larger animals, but the difference is not proportional. It is
the very high heart rate, not a difference in mass, that enables a relatively greater cardiac output
in small animals. This, in turn, helps match cardiac performance with the greater relative
metabolic rate of smaller animals.
Chapter 37
At a PO2 of 35 mmHg and with increased CO2, hemoglobin would be approximately 58% saturated
according to the representative curves shown in Figure 37.10a (in other words, hemoglobin would have
unloaded about 42% of its O2). By contrast, at a PO2 of 30 mmHg but with decreased CO2, hemoglobin
would be about 64% saturated (in this case, hemoglobin would have unloaded only 36% of its O2). This
difference illustrates the significance of CO2 in promoting the release of O2 from hemoglobin. Rabbits
are larger than mice and smaller than humans, of course, and therefore you would predict that their
saturation curves would be between those of mice and humans. Assuming an average sized rabbit of
about 2-3 kg, hemoglobin would be about 45% saturated at a PO2 of 30 mmHg.
Chapter 42
Quantitative Analysis
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Chapter 44
Quantitative Analysis
200; 100%
Chapter 45
Quantitative Analysis
a. 0.693 and 0.638 respectively. Increased numbers of species does not automatically lead
to increased species diversity.
b. 9.815 and 7.668, a 28% difference.