BIOLOGY LL VERSION + MINDTAP V2.0(2 TER
11th Edition
ISBN: 9780357471012
Author: Solomon
Publisher: CENGAGE L
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Chapter 14, Problem 6TYU
Summary Introduction
Introduction: Genes are sets of
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8. Aerobic respiration of a 5 mM solution of tripeptide that is composed of the following three amino acids;
alanine, leucine and isoleucine. Alanine breaks down to pyruvate, leucine breaks down to Acetyl-CoA and
isoleucine breaks down to succinyl-CoA.
Alanine
NADH
FADH2
OP ATP
SLP ATP
Total ATP
Leucine
Isoleucine
Totals
Show your work using dimensional analysis here:
4
9. Aerobic respiration of one lipid molecule. The lipid is composed of one glycerol molecule connected to two
fatty acid tails. One fatty acid is 12 carbons long and the other fatty acid is 18 carbons long in the figure
below. Use the information below to determine how much ATP will be produced from the glycerol part of
the lipid. Then, in part B, determine how much ATP is produced from the 2 fatty acids of the lipid. Finally
put the NADH and ATP yields together from the glycerol and fatty acids (part A and B) to determine your
total number of ATP produced per lipid. Assume no other carbon source is available.
fatty acids
glycerol
18 carbons
12 carbons
0=
influences of environment on the phenotype.
Chapter 14 Solutions
BIOLOGY LL VERSION + MINDTAP V2.0(2 TER
Ch. 14.1 - Explain why bacterial and eukaryotic cells have...Ch. 14.1 - Prob. 1CCh. 14.1 - Prob. 2CCh. 14.2 - Prob. 2LOCh. 14.2 - Distinguish among inducible, repressible, and...Ch. 14.2 - Differentiate between positive and negative...Ch. 14.2 - Prob. 5LOCh. 14.2 - Prob. 1CCh. 14.2 - What structural features does the trp operon share...Ch. 14.2 - Prob. 3C
Ch. 14.2 - Prob. 4CCh. 14.3 - Prob. 6LOCh. 14.3 - Give examples of some of the ways eukaryotic...Ch. 14.3 - Prob. 8LOCh. 14.3 - Prob. 9LOCh. 14.3 - Prob. 10LOCh. 14.3 - Prob. 1CCh. 14.3 - Prob. 2CCh. 14.3 - Prob. 3CCh. 14.3 - Prob. 4CCh. 14.3 - Prob. 5CCh. 14 - The regulation of most bacterial genes occurs at...Ch. 14 - Prob. 2TYUCh. 14 - Prob. 3TYUCh. 14 - Prob. 4TYUCh. 14 - Inactive genes tend to be found in (a) highly...Ch. 14 - Prob. 6TYUCh. 14 - Which of the following is characteristic of genes...Ch. 14 - Through alternative splicing, eukaryotes (a)...Ch. 14 - A mutation that inactivates the repressor gene of...Ch. 14 - Which of the following is an example of positive...Ch. 14 - Prob. 11TYUCh. 14 - PREDICT Compare the types of bacterial genes...Ch. 14 - INTERPRET DATA Develop a simple hypothesis that...Ch. 14 - Prob. 14TYUCh. 14 - Prob. 15TYUCh. 14 - EVOLUTION LINK Suggest why evolution resulted in...Ch. 14 - Prob. 17TYU
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- You aim to test the hypothesis that the Tbx4 and Tbx5 genes inhibit each other's expression during limb development. With access to chicken embryos and viruses capable of overexpressing Tbx4 and Tbx5, describe an experiment to investigate whether these genes suppress each other's expression in the limb buds. What results would you expect if they do repress each other? What results would you expect if they do not repress each other?arrow_forwardYou decide to delete Fgf4 and Fgf8 specifically in the limb bud. Explain why you would not knock out these genes in the entire embryo instead.arrow_forwardYou implant an FGF10-coated bead into the anterior flank of a chicken embryo, directly below the level of the wing bud. What is the phenotype of the resulting ectopic limb? Briefly describe the expected expression domains of 1) Shh, 2) Tbx4, and 3) Tbx5 in the resulting ectopic limb bud.arrow_forward
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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.arrow_forwardWith 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-haemoglobinarrow_forwardWith 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-haemoglobinarrow_forward
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