Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Chapter 23, Problem 19TYK
Summary Introduction
Introduction: The
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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 23 Solutions
Study Guide for Campbell Biology
Ch. 23 - a. What is a major source of genetic variation for...Ch. 23 - In a population of 200 mice, 98 are homozygous...Ch. 23 - Use the allele frequencies you determined in...Ch. 23 - Practice using the Hardy-Weinberg equation so that...Ch. 23 - Prob. 5IQCh. 23 - Why hasnt the highly deleterious sickle-cell...Ch. 23 - a. What is Hardy-Weinberg equilibrium? b. Define...Ch. 23 - Prob. 2SYKCh. 23 - Prob. 1TYKCh. 23 - Prob. 2TYK
Ch. 23 - Prob. 3TYKCh. 23 - Prob. 4TYKCh. 23 - Mice have an estimated 1,000 olfactory receptor...Ch. 23 - Prob. 6TYKCh. 23 - If a population has the following genotype...Ch. 23 - In a population with two alleles, B and b, the...Ch. 23 - Prob. 9TYKCh. 23 - Prob. 10TYKCh. 23 - In a random sample of a population of shorthorn...Ch. 23 - Genetic drift is likely to be seen in a population...Ch. 23 - Porphyria variegata is a genetic disease...Ch. 23 - Cystic fibrosis is a very serious genetic disorder...Ch. 23 - Prob. 15TYKCh. 23 - Prob. 16TYKCh. 23 - Prob. 17TYKCh. 23 - Prob. 18TYKCh. 23 - Prob. 19TYKCh. 23 - Prob. 20TYKCh. 23 - Prob. 21TYK
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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
- Design a grafting experiment to determine if limb mesoderm determines forelimb / hindlimb identity. Include the experiment, a control, and an interpretation in your answer.arrow_forwardThe Snapdragon is a popular garden flower that comes in a variety of colours, including red, yellow, and orange. The genotypes and associated phenotypes for some of these flowers are as follows: aabb: yellow AABB, AABb, AaBb, and AaBB: red AAbb and Aabb: orange aaBB: yellow aaBb: ? Based on this information, what would the phenotype of a Snapdragon with the genotype aaBb be and why? Question 21 options: orange because A is epistatic to B yellow because A is epistatic to B red because B is epistatic to A orange because B is epistatic to A red because A is epistatic to B yellow because B is epistatic to Aarrow_forwardA 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_forward
- 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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