Biology: Life on Earth with Physiology Plus Mastering Biology with Pearson eText -- Access Card Package (11th Edition)
11th Edition
ISBN: 9780133910605
Author: Gerald Audesirk, Teresa Audesirk, Bruce E. Byers
Publisher: PEARSON
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Chapter 45, Problem 4RQ
Summary Introduction
To explain:
The term pollen grain and also determine how it is formed.
Introduction:
In
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Using quail and chicken embryos, what kind of experiment would you conduct to test if rib forming somites have their axial identity specified before segmentation? How do we know this phenotype is due to axial identity being specified before segmentation and not due to our experimental method?
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
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Chapter 45 Solutions
Biology: Life on Earth with Physiology Plus Mastering Biology with Pearson eText -- Access Card Package (11th Edition)
Ch. 45.1 - Prob. 1CYLCh. 45.1 - diagram the life cycles of ferns and flowering...Ch. 45.2 - Prob. 1CSCCh. 45.2 - diagram the structure of a complete flower and...Ch. 45.2 - During finals week in the spring semester, a...Ch. 45.2 - Prob. 1TCCh. 45.2 - Prob. 2CYLCh. 45.2 - explain the processes of pollination and double...Ch. 45.3 - explain how the parts of a flower develop into the...Ch. 45.3 - Prob. 1HYEW
Ch. 45.3 - describe the differences between monocot and dicot...Ch. 45.4 - explain why many seeds undergo dormancy before...Ch. 45.4 - Prob. 2CYLCh. 45.5 - The warmth of hot flowers attracts pollinators and...Ch. 45.5 - Prob. 1CYLCh. 45.5 - Prob. 1TCCh. 45.5 - Prob. 2TCCh. 45.6 - Prob. 1CSCCh. 45.6 - Prob. 1CYLCh. 45.6 - Why doesnt mistletoe simply drop its seeds?Ch. 45.6 - describe how fruit structures aid in seed...Ch. 45.6 - Prob. 2TCCh. 45.6 - Heat-producing flowers are rare, and many are...Ch. 45 - Prob. 1ACCh. 45 - Prob. 1FIBCh. 45 - Prob. 1MCCh. 45 - Diagram the general plant life cycle. Which stages...Ch. 45 - Prob. 2ACCh. 45 - In a flowering plant, the male gametophyte is the...Ch. 45 - Prob. 2MCCh. 45 - Prob. 2RQCh. 45 - Prob. 3ACCh. 45 - Prob. 3FIBCh. 45 - Prob. 3MCCh. 45 - Prob. 3RQCh. 45 - Prob. 4FIBCh. 45 - Which of the following is True? a. Moth-pollinated...Ch. 45 - Prob. 4RQCh. 45 - Prob. 5FIBCh. 45 - Prob. 5MCCh. 45 - Prob. 5RQCh. 45 - Describe the characteristics you would expect to...Ch. 45 - Prob. 7RQCh. 45 - Describe three mechanisms whereby seed dormancy is...Ch. 45 - Prob. 9RQCh. 45 - Describe three types of fruits and the mechanisms...
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- If using animals in medical experiments could save human lives, is it ethical to do so? In your answer, apply at least one ethical theory in support of your position.arrow_forwardYou 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_forward
- You 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_forwardDesign 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_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_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_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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