LIFE:SCIENCE OF BIOL.(LL) >CUSTOM<
11th Edition
ISBN: 9781319209957
Author: Sadava
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 10, Problem 2Q
Summary Introduction
To review:
The difference in the rate of photosynthesis in shade-tolerant and shade-intolerant plants, and the reason for this difference.
Given:
The graphs representing the rate of photosynthesis as a function of the intensity of light are given below:
Introduction:
A shade-tolerant and a shade-intolerant plant were allowed to grow by 3% and 44% daylight conditions. After approximately 5 weeks, a leaf from both the plants was exposed to different light intensities and the rate of photosynthesis in its cells was measured.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The 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 A
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.
Chapter 10 Solutions
LIFE:SCIENCE OF BIOL.(LL) >CUSTOM<
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- 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-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_forwardWhich ONE of the following is FALSE regarding haemoglobin? It has two alpha subunits and two beta subunits. The subunits are joined by disulphide bonds. Each subunit covalently binds a haem group. Conformational change in one subunit can be transmitted to another. There are many variant ("mutant") forms of haemoglobin that are not harmful.arrow_forward
- Which ONE of the following is FALSE regarding haemoglobin? It has two alpha subunits and two beta subunits. The subunits are joined by disulphide bonds. Each subunit covalently binds a haem group. Conformational change in one subunit can be transmitted to another. There are many variant ("mutant") forms of haemoglobin that are not harmful.arrow_forwardDuring a routine medical check up of a healthy man it was found that his haematocrit value was highly unusual – value of 60%. What one of the options below is the most likely reason? He will have a diet high in iron. He is likely to be suffering from anaemia. He lives at high altitude. He has recently recovered from an accident where he lost a lot of blood. He has a very large body size.arrow_forwardExplain what age of culture is most likely to produce an endospore?arrow_forward
- Explain why hot temperatures greater than 45 degrees celsius would not initiate the sporulation process in endospores?arrow_forwardEndospore stain: Consider tube 2 of the 7-day bacillus culture. After is was heated, it was incubated for 24 hours then refrigerated. Do you think the cloudiness in this tube is due mostly to vegetative cells or to endospores? Explain your reasoningarrow_forwardReactunts C6H12O6 (Glucose) + 2NAD+ + 2ADP 2 Pyruvic acid + 2NADH + 2ATP a. Which of the above are the reactants? b. Which of the above are the products? c. Which reactant is the electron donor? GHz 06 (glucose) d. Which reactant is the electron acceptor? NAD e. Which of the products have been reduced? NADH f. Which of the products have been oxidized? g. Which process was used to produce the ATP? h. Where was the energy initially in this chemical reaction and where is it now that it is finished? i. Where was the carbon initially in this chemical reaction and where is it now that it is finished? j. Where were the electrons initially in this chemical reaction and where is it now that it is finished? 3arrow_forward
- There is ________ the concept of global warming. Very strong evidence to support Some strong evidence to support Evidence both supporting and against Evidence againstarrow_forwardHow many types of reactions can an enzyme perform?arrow_forwardYour goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Biology: The Unity and Diversity of Life (MindTap...BiologyISBN:9781305073951Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa StarrPublisher:Cengage Learning
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)
Biology
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Cengage Learning
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Biology: The Unity and Diversity of Life (MindTap...
Biology
ISBN:9781305073951
Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr
Publisher:Cengage Learning