BIOLOGY LL VERSION + MINDTAP V2.0(2 TER
11th Edition
ISBN: 9780357471012
Author: Solomon
Publisher: CENGAGE L
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Chapter 15.1, Problem 1C
Summary Introduction
To write: The sequence and the position of the cut for the complementary strand of the DNA sequence 
.
Concept introduction: DNA is a double-stranded molecule consisting of two strands of
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Chapter 15 Solutions
BIOLOGY LL VERSION + MINDTAP V2.0(2 TER
Ch. 15.1 - Prob. 1LOCh. 15.1 - Explain how gel electrophoresis is used to...Ch. 15.1 - Describe how PCR is used to amplify a specific...Ch. 15.1 - Compare the possible differences between a...Ch. 15.1 - Prob. 1CCh. 15.1 - Different forms of a protein are produced in the...Ch. 15.1 - What advantages does the PCR method have over gene...Ch. 15.2 - Describe the features of a typical CRISPR locus in...Ch. 15.2 - Explain the function of CRISPR in bacterial cells.Ch. 15.2 - Compare CRISPR-based endonucleases with...
Ch. 15.2 - Prob. 8LOCh. 15.2 - Prob. 1CCh. 15.2 - Prob. 2CCh. 15.2 - Prob. 3CCh. 15.3 - Prob. 9LOCh. 15.3 - Prob. 10LOCh. 15.3 - Discuss how qPCR, DNA microarrays (DNA chips), and...Ch. 15.3 - Explain how you would compare the expression of a...Ch. 15.3 - Prob. 2CCh. 15.4 - Describe how genome-wide association studies have...Ch. 15.4 - Explain how targeted gene silencing and knockout...Ch. 15.4 - Prob. 1CCh. 15.5 - Describe at least one important application of DNA...Ch. 15.5 - Prob. 1CCh. 15.5 - What are short tandem repeats (STRs), and why are...Ch. 15.5 - Why do gene targeting and mutagenesis screening in...Ch. 15.6 - Prob. 15LOCh. 15.6 - Prob. 16LOCh. 15.6 - Prob. 1CCh. 15.6 - Prob. 2CCh. 15.7 - Describe at least two safety issue associated with...Ch. 15.7 - What are some of the environment concerns...Ch. 15 - A plasmid (a) can be used as a DNA vector (b) is a...Ch. 15 - DNA molecules with complementary sticky ends...Ch. 15 - Prob. 3TYUCh. 15 - Which technique rapidly replicated specific DNA...Ch. 15 - Prob. 5TYUCh. 15 - A cDNA clone contains (a) introns (b) exons (c)...Ch. 15 - Prob. 7TYUCh. 15 - Gel electrophoresis separates nucleic acids on the...Ch. 15 - A CRISPR locus in a bacterium contains (a) short...Ch. 15 - DNA molecular with complementary sticky ends...Ch. 15 - These highly polymorphic molecular markers are...Ch. 15 - Prob. 12TYUCh. 15 - Prob. 13TYUCh. 15 - Prob. 14TYUCh. 15 - EVOLUTION LINK DNA technology, such as the...Ch. 15 - SCIENCE, TECHNOLOGY, AND SOCIETY What are some...
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- 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
- 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_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_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
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