
Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 8, Problem 23P
Summary Introduction
To describe:
The explanation that helps in understanding the reason that inserted amino acids do not correspond to normal protein sequence.
Introduction:
Amino acids are vital molecules that produce proteins. The process by which mRNA is transformed into amino acids is known as translation. The amino acids join together with the help of polypeptide bonds to form proteins. A scientist observed that a natural enzyme is composed of 227 amino acids. However, the mutant form of this enzyme contains 312 amino acids. This indicates that the mutant form has 85 extra amino acids.
Expert Solution & Answer

Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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?
You 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.
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.
Chapter 8 Solutions
Genetics: From Genes to Genomes
Ch. 8 - For each of the terms in the left column, choose...Ch. 8 - Match the hypothesis from the left column to the...Ch. 8 - How would the artificial mRNA 5GUGUGUGU . . . 3 be...Ch. 8 - An example of a portion of the T4 rIIB gene in...Ch. 8 - Consider Crick and Brenners experiments in Fig....Ch. 8 - The HbSsickle-cell allele of the human -globin...Ch. 8 - The following diagram describes the mRNA sequence...Ch. 8 - The amino acid sequence of part of a protein has...Ch. 8 - The results shown in Fig. 8.5 may have struck you...Ch. 8 - Identify all the amino acid-specifying codons in...
Ch. 8 - Before the technology existed to synthesize RNA...Ch. 8 - A particular protein has the amino acid sequence...Ch. 8 - How many possible open reading frames frames...Ch. 8 - Prob. 14PCh. 8 - Charles Yanofsky isolated many different trpA-...Ch. 8 - The sequence of a segment of mRNA, beginning with...Ch. 8 - You identify a proflavin-generated allele of a...Ch. 8 - Using recombinant DNA techniques which will be...Ch. 8 - Describe the steps in transcription that require...Ch. 8 - Chapters 6 and 7 explained that mistakes made by...Ch. 8 - The coding sequence for gene F is read from left...Ch. 8 - If you mixed the mRNA of a human gene with the...Ch. 8 - Prob. 23PCh. 8 - The Drosophila gene Dscam1 encodes proteins on the...Ch. 8 - Describe the steps in translation that require...Ch. 8 - Locate as accurately as possible the listed items...Ch. 8 - Concerning the figure for Problem 26: a. Which...Ch. 8 - a. Can a tRNA exist that has the anticodon...Ch. 8 - For parts a and b of Problem 28, consider the DNA...Ch. 8 - Remembering that the wobble base of the tRNA is...Ch. 8 - Prob. 31PCh. 8 - The yeast gene encoding a protein found in the...Ch. 8 - The sequence of a complete eukaryotic gene...Ch. 8 - Arrange the following list of eukaryotic gene...Ch. 8 - Prob. 35PCh. 8 - The human gene for 2 lens crystallin has the...Ch. 8 - In prokaryotes, a search for genes in a DNA...Ch. 8 - a. The genetic code table shown in Fig. 8.2...Ch. 8 - a. Very few if any eukaryotic genes contain tracts...Ch. 8 - Explain how differences in the initiation of...Ch. 8 - Do you think each of the following types of...Ch. 8 - Null mutations are valuable genetic resources...Ch. 8 - The following is a list of mutations that have...Ch. 8 - Considering further the mutations described in...Ch. 8 - Adermatoglyphia described previously in Problem 18...Ch. 8 - Prob. 46PCh. 8 - You learned in Problem 21 in Chapter 7 that the...Ch. 8 - When 1 million cells of a culture of haploid yeast...Ch. 8 - Why is a nonsense suppressor tRNATyr, even though...Ch. 8 - A mutant B. adonis bacterium has a nonsense...Ch. 8 - You are studying mutations in a bacterial gene...Ch. 8 - Another class of suppressor mutations, not...Ch. 8 - Yet another class of suppressor mutations not...Ch. 8 - At least one nonsense suppressing tRNA is known...Ch. 8 - An investigator was interested in studying UAG...Ch. 8 - Prob. 56PCh. 8 - In certain bacterial species, pyrrolysine Pyl,...Ch. 8 - Canavanine is an amino acid similar to arginine...
Knowledge Booster
Similar questions
- 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
- 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_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_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_forward
- Explain what age of culture is most likely to produce an endospore?arrow_forwardExplain 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningBiology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage LearningHuman Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning

Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
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

Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning

Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning