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GENETIC ANALYSIS: AN INTEG. APP. W/MAS
2nd Edition
ISBN: 9781323142790
Author: Sanders
Publisher: Pearson Custom Publishing
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Chapter 10, Problem 8P
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A normal hemoglobin protein has a glutamic acid at position 6; in sickle-cell hemoglobin, this glutamic acid has been replaced by a valine. List all the possible mRNA codons that could be present for each type of hemoglobin. Can a single base change result in a change from Glu to Val in hemoglobin?
Below is the DNA base sequence for the normal protein for normal hemoglobin and the base sequence for (abnormal) sickle cell hemoglobin:
Normal GGG CTT CTT TTT
Sickle GGG CAT CTT TTT
A)Transcribe and translate the normal and sickle cell DNA.
B)Identify this as a point or frameshift mutation. Explain.
Shown in the following table are several amino acid substitutionsin the a and b chains of human hemoglobin. determine how many of them can occur as a result of a single nucleotide change.
Chapter 10 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
Ch. 10 - Define the following terms as described in this...Ch. 10 - 2. Using sickle cell disease as an example,...Ch. 10 -
3. Compare and contrast the contributions of...Ch. 10 - Why do differences in protein electrophoretic...Ch. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - 8. Wildtype βglobin protein is composed of amino...Ch. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - 11. How is an autoradiograph produced from a...Ch. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - The family represented in the pedigree and...Ch. 10 - Suppose the mating couple (I-1 and I-2) shown in...Ch. 10 - What are restriction endonucleases, and why are...Ch. 10 - 18. Following restriction digestion, DNA fragments...Ch. 10 - 19. The doublestranded DNA sequence below is part...Ch. 10 - 20. Restriction enzymes recognize specific...Ch. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - 25. A second strain of dwarf plants has a...Ch. 10 - During gel electrophoresis of linear DNA...Ch. 10 - Prob. 27PCh. 10 - 28. In molecular biology, restriction...Ch. 10 - A complete plant gene containing four introns and...Ch. 10 - Prob. 30PCh. 10 - The map below illustrates three alleles in a...Ch. 10 - Prob. 32PCh. 10 - 33. Northern blot analysis is performed on mRNA...
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- The genetic disorder sickle-cell anemia occurs when the amino acid valine takes the place of glutamate during translation of a hemoglobin chain. Using the table of codons below, determine the mutation in DNA that produces this disorder. 1st position ✓ U C A G Select one: U C serine phenylalanine phenylalanine serine leucine serine leucine serine leucine leucine leucine leucine isoleucine isoleucine isoleucine methionine Table of mRNA Codons 2nd position valine valine valine valine proline proline proline proline alanine alaninc alanine alanine A tyrosine tyrosine a. CUC changes to C AG b. GAA changes to GUU c. CTT changes to CAT d. C A G changes to CTC stop stop threonine asparagine threonine asparagine threonine threonine histidine histidine arginine arginine glutamine arginine glutamine arginine lysine lysine G cysteine cysteine stop tryptophan aspartate aspartate glutamate glutamate serine serine arginine arginine glycine glycine glycine glycine 3rd position DCMO U С A G U C A G…arrow_forwardHurler syndrome is due to a mutation in a gene that encodes aprotein called α-l-iduronidase. This protein functions withinlysosomes as an enzyme that breaks down mucopolysaccharides(a type of polysaccharide that has many acidic groups attached).When this enzyme is defective, excessive amounts of the mucopolysaccharides dermatan sulfate and heparin sulfate accumulatewithin the lysosomes, especially in liver cells and connectivetissue cells. This accumulation leads to symptoms such as anenlarged liver and spleen, bone abnormalities, corneal clouding,heart problems, and severe neurological problems. The pedigreebelow contains three members affected with Hurler syndrome,indicated with black symbols. Based on this pedigree, does thissyndrome appear to follow autosomal recessive, autosomaldominant, X-linked recessive, or X-linked dominant inheritance?Explain your reasoning.arrow_forwardCystic fibrosis (CF) is an inherited disorder caused by different types of mutations, many of which prevent ions from moving across cell membranes. Normally there are channel proteins that allow passage of the ions, but in patients with one kind of CF these proteins seem odd. Closer examination shows that these proteins display the correct amino acid sequence. However, they fail to do their job. A) Given that the primary structure of the protein is correct, what can you infer about the DNA sequence for the gene coding this protein on this patient, is there a mutation? Explain. B) Why is the primary structure insufficient to guarantee the proper function of the protein?arrow_forward
- A heptapeptide when treated with trypsin produced two peptides. T1 (D, G, Y) and T2 (K, F, V, A). When the heptapeptide was treated with chymotrypsin, three peptides were produced: CT1 (K,,Y, G), CT2 (F,A, V), and CT3 (D). The sequences of these peptides is not known, however. When the peptide was treated with Sanger’s Reagent and hydrolyzed, DNP-K and DNP-A were recovered. What is the amino acid sequence of the heptapeptide?arrow_forwardGiven the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs (what letters changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions (with spaces), and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - CCTCGTTATGTG - 5' Mutated DNA Sequence: 3' - CCTCGTTATTTG - 5'arrow_forwardIdentify the following mutations and describe what the possible effect on the protein will be. (4) 5’GAT TTT AGC TTA GCC CAT 3’ 5’ GAT TAG CTT AGC CCA T 3’ 3’CTA AAA TCG AAT CGG GTA 5’ 3’ CTA ATC GAA TCG GGT A 5’ 5’ GAT TTT AGC TTA CCC CAT 3’ 5’ GAT TTT AGC TAA CCC CAT 3’ 3’ CTA AAA TCG AAT GGG GTA 5’ 3’ CTA AAA TCG ATT GGG GTA 5’arrow_forward
- A scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. What type(s) of mutation may have occurred in the GS1 gene in strain A?arrow_forwardA scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. The scientist determines the nucleotide sequence of the coding strand of the GS1 gene from strain A. It is identical to the GS1 sequence from the wild type gene except for a single change occurring approximately 1⁄3 of the way into the GS1 open reading frame. A small region of the GS1 sequence (including the site where the mutation occurs) from the wild type and mutant strains is shown below. Wild type TGTCCTCGGCCACAAGTTCTCTATC Strain A TGTCCTCGGCCACTAGTTCTCTATC How has this…arrow_forwardA scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. Sequencing of the GS1 gene from strain B shows that it is identical to the wild type gene except for a single alteration (the replacement of one nucleotide by another). How might this account for the features of the GS1 protein produced by strain B?arrow_forward
- Can you check if A is correct and please help answer the following questions?arrow_forwardShown below are two homologous lengths of the alpha and betachains of human hemoglobin. Consult a genetic code dictionary and determine how many amino acid substitutionsmay have occurred as a result of a single nucleotidesubstitution. For any that cannot occur as a result of a singlechange, determine the minimal mutational distance. Alpha: ala val ala his val asp asp met proBeta: gly leu ala his leu asp asn leu lysarrow_forwardIf an extra nucleotide is inserted in the first exon of the beta globin gene, what effect will it have on the amino acid sequence of the globin polypeptides? Will the globin most likely be fully functional, partly functional, or nonfunctional? Why?arrow_forward
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