Biological Science (6th Edition)
6th Edition
ISBN: 9780321976499
Author: Scott Freeman, Kim Quillin, Lizabeth Allison, Michael Black, Emily Taylor, Greg Podgorski, Jeff Carmichael
Publisher: PEARSON
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Chapter 16, Problem 13PIAT
A small portion of the human transport protein amino acid sequence is shown below. The upper sequence is associated with darker skin, and the lower sequence is associated with lighter skin. What DNA base-pair change created the light-skin form of the human protein from the gene that coded for the dark-skin form?
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Chapter 16 Solutions
Biological Science (6th Edition)
Ch. 16 - Prob. 1TYKCh. 16 - 2. Which of the following is an important...Ch. 16 - Prob. 3TYKCh. 16 - Prob. 4TYKCh. 16 - Prob. 5TYUCh. 16 - A minimal genetic code requires only 21 codons–one...Ch. 16 - Prob. 7TYUCh. 16 - Prob. 8TYUCh. 16 - MODEL Draw a hypothetical metabolic pathway in...Ch. 16 - Prob. 10TYPSS
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- The image below shows the base cytosine and a methylated form of cytosine that occurs frequently in the human genome. Use your knowledge of DNA structure to answer the following questions: a) Does methylation of cytosine affect its ability to base-pair with guanine? Explain your answer. b) Would methylation of cytosine affect the binding of a protein that interacts with a C-G base-pair in the major groove?arrow_forwardThe human genome contains thousands of sequences known as small open reading frames, some of which encode proteins of about 30 amino acids. What is the minimum number of nucleotides required to encode such a protein?arrow_forwardHere is a eukaryotic gene. The numbers given are base pairs of exon and intron. How long in bases will the pre mRNA transcript be? Explain briefly. What is the maximum number of amino acids that could make up the protein product from the final mRNA? Explain briefly.arrow_forward
- In the table below, there are four versions of gene A, one of which is normal, and the other three which contain mutations that make the gene product nonfunctional. Focus on the shaded region of the sequence. Use the genetic code table to answer the question. How would you describe Mutation #2? Partial DNA sequence for gene A ("..." indicates many nucleotides of sequence not shown) 5' ... ATG GTG AGC AAG GAG GAG CTG TTC ACC TGT AAA TAG ... Normal Mutation #1 5' ... ATG GTG AGC AAG GAG AAG CTG TTC ACC TGT AAA TAG ... Mutation #2 5' ... ATG GTG AGC AAG TAG GAG CTG TTC ACC TGT AAA TAG ... Mutation #3 5' ... ATG GTG AGC AAG GAG CTG TTC ACC TGT AAA TAG ... Silent mutation Nonsense mutation Frameshift mutations Missense mútationarrow_forwardA 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?arrow_forwardThe E. coli DNA molecule contains 4.70 x 105 base pairs. Determine the number of codons that can be present in this DNA molecule. number of codons: 1.57 x106 codons Assuming that the average protein in E. coli consists of a chain of 420 amino acids, calculate the maximum number of proteins that can be coded by an E. coli DNA molecule. number of proteins: proteinsarrow_forward
- The following nucleotide sequence is found on the template strand of DNA. First, determine the amino acids of the protein encoded by this sequence by using the genetic code provided in Figure 15.10. Then give the altered amino acid sequence of the protein that will be found in the following mutations: Q.Mutant 3: A one-nucleotide deletion at nucleotide 7arrow_forwardWhich of the following mutations in the protein-coding region of a gene is more likely to lead to complete loss of function of the encoded protein: an insertion of six nucleotides or a deletion of two nucleotides? Briefly explain your answer.arrow_forwardGiven the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs ( what letter changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions 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' - AGGCTCGCCTGT - 5' Mutated DNA Sequence: 3' - AGTCTCGCCTGT - 5'arrow_forward
- Human wildtype and mutant alleles are identical in sequence except for a single base-pair substitution that changes one nucleotide towards the end of intron 2. The wildtype and mutant sequences of the affected portion of the mRNA are listed in the following table. Explain how a single base substitution could alter the reading frame, which could result in a physiological disorder?arrow_forwardBelow is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.arrow_forwardAs described earlier, DNA damage can cause deletion or insertion of base pairs. If a nucleotide base sequence of a coding region changes by any number of bases other than three base pairs, or multiples of 3, a frameshift mutation occurs. Depending on the location of the sequence change, such mutations can have serious effects. The following synthetic mRNA sequence codes for the beginning of a polypeptide: 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCUUAUC-AUGUUU-3′ First, determine the amino acid sequence of the polypeptide. Then determine the types of mutation that have occurred in the following altered mRNA segments. What effect do these mutations have on the polypeptide products? a. 5′-AUGUCUCCUACUUGCUGACGAGGGAAGGAGGUGGCUUAUCA-UGUUU-3′ b. 5′-AUGUCUCCUACUGCUGACGAGGGAGGAGGUGGCUUAUCAU-GUUU-3′ c. 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCCCUUAUC-AUGUUU-3′ d. 5′-AUGUCUCCUACUGCUGACGGAAGGAGGUGGCUUAUCAU-GUUU-3′arrow_forward
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