Biology
12th Edition
ISBN: 9780134813448
Author: Audesirk, Teresa, Gerald, Byers, Bruce E.
Publisher: Pearson,
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Chapter 13.4, Problem 1CSC
Summary Introduction
To describe: The way in which functional chloride channel (CFTR protein) can produce CFTR alleles that cause cystic fibrosis.
Introduction: Cystic fibrosis is a genetic disorder that causes severe damage to the lungs, pancreas,
Summary Introduction
To describe: The way in which a gene can affect the timing and rate of transcription and translation.
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The figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides.
The figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two
mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible
for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided.
Hint: This organism produces a poly-A tail of 120 nucleotides.
99
62
120
84
102
27
117
Gene X
E1
11
E2
12
ЕЗ
13
E4
Exon (E)
Intron (I)
The mRNA sequence 5' AUG AAA CAG GGA UAA 3' encodes a particular peptide of interest to your research team. You have identified a new alternate allele of the sequence 5' AUG AAG CAG GGA UAA 3'. What type of mutation does this alternate sequence illustrate?
Chapter 13 Solutions
Biology
Ch. 13.1 - describe three types of RNA that play roles in...Ch. 13.1 - Prob. 2CYLCh. 13.1 - Prob. 3CYLCh. 13.2 - Prob. 1TCCh. 13.2 - Prob. 1CYLCh. 13.2 - Prob. 2CYLCh. 13.2 - describe an example of post-transcription...Ch. 13.3 - Prob. 1TCCh. 13.3 - Prob. 1CSCCh. 13.3 - Prob. 1CYL
Ch. 13.3 - Prob. 2CYLCh. 13.3 - Prob. 3CYLCh. 13.3 - Prob. 4CYLCh. 13.4 - Prob. 1CSCCh. 13.4 - describe three different types of mutations?Ch. 13.4 - Prob. 2CYLCh. 13.5 - Prob. 1HYEWCh. 13.5 - Envision yourself as a physician. A mother,...Ch. 13.5 - Prob. 2TCCh. 13.5 - Prob. 1CYLCh. 13.5 - Prob. 2CYLCh. 13.5 - Prob. 3CYLCh. 13.5 - Prob. 4CYLCh. 13.5 - Prob. 1CTCh. 13 - Prob. 1MCCh. 13 - Which of the following is not true of RNA? a. It...Ch. 13 - Prob. 3MCCh. 13 - Prob. 4MCCh. 13 - Prob. 5MCCh. 13 - Synthesis of RNA from the instructions in DNA is...Ch. 13 - Prob. 2FIBCh. 13 - Prob. 3FIBCh. 13 - Prob. 4FIBCh. 13 - Prob. 5FIBCh. 13 - If a nucleotide is replaced by a different...Ch. 13 - Prob. 1RQCh. 13 - Name the three types of RNA that are essential to...Ch. 13 - Prob. 3RQCh. 13 - Prob. 4RQCh. 13 - Prob. 5RQCh. 13 - Prob. 6RQCh. 13 - Prob. 7RQCh. 13 - Define mutation. Describe four different effects...Ch. 13 - Prob. 1ACCh. 13 - Prob. 2AC
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- The following four mutations have been discovered in a gene that has more than 60 exons and encodes a very large protein of 2532 amino acids. Indicate which mutation would likely cause a detectable change in the size of the mRNA and/or the size of the protein product. Consider a detectable change to be >10% of the wild-type size. A table of the genetic code is shown below. First letter 0 00 U O A บบบ UUC UUA UUG U CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu >Leu AUU AUC lle AUA AUG Met >Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Second letter C Ser Pro Thr Ala CAU CAC CAA CAG UAU UGU Tyr UAC UGC UAA Stop UGA UAG Stop UGG AAU AAC AAA AAG A GAU GAC GAA GAG His Gin Asn Lys Asp G Glu CGU CGC CGA CGGJ AGU AGC AGA AGG GGU GGC GGA GGG O AAG576UAG (changes codon 576 from AAG to UAG) Cys Stop Trp O GUG326AUG (changes codon 326 from GUG to AUG) Arg Ser Arg Gly DUAG DUA G DCAG DO AG deletion of codon 779 insertion of 1000 base pairs into the sixth intron (this particular…arrow_forwardThe following double stranded segment of DNA is part of a protein coding gene. The segments in uppercase letters (ACTG) represent the exons. The segments in lowercase letters (acgt) represent introns. The lower strand is the template strand that is used by the RNA polymerase to make an RNA transcript. Draw or write-out a) the sequence of the primary transcript and b) the mature mRNA resulting from this stretch of DNA.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
- Tay Sachs disease is an autosomal recessive disease in which a protein – Hex A - is abnormal. To make the Hex A protein: The promoter and transcription termination sites are 33,000 base pairs apart. The Hex A protein has 600 amino acids 5’ and 3’ UTR’s are each 500 bp long. a)How many base pairs would you expect in the final mRNA? Show your work b)How many bases were spliced out? Show your workarrow_forwardOchre and amber are two distinct nonsense mutations. Before the genetic code was worked out, Sydney Brenner, Anthony O. Stretton, and Samuel Kaplan applied different types of mutagens to bacteriophages in an attempt to determine the bases present in the codons responsible for amber and ochre mutations. They knew that the ochre and amber mutations were suppressed by different types of suppressor mutations, which demonstrated that each is a different stop codon. They obtained the following results: (1) A single-base substitution could convert an ochre mutation into an amber mutation. (2) Hydroxylamine induced both ochre and amber mutations in wildtype phages. (3) 2-Aminopurine caused ochre to mutate to amber. (4) Hydroxylamine did not cause ochre to mutate to amber. These data do not allow the complete nucleotide sequence of the amber and ochre codons to be worked out, but they do provide some information about the bases found in the nonsense mutations. a. What conclusions about the…arrow_forwardOchre and amber are two distinct nonsense mutations. Before the genetic code was worked out, Sydney Brenner, Anthony O. Stretton, and Samuel Kaplan applied different types of mutagens to bacteriophages in an attempt to determine the bases present in the codons responsible for amber and ochre mutations. They knew that the ochre and amber mutations were suppressed by different types of suppressor mutations, which demonstrated that each is a different stop codon. They obtained the following results: (1) A single-base substitution could convert an ochre mutation into an amber mutation. (2) Hydroxylamine induced both ochre and amber mutations in wildtype phages. (3) 2-Aminopurine caused ochre to mutate to amber. (4) Hydroxylamine did not cause ochre to mutate to amber. These data do not allow the complete nucleotide sequence of the amber and ochre codons to be worked out, but they do provide some information about the bases found in the nonsense mutations. Q. Of the three nonsense codons…arrow_forward
- Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. Question 6: How many amino acids are present in the wild-type human B-globin protein? = exons = introns Transcription termination site (also poly A site) Promoter Start of transcription 3' 5' ATG 50 TẠC TAA 126 132 |ATT 90 130 222 850 3 5' Start codon Stop codon А. 438 В. 146 C. 620 D. 206 © 2013 John Wiley & Sons, Inc. All rights reserved.arrow_forwardHemophilia in the Russian royal family was caused by defective protein involved in blood clotting (factor IX). This defective protein was caused by a mutation that altered the splicing of the exons. This genetic change in the splicing pattern created a new stop codon in the mRNA for factor IX. Is it likely that the mutation has altered at least one base at the exon-intron boundary in the wild-type pre-mRNA? Why or why not?arrow_forwardHemophilia in the Russian royal family was caused by defective protein involved in blood clotting (factor IX). This defective protein was caused by a mutation that altered the splicing of the exons. This genetic change in the splicing pattern created a new stop codon in the mRNA for factor IX. What effect might this new stop codon have on the primary and tertiary levels of the mutant factor IX protein (compared to the native or wild-type protein).arrow_forward
- Would a gain of function mutaion that occurs in the first exon of a gene with twelve exons more likely be missense or nonsense? Briefly explain your choice. List one disease that we studied in class that is due to a gain of function.arrow_forwardYou take DNA samples from a family with a history of the genetic disease spinal muscular atrophy, which results from many different mutations in the SMA gene. One mutation in the SMA gene is a 240-bp in-frame deletion mutation in the middle of an exon. Data from this family shows you that several individuals from the family have this mutation. What do you predict that you would find when comparing the mRNA and protein products of the mutated and unmutated SMA gene? Select all that apply. 1. The mRNA from the SMA is more stable than unmutated SMA mRNA. 2. The pre-mRNA from the SMA mutant is shorter than the unmutated SMA pre-mRNA. 3. The mature mRNA from the SMA mutant is longer than the unmutated SMA mRNA. 4. The protein from the mutated SMA is shorter than the unmutated SMA protein.arrow_forwardIn 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_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY