The human CFTR gene encodes a chloride ion channel; loss of function mutations in this gene cause cystic fibrosis. A wide variety of mutations causing partial or complete loss of function of this gene are found in the human population. Which of the following mutations is most likely to result in failure of the mutant allele to be transcribed?

The human CFTR gene encodes a chloride ion channel; loss of function mutations in this gene cause cystic fibrosis. A wide variety of mutations causing partial or complete loss of function of this gene are found in the human population. Which of the following mutations is most likely to result in failure of the mutant allele to be transcribed?

Name: 23 The human CFTR gene encodes a chloride ion channel; loss of functio
Uploaded: 2024-03-20T06:42:41.000Z
Channel: Bartleby
Description: 23 The human CFTR gene encodes a chloride ion channel; loss of functionmutations in this gene cause cystic fibrosis. A wide variety of mutations causingpartial or complete loss of function of this gene are found in the humanpopulation. Which of the f

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 10QP: If the coding region of a gene (the exons) contains 2,100 base pairs of DNA, would a missense...

See similar textbooks

Similar questions

The hunchback gene, a gene necessary for proper patterning of the Drosophila embryo, is translationallyregulated. The position of the coding region withinthe transcript is known. How could you determine ifthe sequences within the 5′ UTR or 3′ UTR, or both,are necessary for proper regulation of the mRNA’stranslation?
Lack of phosporylation of the C-terminal domian of Pol Il will result in which of the following? O Assembly of the full set of general transcription factors plus Pol II at the promoter, but unphosphorylated Pol II cannot leave the promoter region No assembly of the general transcription factors at a promoter No binding of Pol II at a promoter The unphosphorylated Pol II can initiate and elongate but cannot terminate
ennar region of gene X, which determines the length of the tail in mice, is mutated so that transcription factors bind it at a much higher affinity compared to the wild-type sequence. What is the most likely phenotypic outcome? Tail length will not change because the enhancer is a non-coding sequence Tail length will increased due to increased activity of the gene's promoter Tail length will decreased because any mutation will cause a loss-of-function of these regulatory regions Not just the tail will be enlarged because increased activity of the enhancer will impact many genes
A strain of Arabidopsis thaliana possesses a mutation in the APETALA2 gene. As a result of this mutation, much of the 3′ UTR of the mRNA transcribed from the gene is deleted. What is the most likely effect of this mutation on the expression of the APETALA2 gene?
The following is a list of mutations that have beendiscovered in a gene that has more than 60 exons andencodes a very large protein of 2532 amino acids.Indicate whether or not each mutation could cause adetectable change in the size or the amount of mRNAand/or a detectable change in the size or the amountof the protein product. (Detectable changes in size oramount must be greater than 1% of normal values.)What kind of change would you predict?a. Lys576Val (changes amino acid 576 from lysineinto valine)b. Lys576Argc. AAG576AAA (changes codon 576 from AAG toAAA)d. AAG576UAGe. Met1Arg (at least two possible scenarios exist forthis mutation)f. promoter mutationg. one base pair insertion into codon 1841h. deletion of codon 779i. IVS18DS, G–A, + 1 (this mutation changes thefirst nucleotide in the eighteenth intron of the gene,causing exon 18 to be spliced to exon 20, thusskipping exon 19)j. deletion of the poly-A addition sitek. G-to-A substitution in the 5′ UTRl. insertion of 1000 base…
The method of Northern blotting is used to determine the amountand size of a particular RNA transcribed in a given cell type.Alternative splicing (discussed in Chapter 14) produces mRNAsof different lengths from the same gene. The Northern blot shownhere was obtained using a DNA probe that is complementary tothe mRNA encoded by a particular gene. The mRNA in lanes 1through 4 was isolated from different cell types, and equal amountsof total cellular mRNA were added to each lane. Explain these results.
The following mutants are observed in a gene coded for a ribosomal protein. The mutations are highlighted: Original mRNA: 5’- AUG GGC UCG ACC UUG -3’ Mutant mRNA 1: 5’- AUG GGC UAG ACC UUG -3’ Mutant mRNA 2: 5’- AUG GGC UCG ACU UUG -3’ Mutant mRNA 3: 5’- AUG GGC UCG AGC UUG -3’ Mutant mRNA 4: 5’- AUG GGC UCG AUC UUG -3’ A. Write out the protein sequence for the original protein. 1. Three letter sequence (write with one space in between, i.e. Val Leu Gly) 2. One letter sequence (write out with one space between, i.e. V L G) B. Characterize the mutants (#1-4) as silent, nonsense, or missense mutations.
Genes can be transcribed into mRNA, in the case of protein coding genes, or into RNA, in the case of genes such as those that encode ribosomal or transfer RNAs. Define a gene. For the following characteristics, state whether they apply to (a) continuous, (b) simple, or (c) complex transcription units.i. Found in eukaryotesii. Contain intronsiii. Capable of making only a single protein from a given gene
Like Hurler syndrome, Fabry disease involves an abnormal accumulationof substances within lysosomes. However, the lysosomesof individuals with Fabry disease show an abnormal accumulationof lipids. The defective enzyme is α-galactosidase A, which is alysosomal enzyme that functions in lipid metabolism. The enzymaticdefect causes cell damage, especially to the kidneys, heart,and eyes. The gene that encodes α-galactosidase A is found on theX chromosome. Let’s suppose a phenotypically unaffected coupleproduces two sons with Fabry disease and one phenotypicallyunaffected daughter. What is the probability that the daughter willhave an affected son?
In this chapter you were introduced to nonsense suppressor mutations in tRNA genes. However, suppressormutations also occur in protein-coding genes. Using thetertiary structure of the β subunit of hemoglobin shownin Figure 9-3(c), explain in structural terms how a mutation could cause the loss of globin protein function. Nowexplain how a mutation at a second site in the same protein could suppress this mutation and lead to a normalor near-normal protein.
IS. Alternative splicing has been estimated to occur in more than 95% of multi-exon genes. Which of the following is not an evolutionary advantage of alternative splicing? Alternative splicing increases diversity without increasing genome size Different gene isoforms can be expressed in different tissues Alternative splicing creates shorter mRNA transcripts Different gene isoforms can be expressed during different stages of development.
The pre-mRNA transcript and protein made by several mutant genes were examined. The results are given below. Determine where in the gene a likely mutation lies: the promoter region, exon, intron, cap on mRNA, or ribosome binding site. a. normal-length transcript, normal-length nonfunctional protein b. normal-length transcript, no protein made c. normal-length transcript, normal-length mRNA, short nonfunctional protein d. normal-length transcript, longer mRNA, shorter nonfunctional protein e. transcript never made