Genetic Analysis: An Integrated Approach (2nd Edition)
Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
bartleby

Concept explainers

Question
Book Icon
Chapter 10, Problem 32P
Summary Introduction

To analyze:

Based on the result of western blot given below:

Genetic Analysis: An Integrated Approach (2nd Edition), Chapter 10, Problem 32P

  1. Identify the protein change resulting in two bands observed in the western blot for organisms having Dd2 genotype.

  2. The absence of detectable protein for organisms with the d1d1 genotype.

  3. The reason behind presence of just one detectable protein band for organisms having d1d2 and d2d2 genotypes.

  4. Based on evaluation of western blot analysis, project the nature of the mutations generating d1 and d2. In other words, changes at the DNA level led to production of these mutations.

Introduction:

Genes encode the functional protein and enzymes; these protein and enzymes perform specific functions to express specific phenotypes. The experimental approach, western blotting, is used to observe the protein level expression pattern of a gene. An individual has two alleles for one gene whereas when multiple alleles for one gene are present in a population, the population is termed as polymorphic population. Dominant allele (D) encodes the functional enzyme but recessive allele (d) does not encode a functional protein. The band pattern for a heterozygous individual is observed in lane 2. The dominant allele suppresses the effect of a recessive allele.

D

Another recessive allele (d2) encodes the protein that inhabits the expression of the dominant allele (D). The band pattern of recessive allele (d2) is observed in western blotting.

Blurred answer
Students have asked these similar questions
Leber Congenital Amaurosis (LCA) causes progressive vision loss due to defects in the gene that encodes RPE65 isomerase.  Affected individuals are homozygous recessive for mutant alleles of the RPE65 gene.  You are trying to determine the molecular nature of the mutations in three individuals with LCA.  For ease of analysis, you may assume that each individual is homozygous for the same mutant allele (though the three individuals have different mutations than each other).   You use the polymerase chain reaction to amplify DNA from each patient and you determine the sequence of the DNA and compare it to unaffected individuals.  You identify the following differences.  Note that the non-template strand of DNA is given and the changes are highlighted using red boldface.  You can assume that the sequences are in the first reading frame (eg. the first three nucleotides of each sequence is a codon).   The coding region of the gene is 1602 bp and the position of the sequences shown below is…
The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.
The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●
Knowledge Booster
Background pattern image
Biology
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Text book image
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Text book image
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Text book image
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Text book image
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Text book image
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education