The following diagram shows the genetic map of an individual in the region of a gene that has a dominant allele D which causes Marfan's syndrome. The DNA probe hybridizes to a segment of DNA that is known to be about 4 map D d units away from the D/d gene. probe If this person has a child with a normal individual who has been shown to be homozygous for the short DNA fragment and the child is also heterozygous DNA restriction fragments, what is the likelihood that the child will have have Marfan's syndrome? ( Please express your answer as a percentage.)

The following diagram shows the genetic map of an individual in the region of a gene that has a dominant allele D which causes Marfan's syndrome. The DNA probe hybridizes to a segment of DNA that is known to be about 4 map D d units away from the D/d gene. probe If this person has a child with a normal individual who has been shown to be homozygous for the short DNA fragment and the child is also heterozygous DNA restriction fragments, what is the likelihood that the child will have have Marfan's syndrome? ( Please express your answer as a percentage.)

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter13: Modern Understandings Of Inheritance

Section: Chapter Questions

Problem 2VCQ: Figure 13.4 Which of the following statements is true? Recombination of the body color and red/...

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For each of the following genetic topics, indicate whether it focuses on transmission genetics, molecular genetics, or population genetics. a. Analysis of pedigrees to determine the probability of someone inheriting a traitb. Study of people on a small island to determine why a genetic form of asthma is prevalent on the islandc. Effect of nonrandom mating on the distribution of genotypes among a group of animals d. Examination of the nucleotide sequences found at the ends of chromosomese. Mechanisms that ensure a high degree of accuracy in DNA replicationf. Study of how the inheritance of traits encoded by genes on sex chromosomes (sex-linked traits) differs from the inheritance of traits encoded by genes on nonsex chromosomes (autosomal traits)
Aliens with orang eye color allele (o) is recessive to the dominant black eye color allele (O). The locus of the orange gene from 10 pure breed orange eyes aliens and 10 pure breed black eyes aliens. You notice a difference in the DNA sequence linked to each allele of the orange gene and you decide to use it as a physical marker to follow recombination between this sequence linked to the orange gene. These sequences consist of short tandem repeat (STR) with two different number of repeats, each associated with one of the two orange gene alleles. a PCR test distinguishes the 10 repeat STR and the 6 repeat STR associated with the O and o alleles respectively. Using primers on each side of the STRs you can amplify by PCR this sequence and visualizing the size of the 10 repeat and 6 repeat STRs in an electrophoresis gel. You can follow the two STR sequences (10 and 6 repeats) linkage and recombination frequencies with the orange gene locus. The black eye aliens yield a PCR fragment that is…
. Consider the genotypes of two lines of chickens: thepure-line mottled Honduran is i/i ; D/D ; M/M ; W/W, andthe pure-line leghorn is I/I ; d/d ; m/m ; w/w, whereI = white feathers, i = colored feathersD = duplex comb, d = simplex combM = bearded, m = beardlessW = white skin, w = yellow skinThese four genes assort independently. Starting withthese two pure lines, what is the fastest and mostconvenient way of generating a pure line that has coloredfeathers, has a simplex comb, is beardless, and has yellowskin? Make sure that you showa. the breeding pedigree.b. the genotype of each animal represented.c. how many eggs to hatch in each cross, and why thisnumber.d. why your scheme is the fastest and the mostconvenient
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In the Human Genome Project, researchers have collectedlinkage data from many crosses in which the male washeterozygous for molecular markers and many crosses wherethe female was heterozygous for the markers. The distancebetween the same two markers, computed in map units, isdifferent between males and females. In other words, thelinkage maps for human males and females are not the same.Propose an explanation for this discrepancy. Do you think thesizes of chromosomes (excluding the Y chromosome) in humanmales and females are different? How could physical mappingresolve this discrepancy?
. In 1932, H. J. Muller suggested a genetic test to determine whether a particular mutation whose phenotypiceffects are recessive to wild type is a null (amorphic)allele or is instead a hypomorphic allele of a gene.Muller’s test was to compare the phenotype of homozygotes for the recessive mutant alleles to the phenotype of a heterozygote in which one chromosomecarries the recessive mutation in question and thehomologous chromosome carries a deletion for aregion including the gene.In a study using Muller’s test, investigators examined two recessive, loss-of-function mutant alleles ofrugose named rg41 and rgγ3. The eye morphologiesdisplayed by flies of several genotypes are indicated inthe following table. Df(1)JC70 is a large deletion thatremoves rugose and several genes to either side of it.Genotype Eye surface Cone cells per ommatidiumwild type smooth 4rg41/rg41 mildly rough 2–3rg41/Df(1)JC70 moderately rough 1–2rg γ3/rg γ3 very rough 0–1rg γ3/Df(1)JC70 very rough 0–1a. Which allele…
In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor recessive gene. If 2 normal parents had a daughter with the symptoms ofthis disease, and a normal son, what is the probability that he might be acarrier of the recessive gene?Express answer in fraction form.
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