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D A EI stoR Search Resul D strength and condi Ahica blogs Field of Hope special-Online-Sum KERRID Help Save & Exit etics Dihybrid cross with autosomal and sex-linked traits, part 1 At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision A heterozygous brown-eyed woman who is a carrier of colorblindness has children with a blue-eyed man who is not colorblind. What is the probability that their first child will be a blue-eyed female who has normal color vision? (Enter the probability as a percent Enter the number only without the percent sig. For example, enter 100% as 100 and enter 125% as 12.5) Fill in the biank Prey 26 f 26 5 6 8

Explanation / Answer

Q) In question when a heterozygous brown eyed female who is carrier of colorblindness marries a man who is blue eyed and not colorblind , the probability of their first child to be blue eyed, normal color vision female.

Answer-

In this question Brown color is dominant that is "B" and Blue color is recesssive that is "b", it is an autosomal inheritance .

For colorblindness "Xc " gene is responsible which is a recessive characteristics and "X " is normal gene which is dominant .

colorblindness is a sex linked X linked recessive trait. A person to have colorblindness got to have two genes carrying the colorblindness .

B- Brown(dominant) , b - blue (recessive)

Xc - Colorblind (recessive), X- Normal(dominant)

Heterozygous brown eyed female who is carrier of colorblindness ( Xc X Bb) when marries Blue eyed normal man ( XY bb)

Xc X Bb gives rise to gametes Xc B , Xc b, XB, Xb

And XYbb gives rise to gametes XB,Xb,YB,yb

The punnett square of dihybrid cross of (Xc X Bb) and (XY bb) will be

so from the punnett square it is concluded that having a female child with blue eyes and normal color vision has 1/4 or 25 % chance .

The answer is 25 .

IInd method to calculate the probablity:-

in another process by applying the product rule of probability

The chance of having a blue eyed female which is a autosomal trait , when a heterozygous brown eyed female marries a blue eyed male the chance of having a female with blue eyes are

when Bb(female) crossed with bb (male) the chance of having a bb(female is )

There is 50% chance or 1/2 chance of a female having blue eyes .

And when a female who is a carrier of colorblind (Xc X)marries a normal male (XY) than the chance of having a female with normal vision are

so the chance of a female having normal vision are 50% or 1/2

By applying the product rule to have a female with normal vision and blue eyes (Xc Xbb or XXbb) is

(Probability of bb) * (probablity of XcX or XX) = 1/2 * 1/2= 1/4 or 25% or 25

so as per question guideline answer will be 25.

(Note- XcX and XX are both having normal vision bbecause a female to be colorblind it requires to have two Xc genes, as colorblindnes is a X -linked recessive trait, and Xc gets dominated by X gene which is normal.)

XcB Xcb XB Xb Xb XcXBb XcXbb XXBb XXbb Xb XcXBb XcXbb XXBb XXbb Yb XcYBb XcYbb XYBb XYbb Yb XcYBb XcYbb XYBb XYbb

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