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Lena [83]
3 years ago
6

A man who is heterozygous for tongue rolling and has a normal vision and a woman who cannot roll her tongue and is heterozygous

for colorblindness have a child. What is the probability that it will be able to roll his/her tongue and have normal vision? Tongue-rolling is due to a dominant allele. Color Blindness is caused by an X-linked recessive gene. ...?

Biology
2 answers:
liubo4ka [24]3 years ago
7 0
RrXY(man) x rrXx(woman) 

<span>since the father is heterozygous for the character(rolling tongue) there is 50% chance that it will be passed onto the child. </span>
<span>considering color blindness </span>
<span>XY times Xx </span>
<span>
The off-springs will be </span>
<span>XX(normal) </span>
<span>Xx(normal) </span>
<span>XY(normal) </span>
<span>xY(colour blind) </span>
<span>
So chances of the child being normal is 75% </span>
<span>Hence,
Combining the possibilities </span>
<span>we get,
1/2 x 3/4 = 3/8</span>
Montano1993 [528]3 years ago
3 0
The answer is 3/8

Let T be a dominant allele for tongue rolling and let t be a recessive allele.
Tongue-rolling is a dominant trait. That means that an individual with at least one dominant allele will have the trait:
TT - dominant homozygous individuals able to roll tongue
Tt - heterozygous individuals able to roll tongue
tt - recessive homozygous individuals able to roll tongue
A heterozygous man (Tt) and recessive homozygous female (tt) will have 2 of 4 children that are able to roll tongue (Tt). Take a look at the attached image: 2 of 4 children (marked yellow on the image) are able to roll tongue (Tt) and 2 of for children are not (tt).
The probability that a child will be able to roll his/her tongue is 2/4 = <span>1/2

</span>Now, let's see what happens with color blindness as an X-linked recessive trait. Let X_V be a dominant allele and X_v be a recessive allele. The genotypes will be as following:
X_VX_V - a female with normal vision
X_VX_v - a heterozygous female with normal vision
X_vX_v - a female with colorblindness
X_VY - a male with normal vision
X_vY - a male with colorblindness
Take look at the attached image. A man with normal vision (X_VY) and a woman heterozygous for colorblindness (X_VX_v) will have 3 of 4 children that have normal vision (marked yellow on the image) and 1 of 4 children with color blindness.
The probability that the child has normal vision is 3/4.

Now, we have two probabilities:
The probability that a child will be able to roll his/her tongue is 1/2.
The probability that the child has normal vision is 3/4.
Since we want both of these traits to occur together, we will just multiply their possibilities:
1/2 * 3/4 = 3/8

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