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5. A man heterozygous for a mutation in a gene that causes sickle cell disease (an autosomal recessive disease, i.e. not sex linked) who is color blind (a recessive X- linked condition) is expecting a child with a woman who is also heterozygous for the sickle cell disease gene. The woman has normal vision, but her father was colorblind. What is the probability that their child (of either sex) will have sickle cell disease AND be colorblind?

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Oseni

The probability that their child will have sickle cell disease and be colorblind would be 1/8.

Both diseases are independent of each other.

So, considering the sickle cell alone: Let A be normal cell and S for sickle cell. A heterozygous man and woman would both be AS.

                                     

                                    AS     x     AS

                                 AA   AS   AS   SS

Probability of their child having sickle cell = 1/4

Considering the color blindness alone, let a represents color blindness allele and A for normal eyes allele. A colorblind man would have [tex]X^aY[/tex] genotype while a woman with normal vision but with a colorblind father would be heterozygous with [tex]X^AX^a[/tex] genotype.

           [tex]X^aY[/tex]     x     [tex]X^AX^a[/tex]

          [tex]X^AX^a[/tex]   [tex]X^aX^a[/tex]  [tex]X^AY[/tex]  [tex]X^aY[/tex]

Probability of their child being colorblind = 1/2

The probability of their child having sickle cell and being colorblind = probability of having sickle cell x Probability of colorblind

                     = 1/4 x 1/2

                           = 1/8

More on probability can be found here: https://brainly.com/question/851793

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