d either the phenotype A-B- or the genotype iable V. PROBABILITY AND GENETIC COU

Question

d either the phenotype A-B- or the genotype iable V. PROBABILITY AND GENETIC COUNSELING In many genetic counseling situations, the counselor will prepare a pedigree for the family or families seeking advice. Insofar as possible the counselor will determine the phenotype and the genotype for each person in the pedigree with respect to the trait in question (e.g., albinism). The counselor can then apply probabiliry principles to determine the probability that a child having a particular abnor mality will be produced among the offspring of a certain marriage. To illustrate the application of these concepts, consider the pedigree shown in Figure 21. (For the purposes of this problem, your instructor will indicate which members of the pedigree express the genetic abnormality,) Unless there is evidence to the contrary, assume that individuals who have mar- ried into this family do not carry the recessive gene for the trait. For a specific example. you might assume that four members of this pedigree are albinos: the woman in the first generation, her second daughter (the mother of individuals 5,6, 7, and 8), and individuals 4 and 11 in the third generation. Now, assume that you are a genetic counselor and that individuals 6 and 12 in the third generation of this pedigree come to you and ask, "What is the prob- ability that if we mary and have a family, an albino child will be born to us? The counselor must determine the probabiliry that individuals 6 and 12 are heterozygous carriers of the recessive gene for albinism. The counselor must also consider the probability of two heterozy gous carriers producing homozygous recessive child. First of all, the mother of individual 6S an al- bino (cc), which means that 6 is of necessity (probability 1) a heterozygote. The father of individual 12 must be heterozygous (Cc) since bis mother is an albino.-Although individual 12 is not an albino, he has a 1/2 chance of being heterozygous, depending on which allele (C or o) he inherited from his father. Finally, rwo heterozygotes (Ce x Cc) have 1 chance in 4 of producing an albino (cc) child probability of their having an albino child will be 1 x 1/2 x 1/4 1/8. This is the probability that the three independent events will occur simultaneously A slightly more confusing situation arises in determining the probability that an unaffected child of two known-heterozygous parents is itself heterozygous. For example, in the pedigree shown, we know with certainty that the parents of individual 10 must be heterozygous. The question, then, is What is the probabilty that one of their phenoty pically normal children is heterozygous? We know that in the mating Cc x Ce the expected ratio of offspring is 1CC 2Cc:lcc, that is, among the expected 5 6 7 9 0 12 13 14 16 FIGURE 2.1 Huan pedigree showing four geperations represeni mal Circles represeni females, squares

Explanation / Answer

In this case, the genotype of 4 is essentially cc. And 7 is a heterozygote since his mother is albino. Hence his genotype would be Cc. So probability of getting c allele from 4 is 1 and probability of getting a c allele from 7 is ½. Hence, probability of their offspring to be albino is 1x1/2=1/2.

Here, 5 is again a heterozygous individual with genotype Cc(similar to 7 in A.). 1 can be CC or Cc. Probability of one to be Cc is 2/3. Therefore, probability of their offspring to be albino is 1/2x1/2x2/3=1/6. (1/2 is the probability of getting c allele from 5, 1/2x2/3 is the probability of getting a c allele from 1).

Here 6 is again a heterozygous individual. 13 can be either CC or Cc. of 13 to Cc is 1/2. Hence the probability of their offspring to be an is, 1/2x1/2x1/2=1/8. (1/2 is the probability of getting c allele from 6, 1/2x1/2 is the probability of getting a c allele from 13).

10 can be either CC or Cc. of 10 being Cc is 2/3. of 14 being Cc is 1/2., similar to 13. Hence the of their offspring to be an is 1/2x2/3x1/2x1/2=1/12. (1/2x2/3 is the probability of getting c allele from 10, 1/2x1/2 is the probability of getting a c allele from 14).

Here, we know the of 3 being Cc is 2/3. Now for 17, her father is CC (assumption mentioned the start of the ). Now for 17 to have c allele, her mother to be Cc. of her mother being Cc is 2/3. Hence the of 17 having c allele is 2/3x1/2=1/3. Now the that the offspring of 3 and 17 shall be albino is 2/3x1/2x1/3x1/2=1/18(2/3x1/2 is the probability of getting c allele from 3, 1/2x1/3 is the probability of getting a c allele from 17).

Again, for 3, the of offspring getting c allele from 3 is 2/3x1/2=1/3. heterozygous is 1/2. of glele from 15 is 1/2x1/2=1/4. Hence r offspring will be albino is 1/3x1/4=1/12.

In this case, we know 8 is heterozygous. that 16 has the c allele inherited from 8 is 1/2. Hence the that the offspring will have c allele from 16 is 1/2x1/2=1/4. Now for 17, her father is CC (assumption mentioned the start of the ). Now for 17 to have c allele, her mother to be Cc. the of her mother being Cc is 2/3. Hence the of 17 having c allele is 2/3x1/2=1/3. Hence the that their child will be albino is 1/4x1/3=1/12.

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