Continue to assume that the same four individuals in the pedigree are albinos, and calculate the probability of the albino trait appearing in the offspring if the following cousins and second cousins should marry and reproduce: a 4x5 b. 6 x 10 c. 7x14

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ACTIVITY 3. Probability and Genetic Counseling
In many genetic counseling situations, the counselor will prepare a pedigree for the family or families
seeking advice. As much 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 probability principles to determine the probability that a child having a particular abnormality will
be produced among the offspring of a certain marriage.
To illustrate the application of these concepts, consider the pedigree shown in Figure 4. (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 married into this
family do not carry the recessive gene for the trait.
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□OD
5 6 7
16
17
9 10 11
12 13 14 15
Fig. 4. Human pedigree showing four generations. Circles represent females; squares
represent males.
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 probability that if
we marry and have a family, an albino child will be born to us?"
The counselor must determine the probability that individuals 6 and 12 are heterozygous carriers of
the recessive gene for albinism. The counselor must also consider the probability of two heterozygous
carriers producing a homozygous recessive child. First of all, the mother of individual 6 is an albino (cc),
which means that 6 must be (probability = 1 or 100 %) a heterozygote. The father of individual 12 must
be heterozygous (Cc) since his mother is an albino. Although individual 12 is not an albino, he has a ½
chance of being heterozygous, depending on which allele (C or c) he inherited from his father. Finally,
two heterozygotes (Cc x Cc) have a % chance of producing an albino (cc) child. The genetic counselor
may advise individuals 6 and 12 that if they were to have children, the probability of their having an
albino child will be 1 x ½ x % = %. This is the probability that the three independent events will occur
simultaneously.
Transcribed Image Text:ACTIVITY 3. Probability and Genetic Counseling In many genetic counseling situations, the counselor will prepare a pedigree for the family or families seeking advice. As much 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 probability principles to determine the probability that a child having a particular abnormality will be produced among the offspring of a certain marriage. To illustrate the application of these concepts, consider the pedigree shown in Figure 4. (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 married into this family do not carry the recessive gene for the trait. это □OD 5 6 7 16 17 9 10 11 12 13 14 15 Fig. 4. Human pedigree showing four generations. Circles represent females; squares represent males. 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 probability that if we marry and have a family, an albino child will be born to us?" The counselor must determine the probability that individuals 6 and 12 are heterozygous carriers of the recessive gene for albinism. The counselor must also consider the probability of two heterozygous carriers producing a homozygous recessive child. First of all, the mother of individual 6 is an albino (cc), which means that 6 must be (probability = 1 or 100 %) a heterozygote. The father of individual 12 must be heterozygous (Cc) since his mother is an albino. Although individual 12 is not an albino, he has a ½ chance of being heterozygous, depending on which allele (C or c) he inherited from his father. Finally, two heterozygotes (Cc x Cc) have a % chance of producing an albino (cc) child. The genetic counselor may advise individuals 6 and 12 that if they were to have children, the probability of their having an albino child will be 1 x ½ x % = %. 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
probability that one of their phenotypically normal children is heterozygous?" We know that in the
mating Ccx Cc, the expected ratio of offspring is 1 CC: 2 Cc: 1 cc, that is, among the expected three normal
1 CC+2 Cc offspring, two may be expected to be heterozygous. Thus, the probability is 2/3 that the normal
offspring of heterozygous parents will themselves be heterozygous.
Continue to assume that the same four individuals in the pedigree are albinos, and calculate the
probability of the albino trait appearing in the offspring if the following cousins and second cousins should
marry and reproduce:
a. 4x5
b. 6 x 10
c. 7x14
d. 2x10
Transcribed Image Text: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 probability that one of their phenotypically normal children is heterozygous?" We know that in the mating Ccx Cc, the expected ratio of offspring is 1 CC: 2 Cc: 1 cc, that is, among the expected three normal 1 CC+2 Cc offspring, two may be expected to be heterozygous. Thus, the probability is 2/3 that the normal offspring of heterozygous parents will themselves be heterozygous. Continue to assume that the same four individuals in the pedigree are albinos, and calculate the probability of the albino trait appearing in the offspring if the following cousins and second cousins should marry and reproduce: a. 4x5 b. 6 x 10 c. 7x14 d. 2x10
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