Please answer the questions provided in the image. ACTIVITY 3. Probability and Genetic CounselingIn many genetic counseling situations, the counselor will prepare a pedigree for the family or familiesseeking advice. As much as possible, the counselor will determine the phenotype and the genotype foreach person in the pedigree with respect to the trait in question (e.g., albinism). The counselor can thenapply probability principles to determine the probability that a child having a particular abnormality willbe produced among the offspring of a certain marriage.To illustrate the application of these concepts, consider the pedigree shown in Figure 4. (For the purposesof this problem, your instructor will indicate which members of the pedigree express the geneticabnormality.) Unless there is evidence to the contrary, assume that individuals who have married into thisfamily do not carry the recessive gene for the trait.это□OD5 6 716179 10 1112 13 14 15Fig. 4. Human pedigree showing four generations. Circles represent females; squaresrepresent males.For a specific example, you might assume that four members of this pedigree are albinos: the womanin the first generation, her second daughter (the mother of individuals 5, 6, 7, and 8), and individuals4 and 11 in the third generation. Now, assume that you are a genetic counselor and that individuals 6and 12 in the third generation of this pedigree come to you and ask, "What is the probability that ifwe 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 ofthe recessive gene for albinism. The counselor must also consider the probability of two heterozygouscarriers 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 mustbe heterozygous (Cc) since his mother is an albino. Although individual 12 is not an albino, he has a ½chance 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 counselormay advise individuals 6 and 12 that if they were to have children, the probability of their having analbino child will be 1 x ½ x % = %. This is the probability that the three independent events will occursimultaneously. A slightly more confusing situation arises in determining the probability that an unaffected child of twoknown- heterozygous parents is itself heterozygous. For example, in the pedigree shown, we know withcertainty that the parents of individual 10 must be heterozygous. The question, then, is "What is theprobability that one of their phenotypically normal children is heterozygous?" We know that in themating Ccx Cc, the expected ratio of offspring is 1 CC: 2 Cc: 1 cc, that is, among the expected three normal1 CC+2 Cc offspring, two may be expected to be heterozygous. Thus, the probability is 2/3 that the normaloffspring of heterozygous parents will themselves be heterozygous.Continue to assume that the same four individuals in the pedigree are albinos, and calculate theprobability of the albino trait appearing in the offspring if the following cousins and second cousins shouldmarry and reproduce:a. 4x5b. 6 x 10c. 7x14d. 2x10

Name: Please answer the questions provided in the image. ACTIVITY 3. Probabi
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Description: Please answer the questions provided in the image. ACTIVITY 3. Probability and Genetic CounselingIn many genetic counseling situations, the counselor will prepare a pedigree for the family or familiesseeking advice. As much as possible, the counselor

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Please answer the questions provided in the image. ACTIVITY 3. Probability and Genetic CounselingIn many genetic counseling situations, the counselor will prepare a pedigree for the family or familiesseeking advice. As much as possible, the counselor will determine the phenotype and the genotype foreach person in the pedigree with respect to the trait in question (e.g., albinism). The counselor can thenapply probability principles to determine the probability that a child having a particular abnormality willbe produced among the offspring of a certain marriage.To illustrate the application of these concepts, consider the pedigree shown in Figure 4. (For the purposesof this problem, your instructor will indicate which members of the pedigree express the geneticabnormality.) Unless there is evidence to the contrary, assume that individuals who have married into thisfamily do not carry the recessive gene for the trait.это□OD5 6 716179 10 1112 13 14 15Fig. 4. Human pedigree showing four generations. Circles represent females; squaresrepresent males.For a specific example, you might assume that four members of this pedigree are albinos: the womanin the first generation, her second daughter (the mother of individuals 5, 6, 7, and 8), and individuals4 and 11 in the third generation. Now, assume that you are a genetic counselor and that individuals 6and 12 in the third generation of this pedigree come to you and ask, &#34;What is the probability that ifwe marry and have a family, an albino child will be born to us?&#34;The counselor must determine the probability that individuals 6 and 12 are heterozygous carriers ofthe recessive gene for albinism. The counselor must also consider the probability of two heterozygouscarriers 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 mustbe heterozygous (Cc) since his mother is an albino. Although individual 12 is not an albino, he has a ½chance 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 counselormay advise individuals 6 and 12 that if they were to have children, the probability of their having analbino child will be 1 x ½ x % = %. This is the probability that the three independent events will occursimultaneously. A slightly more confusing situation arises in determining the probability that an unaffected child of twoknown- heterozygous parents is itself heterozygous. For example, in the pedigree shown, we know withcertainty that the parents of individual 10 must be heterozygous. The question, then, is &#34;What is theprobability that one of their phenotypically normal children is heterozygous?&#34; We know that in themating Ccx Cc, the expected ratio of offspring is 1 CC: 2 Cc: 1 cc, that is, among the expected three normal1 CC+2 Cc offspring, two may be expected to be heterozygous. Thus, the probability is 2/3 that the normaloffspring of heterozygous parents will themselves be heterozygous.Continue to assume that the same four individuals in the pedigree are albinos, and calculate theprobability of the albino trait appearing in the offspring if the following cousins and second cousins shouldmarry and reproduce:a. 4x5b. 6 x 10c. 7x14d. 2x10

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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