Problem 3. A certain organism possesses a pair of each of 5 different genes (whi

Question

Problem 3. A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters) The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or Xx will have brown eyes, whereas one having gene pair xx will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes aA, bB, ce, d D, ee and AA, 89,-DD, ee would have different genotypes but the same phenotype.) In a mating between two organisms, cach one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. Figure 1 shows the possible outcomes irn a mating between organisms having genotypes aA, bB, cC, dD, cE and aa, bB, cc, Dd, ee. In that mating, what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent (b) the second parent (c) either parent (d) neither parent

Explanation / Answer

a)for the A gene he possible genotypes are

aA and aa progeny will have the same phenotype as the first parent with probability 1/2

for the B gene, the possible genotypes are bb bB and BB the progeny will have the same phenotype as the first parent with probability 3/4

P(matches 1st parent’s phenotype) =P(A correct) *P(B correct)*P(C correct)*P(D correct)*P(E correct)

                                                   =1/2* 3/4 *1/2 * 3/4 *1/2

                                                    = 9/128

For the A gene, the progeny will have the same genotype as the first parent with

probability is 1/2

For the B gene, the progeny will have the same genotype as the

first parent with probability 1/2

P(matches 1st parent’s genotype) = P(A correct) *P(B correct)*P(C correct)*P(D correct)*P(E correct)

                                                  = 1/2 * 1/2 * 1/2* 1/2 * 1/2

                                                   =1/32

b)

P (matches 2nd parent’s phenotype) = P(A correct) *P(B correct)*P(C correct)*P(D correct)*P(E correct)

                                                   =1/2* 3/4 *1/2 * 3/4 *1/2

                                                    = 9/128

P (matches 2nd parent’s genotype) =P(A correct) *P(B correct)*P(C correct)*P(D correct)*P(E correct)

                                                  = 1/2 * 1/2 * 1/2* 1/2 * 1/2

                                                   =1/32

C)

P(matches either parent’s phenotype)=P(matches 1st parent’s phenotype) +P(matches 2nd parent’s phenotype)

SINCE EVENTS ARE DISJOINT

                                                        =9/128 + 9/128 = 9/64

P(matches either parent’s genotype)=P(matches 1st parent’s genotype) +P(matches 2nd parent’s genootype)

                                                      =1/32+1/32 = 1/16

D)P(matches neither parent’s phenotype) = 1-P((matches either parent’s phenotype)

                                                             =1-9/64

                                                             =55/64

P(matches neither parent’s genotype) = 1-P(matches either parent’s genotype) =1-1/16 = 15/16

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.