Q. 3.69 3.69. A certain organism possess... [FREE SOLUTION]

Chapter 3: Q. 3.69 (page 104)

3.69. 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 $x X$ , then it will outwardly have the appearance of the $X$ gene. For instance, if 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, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms, each 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 the organism鈥檚 mate. In a mating between organisms having genotypes , bB, cC, dD, eE and aa, bB, cc, Dd, ee 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?

Short Answer

Expert verified

The probability that the progeny will $(i)$ phenotypically

a) $\frac{9}{128}$

b) $\frac{9}{128}$

c) $\frac{9}{64}$

d) $\frac{55}{64}$

The probability that the progeny will localid="1649421184104" $(i i)$ genotypically

a) localid="1649421094188" $\frac{1}{32}$

b) $\frac{1}{32}$

c) $\frac{1}{16}$

d) localid="1649421188700" $\frac{15}{16}$

Step by step solution

Given Data

First Parent : $a A b B c C d D e E$

Second Parent : $a a b B c c d D e e$

Events:

localid="1649682516809" $A_{i} -$ the progeny received gene $A$ from the $i -th$ parent, $i = 1, 2$

localid="1649683306917" $B_{i} -$ the progeny received gene $B$ from the $i -th$ parent, $i = 1, 2$

The mutual independent events are, $A_{1}, A_{2}, B_{1}, B_{2}, 鈥�, E_{2}$

Evaluate:

$(i)$ a) $P (" P h e n o t y p e a s t h e f i r s t p a r e n t ")$

b) $P (" P h e n o t y p e a s t h e s e c o n d p a r e n t ")$

c) $P (" P h e n o t y p e a s e i t h e r p a r e n t ")$

d) $P (" P h e n o t y p e a s n e i t h e r p a r e n t ")$

$(i i)$ a) $P (" G e n o t y p e a s t h e f i r s t p a r e n t ")$

b) $P (" G e n o t y p e a s t h e s e c o n d p a r e n t ")$

c) $P (" G e n o t y p e a s e i t h e r p a r e n t ")$

d) $P (" G e n o t y p e a s n e i t h e r p a r e n t ")$

Mutually Exclusive

failed to have same genotype or phenotype they're mutually exclusive,

$P (" P h e n o t y p e a s e i t h e r p a r e n t ") = P (" P h e n o t y p e a s t h e f i r s t p a r e n t ") + P (" P h e n o t y p e a s t h e S e c o n d p a r e n t ")$

The probability value is,

$P (" P h e n o t y p e a s n e i t h e r p a r e n t ") = 1 - P (" P h e n o t y p e a s e i t h e r p a r e n t ")$

To each pair of alleles, estimate the likelihood from each gender within the offspring:

$P (" A A ") = P (A_{1} A_{2})$

$\begin{aligned} = P (A_{1}) P (A_{2}) \end{aligned}$

localid="1649420710049" $= \frac{1}{2} 脳 0$

$= 0$

$\begin{aligned} P (" a a^{鈥测赌�}) = P (A_{1}^{c} A_{2}^{c}) \\ = P (A_{1}^{c}) P (A_{2}^{c}) \end{aligned}$

$= \frac{1}{2} 脳 1$

$= \frac{1}{2}$

$P (" B B ") = \frac{1}{4} P (" B b ") = \frac{1}{2} P (" b b ") = \frac{1}{4}$

$P (" C C ") = 0 P (" C c ") = \frac{1}{2} P (" c c ") = \frac{1}{2} P (" D D ") = \frac{1}{4} P (" D d ") = \frac{1}{2} P (" d d ") = \frac{1}{4} P (" E E ") = 0 P (" E e ") = \frac{1}{2} P (" e e ") = \frac{1}{2}$

Phenotype (part a and b)

Calculation of

$(i) (a)$ $P (" P h e n o t y e a s t h e f i r s t p a r e n t ")$

$= P [(" A a " 鈭� " A A^{鈥测赌�}) (" B b " 鈭� " B B ") (" C c " 鈭� " C C ") (" D d " 鈭� " D D ") (" E e^{" 鈥�} 鈭� " E E ")]$

$= P (" A a " 鈭� " A A ") P (" B b " 鈭� " B B ") P (" C c " 鈭� " C C ") P (" D d " 鈭� " D D ") P (" E e " 鈭� " E E ")$

$\begin{aligned} = [P (" A a^{鈥测赌�}) + P (" A A^{鈥测赌�})] [P (" B b ") + P (" B B ")] [P (" C c ") + P (" C C^{鈥测赌�})] \\ [P (" D d^{鈥测赌�}) + P (" D D^{鈥测赌�})] [P (" E e^{鈥测赌�}) + P (" E E ")] \end{aligned}$

$= (\frac{1}{2} + 0) (\frac{1}{2} + \frac{1}{4}) (\frac{1}{2} + 0) (\frac{1}{2} + \frac{1}{4}) (\frac{1}{2} + 0)$

$= \frac{9}{128}$

localid="1649422377779" $(i) b) P (" P h e n o t y p e a s t h e s e c o n d p a r e n t ")$

$= P [(" a a^{鈥测赌�}) (" B b " 鈭� " B B ") (" c c ") (" D d " 鈭� " D D ") (" e e ")]$

$= P (" a a ") P (" B b " 鈭� " B B ") P (" c c ") P (" D d " 鈭� " D D ") P (" e e ")$

$\begin{aligned} = [P {(" a a ")}^{}] [P (" B b ") + P (" B B ")] [P (" cc ")] \\ [P (" D d^{鈥测赌�}) + P (" D D^{鈥测赌�})] [P (" e e ")] \end{aligned}$

$= (\frac{1}{2}) (\frac{1}{2} + \frac{1}{4}) (\frac{1}{2}) (\frac{1}{2} + \frac{1}{4}) (\frac{1}{2}) = (\frac{9}{128})$

localid="1649422361835" $(i) c) P (" P h e n o t y p e a s e i t h e r p a r e n t ") = P (" P h e n o t y p e a s t h e f i r s t p a r e n t ") + P (" P h e n o t y p e a s t h e s e c o n d p a r e n t ")$

$= \frac{9}{128} + \frac{9}{128}$

$= \frac{9}{64}$

$P (" p h e n o t y p e as neither parent") = 1 鈭� P (" p h e n o t y p e as either parent")$

$= 1 - \frac{9}{64} = \frac{55}{64}$

Genotype (part c and d)

$I I) a) P (" g e e n o t y p e as the first parent")$

$= P (" A a^{鈥测赌�}) (" B b ") (" C c ") (" D d ") (" E e^{鈥测赌�})$

$= P (" A a ") P (" B b ") P (" C c ") P (" D d ") P (" E e ")$

$= \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} = \frac{1}{32}$

$b) P (" g e n o t y p e as the second parent")$

$= P [(" a a ") (" B b ") (" c c ") (" D d ") (" e e ")]$

$= P (" a a ") P (" B b ") P (" c c ") P (" D d ") P (" e e ")$

$= \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} 脳 \frac{1}{2} = \frac{1}{32}$

$c) P ("genotype as either parent")=$ $P (" g e n o t y p e a s t h e f i r s t p a r e n t ") + (" g e n o t y p e a s t h e s e c o n d p a r e n t ")$

$= \frac{1}{32} + \frac{1}{32} = \frac{1}{16}$

$d) P (" g e n o t y p e as neither parent") = 1 鈭� P (" g e n o t y p e as either parent")$

$= 1 - \frac{1}{16} = \frac{15}{16}$

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Short Answer

Step by step solution

Given Data

Mutually Exclusive

Phenotype (part a and b)

Genotype (part c and d)

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