An Experiment Was Done to Determine the Linkage Relationship of Three

Question

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The Answer is The data clearly show sex linkage due to disproportionate representation of phenotypic classes between the sexes.
a)

ext { P: }

$The data clearly show sex linkage due to disproportionate representation of phenotypic classes between the sexes. a) ext { P: } \frac { a c b ^ { + } } { a c b ^ { + } } X \quad \frac { a ^ { + } c ^ { + } b } { \mathrm { Y } } b) \mathrm { F } _ { 1 } \quad \frac { a c b ^ { + } } { a ^ { + } c ^ { + } b } X \quad \frac { a c b ^ { + } } { \mathrm { Y } } Loci order must be \underline{ ext{ a c b }} because the most rare F2 classes of 38 and 42 flies each (due to D-CO)are a++ and +bc. c)a-b: Look for recombinants \underline{ ext{a b }} (157 + 59)and \underline{ ext{ a }} + \underline{ ext{ b }} + (163 + 61)Total = 440 RF = 440/1000 = 44% a-c: Look for recombinants \underline{ ext{ a c }} + (157 + 38)and \underline{ ext{ a }} + \underline{ ext{ c }} (163 + 42)Total = 400 RF = 400/1000 = 0.40 or 40% b-c: Look for recombinants \underline{ ext{b c }} (59 + 42)and \underline{ ext{ b }} + \underline{ ext{ c }} + (61 + 38)Total = 200 RF = 200/1000 = 0.20 or 20% d)Expected D-CO = 0.4 * 0.2 *1000 = 80 Observed D-CO = 80;therefore,Interference (I)= 0.The correct answer is I = 0.$

\frac { a c b ^ { + } } { a c b ^ { + } } X

\quad

\frac { a ^ { + } c ^ { + } b } { \mathrm { Y } }

$The data clearly show sex linkage due to disproportionate representation of phenotypic classes between the sexes. a) ext { P: } \frac { a c b ^ { + } } { a c b ^ { + } } X \quad \frac { a ^ { + } c ^ { + } b } { \mathrm { Y } } b) \mathrm { F } _ { 1 } \quad \frac { a c b ^ { + } } { a ^ { + } c ^ { + } b } X \quad \frac { a c b ^ { + } } { \mathrm { Y } } Loci order must be \underline{ ext{ a c b }} because the most rare F2 classes of 38 and 42 flies each (due to D-CO)are a++ and +bc. c)a-b: Look for recombinants \underline{ ext{a b }} (157 + 59)and \underline{ ext{ a }} + \underline{ ext{ b }} + (163 + 61)Total = 440 RF = 440/1000 = 44% a-c: Look for recombinants \underline{ ext{ a c }} + (157 + 38)and \underline{ ext{ a }} + \underline{ ext{ c }} (163 + 42)Total = 400 RF = 400/1000 = 0.40 or 40% b-c: Look for recombinants \underline{ ext{b c }} (59 + 42)and \underline{ ext{ b }} + \underline{ ext{ c }} + (61 + 38)Total = 200 RF = 200/1000 = 0.20 or 20% d)Expected D-CO = 0.4 * 0.2 *1000 = 80 Observed D-CO = 80;therefore,Interference (I)= 0.The correct answer is I = 0.$
b)

\mathrm { F } _ { 1 }

\quad

$The data clearly show sex linkage due to disproportionate representation of phenotypic classes between the sexes. a) ext { P: } \frac { a c b ^ { + } } { a c b ^ { + } } X \quad \frac { a ^ { + } c ^ { + } b } { \mathrm { Y } } b) \mathrm { F } _ { 1 } \quad \frac { a c b ^ { + } } { a ^ { + } c ^ { + } b } X \quad \frac { a c b ^ { + } } { \mathrm { Y } } Loci order must be \underline{ ext{ a c b }} because the most rare F2 classes of 38 and 42 flies each (due to D-CO)are a++ and +bc. c)a-b: Look for recombinants \underline{ ext{a b }} (157 + 59)and \underline{ ext{ a }} + \underline{ ext{ b }} + (163 + 61)Total = 440 RF = 440/1000 = 44% a-c: Look for recombinants \underline{ ext{ a c }} + (157 + 38)and \underline{ ext{ a }} + \underline{ ext{ c }} (163 + 42)Total = 400 RF = 400/1000 = 0.40 or 40% b-c: Look for recombinants \underline{ ext{b c }} (59 + 42)and \underline{ ext{ b }} + \underline{ ext{ c }} + (61 + 38)Total = 200 RF = 200/1000 = 0.20 or 20% d)Expected D-CO = 0.4 * 0.2 *1000 = 80 Observed D-CO = 80;therefore,Interference (I)= 0.The correct answer is I = 0.$

\frac { a c b ^ { + } } { a ^ { + } c ^ { + } b } X

\quad

\frac { a c b ^ { + } } { \mathrm { Y } }

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Loci order must be

\underline{ ext{ a c b }}

because the most rare F₂classes of 38 and 42 flies each (due to D-CO)are a++ and +bc.

c)a-b: Look for recombinants

\underline{ ext{a b }}

(157 + 59)and

\underline{ ext{ a }}

⁺

\underline{ ext{ b }}

⁺ (163 + 61)Total = 440
RF = 440/1000 = 44%

a-c: Look for recombinants

\underline{ ext{ a c }}

⁺ (157 + 38)and

\underline{ ext{ a }}

⁺

\underline{ ext{ c }}

(163 + 42)Total = 400
RF = 400/1000 = 0.40 or 40%

b-c: Look for recombinants

\underline{ ext{b c }}

(59 + 42)and

\underline{ ext{ b }}

⁺

\underline{ ext{ c }}

⁺ (61 + 38)Total = 200
RF = 200/1000 = 0.20 or 20%

d)Expected D-CO = 0.4 * 0.2 *1000 = 80
Observed D-CO = 80;therefore,Interference (I)= 0.The correct answer is
"I = 0."

An Experiment Was Done to Determine the Linkage Relationship of Three

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