a. What are the parental phenotypes and what are the recombinant phenotypes? b.

Question

a. What are the parental phenotypes and what are the recombinant phenotypes?

b. Calculate the distances between the 3 traits in map units.

c. Draw a genetic map of these three traits that shows their orientation and distance from each other.

QUESTION 5. A female fruit fly is heterozygous for the following traits, cinnabar eyes, long body, and winged. She is crossed with a homozygous recessive male that has brown eyes, dwarf body, and wingless. They produced the following offspring. (15 pts) Phenotype Number Cinnabar eyes, long body, winged 489 Cinnabar eyes, long body, wingless 202 Cinnabar eyes, dwarf body, winged (double cross-over) 10 Cinnabar eyes, dwarf body, wingless 74 Brown eyes, long body, winged 68 Brown eyes, long body, wingless (double cross-over) 12 Brown eyes, dwarf body, winged 198 Brown eyes, dwarf body, wingless 511 Total 1564 a. What are the parental phenotypes and what are the recombinant phenotypes? (3 pts) Parental: Recombinant: b. Calculate the distances between the 3 traits in map units. (7 pts) c. Draw a genetic map of these 3 traits that shows their orientation and distance from each other. (5 pts) a. What are the parental phenotypes and what are the recombinant phenotypes? b. Calculate the distances between the 3 traits in map units. c. Draw a genetic map of these three traits that shows their orientation and distance from each other.

Explanation / Answer

a) Parental:

cinnabar eye(C), long body(L), winged (W) and brown eye(c), dwarf body(l) , wingless(w)

recombinant: cinnabar eye(C), long body(L), wingless(w)

cinnabar eye(C), dwarf body(l), winged (W)

cinnabar eye(C), dwarf body(l), wingless(w)

brown eye(c), long body(L), winged(W)

brown eye(c), long body(L), wingless(w)

brown eye(c), dwarf body(l), winged(W)

b) Recombinant between cinnabar eye(C), long body(L)

= 10+ 74+68+12 = 164

Recombinant frequency = 164/1564 * 100= 10.48%, so distance between them would be 10.48 map unit

Recombinant between cinnabar eye(C), winged(W)

= 202+74+ 68+198 = 542

Recombinant frequency = 542/1564 *100 = 34.65%, so distance between them would be 34.65 map unit

Recombinant between long body(L), winged(W)

= 202+10+12+198 = 422

Recombinant frequency = 422/1564 *100 = 26.98%, so distance between them would be 26.98 map unit

c) Map

C 10.48   L          26.98            W

Coefficient of coincidence

Number of double cross over = 10+12 = 22

The expected occurrence of double cross over is 0.1048

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