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hunter4565 hunter4565
wrote...
Posts: 10
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8 years ago
You examine 450 progeny from the cross in Part C and classify them according to phenotype. The results are shown in the table below.

!See Below Image!

Use these data to complete the linkage map shown below.
Drag the blue labels to the blue targets to identify the allele loci indicated by vertical dashes. Note that the apricot allele is already labeled.
Drag two of the white labels to the gray targets to identify the distances between loci. Each white label gives the distance in map units. One map unit, or one centiMorgan (cM), is equivalent to a 1% recombination frequency.

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Doctor-2-BDoctor-2-B
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8 years ago
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RF between AB:
39 + 42 + 5 + 4 = 90 / 450 = 20% RF

RF between AC:
17 + 19 + 39 + 42 = 26% RF

RF between BC:
17 + 19 + 5 + 4 = 45 / 450 = 10% RF

That means that the gene order is:

A----------------------B-------------C
        20 mu           10 mu

a) 20 mu
b) 10 mu
c) blunt
d) crimson
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wrote...
8 years ago
THANK YOU!!
wrote...
7 years ago
 AH/ah Cc × ah/ah cc Give the genotypes of progeny plants in this experiment
wrote...
Donated
Valued Member
7 years ago
This really should be its own post/question/thread/etc. but I couldn't help myself - so I answered it. Slight Smile

P1 gamete possibilities:
AHC
AHc
ahC
ahc

P2 gamete possibilities:
ahc

Progeny possibilities:
AaHhCc
AaHhcc
aahhCc
aahhcc



Pretty fly for a SciGuy
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7 years ago
Thank you so much!
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thank you!!
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This is awesome thanks
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Thank you
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9 months ago
Thank you so much
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Thank you
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Thanks!
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thank you
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thank you
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