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• #16654
nnyeon145
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Based on the information in Experiment II, estimate the minimum genetic map distance between the two genetic loci. (Note: One genetic map unit (m.u.)
gives a recombinant frequency (RF) of 1 percent.)

I scanned the whole passage in the attachment.

This is what I did:
B+/ba x ba/ba (parental strands)

If there is no crossover, B+/ba (black), ba/ba (white)
If there is one crossover (between the sister chromatids), BA/ba (albino), b+/ba (brown)

So in the crossover, since crossover was done in between the two alleles, 100+34/200= 67amu. (wrong)

The back of the book started doing crossover between non sister chromatids. What about the crossover between sister chromatids? what does crossover between nonsister chromatids help with figurig out the location of the alleles?

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• #111755
JackBean
Participant

sister chromatids are identical, since they are product of DNA replication ðŸ˜‰

• #114054
Cat
Participant
quote nnyeon145:

Based on the information in Experiment II, estimate the minimum genetic map distance between the two genetic loci. (Note: One genetic map unit (m.u.)
gives a recombinant frequency (RF) of 1 percent.)

I scanned the whole passage in the attachment.

This is what I did:
B+/ba x ba/ba (parental strands)

If there is no crossover, B+/ba (black), ba/ba (white)
If there is one crossover (between the sister chromatids), BA/ba (albino), b+/ba (brown)

So in the crossover, since crossover was done in between the two alleles, 100+34/200= 67amu. (wrong)

Consider heterozygous parent contribution (second parental allele pair is ba):
Parent allele pairs are B+ (black) and ba (white).
Cross-over allele pairs are Ba (white) and b+ (brown).

As you can see, white can be due to parental and cross-over genotypes. Therefore, you need to discard it from calculations. Thus, you are left with black and brown progeny to consider: (34*100)/(34+66) = 34amu.