Genetics problem

[koojo]

"Three genes (A, B, C) are all on the same chromosome. A is 10 map (cM) units to the left of B, and C is 20 map units to the right of B. A triple heterozygote (ABC/abc) is testcrossed to a tripple recessive (abc/abc). What will be the frequency of each gamete?"

 

This is what I figured so far:

 

ABC abc

x

abc abc

 

Gametes:

ABC (parental)

abc (parental)

Abc (single recombination)

aBC (single recomb)

AbC (double recomb)

aBc (double recomb)

ABc (single recomb)

abC (single recomb)

 

If the distance between A and B is 10, and B and C is 20, then if I were to make a theoretical TOTAL population be 1000, then I KNOW that

 

Abc, aBC, AbC, aBc have an overall population size of 100, since you add up the number of offspring that have these gamets and divide by total population size (1000).

 

AbC, ABc, ABc, abC have an overall population size of 200.

 

Therefore, I know the parentals have an overall population of 800.

 

KNOWING THIS, HOW DO I FIND THE FREQUENCY OF EACH GAMETE?

[koojo]

So no one knows genetics?

[Dak]

this looks suspiciously like a uni assighnment

 

ignoring the last bit 'cos it made no sence and plowing ahead, im assuming that your asking stuff about how to work out the offspring ratio for two parents with reguards to three linked (and thus non-mendelian) genes?

 

ok, to start off -- do you know what MUs mean with reguards to the chance of a cross-over occouring inbetween the two gene loci?

 

basically, 1 MU = 1% chance of a cross-over occouring betwix the two loci; or in other words, a 1% chance that the two genes will be inherited seperately (and thus 99% chance they will be inherited together). (thats what makes the genes linked, btw)

 

so, from that you can work out the persentage chance of any given zygote being spawned from the hetrozygouse parent.

 

eg,

 

A-10MU-B--20MU--C

a-10MU-b--20MU--c

 

from that, the possible zygotes are:

 

ABC

ABc

AbC

aBC

aBc

abC

Abc

abc

 

take the first one: ABC. that relys on no cross overs occouring atall, to produce the chromosome:

 

A-B-C

a-b-c

and from that, there is a 50% chance of getting an ABC zygote, and a 50% chance of an abc zygote.

 

so the probability of getting an ABC zygote = the probability of a cross over not occouring/2 = (probability of no cross over twix A & B X probability of no cross over between B & C)/2

 

P(no cross A&B) = 90% (as they are 10MU apart, the prob of a cross = 10% so the prob of no cross = 90%)

 

P(no cross B&C) = 80% (similar reasons)

 

so P(ABC) = (P(no cross A&B) X P(no cross B&C))/2

 

=(90% X 80%)/2

 

= 72%/2

 

=36%

 

So now we have

 

ABC <-- 36%

ABc

AbC

aBC

aBc

abC

Abc

abc

 

the rest can be worked out in a similar fashion, eg:

 

Prob(ABc) = (prob(no cross over between A and B) X prob(a cross over between B & C))/2

 

do that for all of them, and you have the zygote frequency from the hetrozygouse parent. the frequency from the homozygote will be easy:

 

abc <-- 100%

 

simply change the % chance of occourance into a frequency, and bobs yer unkle

 

now, just do a normal cross and incorporate the zygote frequency into the mating frequency, (eg, ABCand bobs yer unkle -- enough number crunching to put a smile on daves face later, and youll have your answre.

[Dak]

bump! made a mistake, which iv now corrected.

 

the added bits are in blue.