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fodog414 fodog414
wrote...
Posts: 279
11 years ago
Genes A, B, and C are linked on a chromosome and found in the order A-B-C. Genes A and B recombine with a frequency of 10%, and genes B and C recombine at a frequency of 21%. For the cross a + b + c /abc + × abc / abc, predict the frequency of progeny. Assume interference is zero.

Part A
Predict the frequency of a+b+c progeny.
Enter your answer to four decimal places (example 0.2356 or 0.2300).

Part B
Predict the frequency of abc+ progeny.

Part C
Predict the frequency of a+ bc+ progeny.

Part D
Predict the frequency of ab+c progeny.

Part E
Predict the frequency of a+b+c+ progeny.

   
Part A
Predict the frequency of a+b+c progeny.
Enter your answer to four decimal places (example 0.2356 or 0.2300).

Part B
Predict the frequency of abc+ progeny.

Part C
Predict the frequency of a+ bc+ progeny.

Part D
Predict the frequency of ab+c progeny.

Part E
Predict the frequency of a+b+c+ progeny.

Part F
Predict the frequency of abc progeny.

Part G
Predict the frequency of a+bc progeny.

Part H
Predict the frequency of ab+c+ progeny.


If anyone can explain one or a few of the problems I think I can get the rest, or all of them with an explanation would be fantastic!
 Slight Smile
Thanks Fodog
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wrote...
Donated
Valued Member
11 years ago
Here is the breakdown:
Without recombination, linked genes would pan out like this:
a + b + c /abc+ × abc / abc
50%
a+b+c/abc
and 50%
abc+/abc

But since that's not the case (i.e. you have recombination), here's how that's gonna go:
to get a+b+c:
You need 0 recombination
(1-.10)(1-.21) = .711
So that'll happen .711 of the time.
That parent would donate that 1/2 of the time, so 0.3555
duddy,  aravartanyan
Pretty fly for a SciGuy
fodog414 Author
wrote...
11 years ago
Alright i'm following you sort of on how you did part A which is Fantastic so thank you!

Im still struggling tho I was thinking for Part B you would only have (1-.21)/2=.395 which is wrong..


If you could explain part B, and say C and E I think I can get it, possibly I won't bother you again? Confounded Face

you don't have to explain this but I think it might make more sense if you did A and B recombine with a

frequency of 10% so does this mean that a+ and b+ =10% and or a and b+=10%

also if B and C recombine to have a frequency of 21% is that b and c= 21% or just b+ and c+=21%

this is what im getting confused on all those kind of variations

Thanks again for any help
wrote...
Donated
Valued Member
11 years ago
I attached a recombination chart.  It will help you understand all this.

Part B
Predict the frequency of abc+ progeny.
Well since this is the other possibility for parent 1, it would be the same as part A.  You would require 0 recombination, with the calculations the same as Part A.  Again, 0.3555

Part C
Predict the frequency of a+ bc+ progeny.
This would require a crossover between A and B (since in the parent, the A and B are the same but now we want them different) and no crossover in B to C:
(.10)*(1-.21) = 0.0790
Again, however, this would only be donated ½ of the time, so the answer is 0.0395

Part D
Predict the frequency of ab+c progeny.
This is the other single recombinant product as part C, and will yield the same answer (a single recombination between A and B and none between B and C): 0.0395
Part E
Predict the frequency of a+b+c+ progeny.
This could be achieved with no recombination between A-B and a recombination between B-C:
(1-.1)(.21) = .189 * ½ = .0945
Part F
Predict the frequency of abc progeny.
Same recombination event as part E
(1-.1)(.21) = .189 * ½ = .0945
Part G
Predict the frequency of a+bc progeny.
Double recombination would require BOTH recombination events to occur.  By the product rule:
(0.1)(0.21) = .021 * ½ = .0105
Part H
Predict the frequency of ab+c+ progeny.
The other double recombinant… same as Part G:
(0.1)(0.21) = .021 * ½ = .0105
 Attached file 
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potentialpod,  xgothicmonsterx
Pretty fly for a SciGuy
fodog414 Author
wrote...
11 years ago Edited: 11 years ago, fodog414
Oh you are fantastic thank YOU!!

They were all correct expect for Part C << not sure why

It's really appreciated thank you again Doctor-2-B

Cheers
Fodog
Post Merge: 11 years ago

Oh the problem with part C was that it was still only being donated at 1/2 so the answer is 0.0395
wrote...
Donated
Valued Member
11 years ago
So all the answers were correct?

If so, pretty awesome.  Just took that final - no grade yet.  Definitely a confidence boost. Slight Smile
Pretty fly for a SciGuy
fodog414 Author
wrote...
11 years ago
Yeah you were solid!! i'm kind of getting confused on why some of them are 1 minus I was thinking if it was a+ to b+ its a 1 minus or b+ to c+ also 1 minus but looking at it thats not the case. How are you distinguishing what you are 1 minus? Is it whenever there is no crossover?
wrote...
11 years ago
Thank you for this! It's really helpful.
wrote...
Donated
Valued Member
11 years ago
@fodog
Realizing I never answered that.  It's like the opposite of the event occurring.

Eg.
What are the odds of NOT rolling a 4 on a die.  P(rolling a 4) = 1/6, so P(not rolling a 4) = 1 - 1/6 = 5/6

ie P(E) + P(not E) = 1 ALWAYS
Pretty fly for a SciGuy
wrote...
11 years ago
@doctor-2-B

Thanks Doc for all the help in my genetics class!! Slight Smile

I'm taking a break to finish all my generals I'll be back in bio next spring for more questions no doubt Smiling Face with Glasses
wrote...
10 years ago
Thank you so much for the detailed answer! I was so confused!
wrote...
4 years ago
Thank you
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