aldol16

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Nov 1, 2015
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Construct a Punnet square. Cross KM, Km, kM, km with Km, Km, km, km (these are all the possibilities). The offspring will thus have: KKMm, KKMm, KkMm, KkMm, KKmm, KKmm, Kkmm, Kkmm, kKMm, kKMm, kkMm, kkMm, kKmm, kKmm, kkmm, kkmm.

Now, anybody without K or with M will be deaf. They are bolded below:

KKMm, KKMm, KkMm, KkMm, KKmm, KKmm, Kkmm, Kkmm, kKMm, kKMm, kkMm, kkMm, kKmm, kKmm, kkmm, kkmm.

Now, that's 10 out of 16, or 5/8.
 
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betterfuture

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Feb 16, 2016
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Isn't there like a additive/multiple rule or something - the shortcut method? The punnet square would take up too much time and can get confusing. I don't know if it applies here or not.
 

aldol16

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I think the 16-size Punnet square is the best and most straight-forward way of doing it but whatever seems easiest for you. I didn't have to go fill everything in. I simply crossed out all rows and columns that had a big M and looked to see which crosses would end up with two small k's. The only time-consuming part was writing out the four possible combinations for each parent.
 

laczlacylaci

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Jun 20, 2016
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upload_2016-8-31_9-9-35.png
I usually just do this, because its faster and I'm lazy af.
You know that in the presence of one big M no matter what other alleles look like, that child will be deaf, so you check columns 1 and 3.
You also know in presence of big A, people will hear fine, so aa means deaf, check all the options where aa appears.. (I used A here, because K and k look identical on paper.)

Count up your boxes and divide by 16.
 
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