# Scored B/B #58

#### betterfuture

2+ Year Member

How do you do this?

I am stuck. Any help? Thanks!

#### aldol16

2+ Year Member
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.

betterfuture
OP
B

#### betterfuture

2+ Year Member
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.

5+ Year Member

#### aldol16

2+ Year Member
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

2+ Year Member

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.

betterfuture