ok i still need help with my punnett squares!!?

Question

this is an example of one of the questions that i dont understand can anyone help me through this problem? ok it tells you that a blue eyed man heterozygous for cystic fibrosis (bb, Cc) and a heterozygous brown eyed women homozygous dominant for cystic fibrosis (Bb, CC). then you have to write all of the possible allele combinations of the offspring. i dont know how to do it can someone help? PLEASE I NEED HELP ON THIS FAST!!! thank you so much

Answer 1

bbCc x BbCC.

Mr bbCc can make two types of gametes; bC and bc. Mrs BbCC can make two types of gametes also; BC and bC.

bC + BC = BbCC
bC + bC = bbCC
bc + BC = BbCc
bc + bC = bbCc

The Punnett Square is hard to make in text form, but I'll try:
___.|__BC__|__bC__|
bC_|_BbCC_|_bbCC_|
bc_|_BbCc__|_bbCc_|

Answer 2

Man = bbCc
His possible gametes are bC bc bC bc.
Match the first b with each c, then match the second b with each c. You might think of this as the FOIL method, just like you do in math - first, outer, inner, last.

Woman = BbCC
Her possible gametes are BC BC bC bC

Put his gametes on the side of the Punnett square.
Put her gametes across the top.

In the first box, you have bC to the left and BC at the top.
That makes an offspring that is bBCC (brown eyes, cystic fibrosis).

Finish the Punnett square and tally up the phenotypes, just like telling the score. If you do this one correctly, you will count up 8 brown-eyed, cystic fibrosis: 8 blue-eyed, cystic fibrosis. This ratio should be reduced (just like a fraction) to 1 brown-eyed, cystic fibrosis: 1 blue-eyed, cystic fibrosis.

I added an answer to your previous question before I saw that you had posted a second question. You might read it to get a little more information. I'll check back on this question in a few minutes to see if you have added any more details.

Answer 3

not totally sure but heterozygonous and homozygonous dominant won't ever make a homozygonous recessive so there are only 2 possible combinations for cystic fibrosis CC and Cc and 2 for brown eyes Bb and bb because a hetero and a homozygonous recessive won't ever make a homozygonous dominant. Four all together Cc CC
__________________

bb bbCc bbCC

Bb BbCc BbCC

Answer 4

Create a 4x4 chart and place the bbCc on the top and the BbCC on the side. Start combining to find your answer.

Answer 5

I can try and help. First you draw a square with four sections. Then you put bb on top side, and Bb on the right side. Then you kind of just multiply the two together, so you'd get:
_b__b_
B |Bb | Bb| Possible combinations: Bb,bb.
----------
b |bb | bb |
----------
Then you do the same for Cc, CC.

Answer 6

Set up your Punet Squares side-by-side for each group of allelles:

First Trait PS

____b____b
B___Bb___Bb
b___bb____bb

Second Trait PS

_____C____c
C____CC___Cc
C____CC___Cc

Now, Take the combinations of each and do it again:

____CC_____Cc_____CC_____Cc
Bb__BbCC___BbCc___BbCC__BbCc
Bb__BbCC___BbCc___BbCC__BbCc
bb__bbCC___bbCc____bbCC__bbCc
bb__bbCC___bbCc____bbCC__bbCc

(The underlines are to keep the chart in order when it prints . . . ANSWERS screws up with spaces).

GENOTYPES: BbCC, bbCC, BbCc, bbCC, bbCc