Anonymous

# Dihybrid crosses? Bleh. Help?

We are working on dihybrid crosses in biology class, and I am just not getting it. Here is what the problem says: In a certain set of lab mice, black hair is dominant to white hair, and coarse hair is dominant to fine hair. Determine the percentage of black, fine-haired offspring that will result if two mice heterozygous for both of these traits are mated. Draw a Punnet Square to support your answer.

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When I did this, I got 2/16 or1/8....is that right?

Update:

I got 3/16 too, nevermind. Thanks for the help. As soon as I can give you the best answer, you'll get it :)

Relevance
• Anonymous

So B is black hair and b is white hair, and C is coarse hair and c is fine hair. The alleles of the mice would be BbCc. So BbCc is broken down to BC Bc bC bc. Multiply these alleles by themselves and you get 16 resulting allele patterns

BC Bc bC bc

BC BBCC BBCc BbCC BbCc

Bc BBCc BBcc BbCc Bbcc

bC BbCC BbCc bbCC bbCc

bc BbCc Bbcc bbCc bbcc

They are:

1 BBCC

2 BBCc

2 BbCC

4 BbCc

1 BBcc

2 Bbcc

1 bbCC

2 bbCc

1 bbcc

So therefore, you have:

9 mice that have black, coarse hair,

3 mice that have black, fine hair,

3 mice that have white, coarse hair,

and 1 mouse with white, fine hair.

Hope that helps

Source(s): Too many years of bio

Sorry, but it does look like you got it...

Heterozygous for both would mean that they each have a dominant and a recessive for each trait.

This is known as a typical Dihybrid Cross. The results of which always follow a phenotype ratio of 9:3:3:1.

Go back and redo your cross. Your answer is one of the four....either 9, 3, 3, or 1 out of 16.

Post again and let me know if you got it....I will watch this for a few minutes to see if you are still working on it.

• Anonymous
4 years ago

Using a Punnet Square you will get the following: 1 RRYY 3 RrYY 3 RrYy 2 RRYy 1 rrYY 2 rrYy 1 rryy 2 Rryy 1 RRyy That equates to: Round Yellow 9/16 Round Green 3/16 Wrinkle Yellow 3/16 Wrinkle Green 1/16 Hope that helps. Good luck

----- BC bC Bc bc

BC BBCC BbCC BBCc BbCc

bC BbCC bbCC BbCc bbCc

Bc BBCc BbCc BBcc Bbcc

bc BbCc bbCc Bbcc bbcc

Black Fine haired off spring are those who have either

BBcc or Bbcc

how many of each combination

1:2 so the total is 3/16

B=Black

b=white

C=corse

c=fine

punnett square

parents both BbCcx BbCc

gametes: BC,Bc,bC,bc for both parents

BC Bc bC bc

BC BBCC BBCc BbCC BbCc

Bc BBCc BBcc BBCc Bbcc

bC BbCC BbCc bbCC bbCc

bc BbCc Bbcc bbCc bbcc

Black and Coarse 9

Black and Fine 3

White and Coarse 3

White and fine 1