Genetics-Two heterozygous brown-eyed (Bb)

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In summary, the probability of three out of five children having blue eyes when the parents are heterozygous for brown eyes is 8.78%. This is calculated using the binomial expansion equation with the probability of blue eyes being 1/4 and the probability of brown eyes being 3/4.
  • #1
jena
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Hi,

My Question:

Two heterozygous brown-eyed (Bb) individual have five children.What ist he probability that three will have blue eyes?

Answer:

Would the possibility be zero for all three since blue eyes is recessive

Thank You:smile:
 
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  • #2
Hi,

I looked over the question again and I know for sure this isn't the right answer. I do know that I should use a binomial expansion, so if the

Probability of blue eyes is 1/4 then I would square this and get

P(blue)=(1/4)^2=.0625*100=6.25%

Is this correct?

Thank You
 
  • #3
Why are you squaring the probability?

This one is also a bit trickier because they aren't just asking the probability of 3 having blue eyes, but 3 out of 5. That will change how you do your calculations somewhat.
 
  • #4
You might try to find the probability of 1/5 of the children having blue eyes. Having 5 kids would do what to that probability?
 
  • #5
The probability of having 1 child having blue eyes is 20% right, so to figure the probability of three children having blue eyes would be

P(blue)=(1/5)^3 or .80 %

Is this correct?
 
  • #6
jena said:
The probability of having 1 child having blue eyes is 20% right

No it is not 20%, why do you think that?
 
  • #7
You had it right the first time when you said the probability of one child having blue eyes is 1/4, or 25%. Now try it.
 
  • #8
I thought it was 20% because 1/5 is .20 multiply that by 100 and you get 20%, but I know what I did wrong this time all I have to do is use the binomial equation.

P=((n!)/(x!(n-x)!))*((p^x)(q^(n-x)))

I used the following steps to come up with an answer

Step 1: Calculate the individual probabilities
• P(blue eyes)= p=1/4
• P(brown eyes)=q=3/4

Step 2: Determine the number of events
• n=total number of children=5
• x= number of brown eyed children=3

Step 3: Substitute the values for p, q, x, and n in the binomial expansion equation(like above)

Finally

P=((5!)/(3!(5-3)!))*(((1/4)^3)((3/4)^(5-3)))
P=8.78%

I hope this time around that's right:smile:
 
  • #9
Yes, it is right!
 

1. What is the probability of offspring having brown eyes from two heterozygous brown-eyed parents?

The probability of offspring having brown eyes is 75%, as the dominant trait for brown eyes (B) is present in both parents.

2. Can two heterozygous parents produce offspring with different eye colors?

Yes, it is possible for two heterozygous brown-eyed parents to produce offspring with different eye colors if the other parent carries a different dominant eye color gene, such as blue (b).

3. What is the genotype of the offspring from two heterozygous brown-eyed parents?

The genotype of the offspring is 25% BB, 50% Bb, and 25% bb.

4. How does one determine the phenotype of the offspring from two heterozygous brown-eyed parents?

The phenotype of the offspring can be determined by looking at the dominant and recessive genes present in their genotype. In this case, the dominant trait for brown eyes (B) will be expressed in 75% of the offspring, while the recessive trait for blue eyes (b) will be expressed in 25% of the offspring.

5. Is it possible for two heterozygous brown-eyed parents to produce offspring with blue eyes?

Yes, it is possible for two heterozygous brown-eyed parents to produce offspring with blue eyes if both parents carry the recessive gene for blue eyes (b) and it is passed down to the offspring.

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