mother and father with heterozygous brown eyes. they have 6 children with blue eyes..1)explain!..2)what are the chances that their next child will be brown-eyed?..c) what are the chances that their child will be blue eyed?
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Heterozygous means that the individual has both types of the gene, in this case one blue allele and one brown allele. So, we can call each parent Bb, with the B meaning brown (because it's dominant) and the b meaning blue (because it's recessive).
So, we have a Bb x Bb cross. Because blue eyes are recessive, you need two blue alleles to display the trait (bb). Any other combination (Bb or BB) will result in brown eyes. The genes that the child gets from each parent is random, and therefore it's just by happen-chance that each child received a b from the mother and a b from the father. This means that each child is bb, and displays blue eyes.
Now for the next child. Let's look at the (sad attempt at a) Punnett square I made below:
__| B | b |
----|-----|-----|
B |BB| Bb|
----|-----|-----|
b | Bb| bb|
I'm not sure if you've studied Punnett squares, as not everyone does, so I'll give you a quick explanation. The single B/b's at the top and side are each of the parents. (Father is across the top, mother is down the side.) The 4 combinations in the middle are the possibility for their children.
As you can see, 3 of the combos have an uppercase B. Since this is the brown (dominant) allele, children with any of these 3 genetic combinations will have brown eyes. Only children with the bb combo will have blue eyes.
So now it's just a matter of counting. The chances of the next child being brown-eyed are 3:1, 3 out of 4, or 75%. The chance of the kid being blue-eyed are 1:3, 1 out of 4, or 25%.
Note: Previous children's eye color does not affect that of future children.
So, we have a Bb x Bb cross. Because blue eyes are recessive, you need two blue alleles to display the trait (bb). Any other combination (Bb or BB) will result in brown eyes. The genes that the child gets from each parent is random, and therefore it's just by happen-chance that each child received a b from the mother and a b from the father. This means that each child is bb, and displays blue eyes.
Now for the next child. Let's look at the (sad attempt at a) Punnett square I made below:
__| B | b |
----|-----|-----|
B |BB| Bb|
----|-----|-----|
b | Bb| bb|
I'm not sure if you've studied Punnett squares, as not everyone does, so I'll give you a quick explanation. The single B/b's at the top and side are each of the parents. (Father is across the top, mother is down the side.) The 4 combinations in the middle are the possibility for their children.
As you can see, 3 of the combos have an uppercase B. Since this is the brown (dominant) allele, children with any of these 3 genetic combinations will have brown eyes. Only children with the bb combo will have blue eyes.
So now it's just a matter of counting. The chances of the next child being brown-eyed are 3:1, 3 out of 4, or 75%. The chance of the kid being blue-eyed are 1:3, 1 out of 4, or 25%.
Note: Previous children's eye color does not affect that of future children.
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The probability of any one of their children having blue eyes is 1/4. The probability of 6 out of 6 children having blue eyes is 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 or 1 in 4,096. The probability that their next child will have brown eyes is the same as the probability for each of their other children - 3/4. The probability of their next child having blue eyes is the same as the probability for each of their other children - 1/4.
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75% chance next kid is brown eyed.
25% of blue
heterozygous brown = carries blue eyed gene.
25% of blue
heterozygous brown = carries blue eyed gene.