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In humans phenylketonuria and Lactose Intolerance are both autosomal recessive traits. If 2 individuals who are heterozygous for both traits marry and have children, what is the chance of having a child with PKU and Lactose Intolerance?

I think it is 1/16 but I'm not really sure.. .Is this correct?

Another problem I'm having difficulty with is...
If a heterozygote for PKU and Lactose Intolerance marries someone with both traits, what is the chance of having a child with both traits?

I know the cross would be PpLl X ppll but I don't know how to work the problem out.
I used
P= non carrier of PKU
p=PKU carrier
L=non Lactose intolerant
l=lactose inolerant gene

2006-11-09 15:42:00 · 2 answers · asked by totallyclueless 1 in Science & Mathematics Biology

2 answers

First you have to make out your alleles. Your first parent PpLl will produce spermatozoids that can be either PL Pl pL pl. As for the mother, she is also heterozygotous for both traits, so she is also PpLl and will produce eggs that are either PL Pl pL or pl. The you make the Punnett square with all those four colomns and four rows. You should get something like
___|_PL__|_Pl_|_pL_|_pl__
PL_|_PPLL|_PPLl_|_PpLL_|_PpLl_
Pl_|PPLl|_PPll_|_PpLl_|_Ppll_
pL_|PpLL_|_PpLl_|_ppLL_|_ppLl_
pl_|PpLl_|_Ppll_|_ppLl_|_ppll_

OKAY NOW...
PPLL means healthy all the way
ppll means has all the diseases
In between they are heterozygotous for one or both disease.

Since you have 16 possibilities, then yes your 1/16 answer is correct for the probabilities of having both diseases

can you do the square for your second problem :-)

2006-11-09 16:08:48 · answer #1 · answered by kihela 3 · 0 0

ok, here is extremely a run down of blood varieties first: a style blood might have A antigens and bring anti-B antibodies. a style blood phenotype is generally represented as AA or Ai genotype. B style blood might have B antigens and bring anti-A antibodies. B style blood phenotype is generally represented as BB or Bi genotype AB style blood might have the two A and B antigens and could produce NO antibodies. AB style blood phenotype is generally represented through fact the AB genotype. O style blood will have no antigens and could produce the two anti-A and anti-B antibodies. O style blood phenotype is generally represented as ii phenotype. (so it has 2 recessive ii alleles) so which you have 2 mothers and fathers who're heterozygous for a style blood phenotype, so which you have the Ai genotype for the two mothers and fathers. Ai x Ai will provide the potential for AA, Ai, and ii offspring. yet! The ratio of those offspring isn't equivalent. that is incredibly useful in case you utilize a punnet sq. while you're new to figuring out genotypes of offspring, yet what you additionally can do is harm down the Ai x Ai. So enable's artwork with the A's first. If we x them mutually, we get one achieveable AA. Then artwork with them one after the other. the 1st parent's A can x the 2d parent's i, and the 2d parent's A can x the 1st parent's i. so which you have 2 achieveable Ai's. Then we can x the two i's mutually to get one achieveable ii. So of the 4 opportunities - a million AA, 2 Ai, and a million ii = 25% possibility of AA, 50% possibility of Ai, and 25% possibility of ii. you like the O style blood threat/possibility, so which you like the ii proportion. So that is 25% possibility that 2 mothers and fathers heterozygous for style A blood might have a toddler with style O blood.

2016-11-23 13:40:52 · answer #2 · answered by ? 4 · 0 0

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