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Phenylketonuria and Galactosemia In humans there is a disease called Phenylketonuria (PKU) which is caused by a recessive allele. People with this disease have a defective enzyme and cannot break down the amino acid phenylalanine. This disease can result in mental retardation or death. Let “E” represent the normal enzyme, and “e” the defective enxyme. Also in humans is a condition called galactose intolerance or galactosemia, which is also caused by a recessive allele. Let “G” represent the normal allele for galactose digestion and “g” represent the allele for galactose intolerance. In both diseases, normal dominates over recessive. Complete a Punnett square for two adults are heterozygous for both traits What is the probability of having a child without either disease? What are the chances of having a child that has only PKU, not galactosemia? What are the chances of having a child with both diseases?

          Phenylketonuria and Galactosemia
In humans there is a disease called Phenylketonuria (PKU) which
is caused by a recessive allele.  People with this disease
have a defective enzyme and cannot break down the amino acid
phenylalanine.  This disease can result in mental retardation
or death.  Let “E” represent the normal enzyme, and “e” the
defective enxyme. 
Also in humans is a condition called
galactose intolerance or galactosemia, which is also caused by a
recessive allele.  Let “G” represent the normal allele for
galactose digestion and “g” represent the allele for galactose
intolerance.  In both diseases, normal dominates over
recessive. 
Complete a Punnett square for two adults are heterozygous for
both traits
What is the probability of having a child without either
disease?
What are the chances of having a child that has only PKU, not
galactosemia?
What are the chances of having a child with both diseases?

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Added by Nathaniel S.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

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Solved by Expert Shaiju T

06/17/2022

Step 1

- Phenylketonuria (PKU) is controlled by alleles E (normal enzyme) and e (defective enzyme). E is dominant over e. - Galactosemia is controlled by alleles G (normal digestion) and g (galactose intolerance). G is dominant over g. - Both parents are heterozyous for Show more…

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Phenylketonuria and Galactosemia In humans there is a disease called Phenylketonuria (PKU) which is caused by a recessive allele. People with this disease have a defective enzyme and cannot break down the amino acid phenylalanine. This disease can result in mental retardation or death. Let “E” represent the normal enzyme, and “e” the defective enxyme. Also in humans is a condition called galactose intolerance or galactosemia, which is also caused by a recessive allele. Let “G” represent the normal allele for galactose digestion and “g” represent the allele for galactose intolerance. In both diseases, normal dominates over recessive. Complete a Punnett square for two adults are heterozygous for both traits What is the probability of having a child without either disease? What are the chances of having a child that has only PKU, not galactosemia? What are the chances of having a child with both diseases?

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Transcript

00:01 Hello everyone for the given question asks.

00:02 In humans there is a disease called phenyl ketone urea pqau which is caused by recessive allele.

00:07 People with these disease have a defective enzyme and cannot break down the amino acid phenylalanine.

00:13 This disease can result in mental retardation or death.

00:17 Let capitally represent a normal enzyme and small e the defective enzyme.

00:21 Also in humans is a condition called lactose intolerance or galactosemia which is also caused by recessive gene.

00:28 Let capital g represent a normal allele for galactose indigestion and small g for galactose intolerance.

00:38 In both alleles, normal dominates over recessive, complete a pennant square for two adults who are hetrozygous for both rates.

00:46 What is the probability of child having without either disease? what is the chances of a child having p .k.

00:57 But not galactosemia and the chances of having child.

01:00 With both the disease.

01:01 So here we have been given a question of thai hybrid cross.

01:07 Now first let us list the parents.

01:10 So parents we have both heterogeneous for both trades.

01:14 So the parents have to be capital e small e, capital g small g crossed with capital e small e.

01:24 Now we have to make gametes...

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