Question

- Phenylketonuria (PKU) is a disease that results from a
recessive gene. Two normal parents produce a child with PKU. After
determining the parents genotype and performing a cross, answer to
the following questions:
**[3 pts –**1 A, 1 B, and 1 Punnett square**.**

- What is the probability of that event?

The probability is 1 out of 4 or 25% chance.

- This couple wishes to have a large family and would like to have 5 children. What is the probability that the next 4 children won’t have PKU?

¼ * ¼ * ¼ * ¼ * ¼ = ? (I don't know if this is the direction to be going in)

Answer #1

Phenylketonuria (PKU) is inherited as autosomal recessive disorder

Let, A = Normal allele, a = Allele containing phenylketonuria trait

As both normal parents have a child with PKU, they must be heterozygous for this trait, i.e., Aa genotype. The cross between them is shown in below Punnett square.

Gametes |
A |
a |

A |
AA (Normal child) | Aa (Normal child) |

a |
Aa (Normal child) | aa (Affected child) |

From the above Punnett square, we
find that **probability of occurring phenylketonuria is 1/4
or 25%**.

So, probability of a child that will not develop phenylketonuria is 3/4. Now, born of a child is independent of born of another child.

So, **probability that the
next 4 children won’t have PKU** = 3/4 x 3/4 x 3/4 x 3/4 =
**81/256**

Phenylkeptonuria (PKU) is a disease that results from a
recessive gene and 1/20,000 in the population know to be carriers.
Two unaffected parents produce a child with PKU. (a) What is the
probability that their next child will have PKU? (b) They want to
have a big family and decide to have 4 more children. What is the
probability that at least one child out of next 4 children will not
show PKU?

Phenylketonuria (PKU) is a heritable condition in humans
involving inability to break down the amino acid phenylalanine
because of lack of a certain enzyme. If not diagnosed and treated
very shortly after birth, PKU’s develop severe mental disabilities
and usually do not reproduce. Almost all PKU children, therefore,
are born to parents who are not PKU’s.
A). Is the gene responsible for PKU dominant or recessive?
Explain how you know.
B) The normal (non-PKU) brother of a PKU seeks the...

In humans, PKU (phenylketonuria) is a disease caused by an
enzyme inefficiency at step 1 (diagram above), and AKU
(alkaptonuria) is due to an enzyme inefficiency in step 4. Both
mutant alleles are recessive to the respective wild type allele,
and the genes involved are located on different chromosomes.
A person with PKU marries a person with AKU. What phenotypes do
you expect for their children?
Justify by providing correctly written genotypes for both
parents and potential offspring.

Question 5
Cystic fibrosis (CF) is a recessive gene disease meaning that a
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means that there is a 25% chance of a child having cystic fibrosis
if both parents are CF carriers Hence the number of children with
CF (in a family whose parents are CF carriers) has a binomial
distribution with n = the number of children in the family and p=
0.25
i. In a family...

Question 5
Cystic fibrosis (CF) is a recessive gene disease meaning that a
child has to inherit a defective CF gene from each parent. This
means that there is a 25% chance of a child having cystic fibrosis
if both parents are CF carriers Hence the number of children with
CF (in a family whose parents are CF carriers) has a binomial
distribution with n = the number of children in the family and p=
0.25
i. In a family...

Albinism, a lack of pigmentation in humans, results from an
autosomal recessive gene. Two parents with normal pigmentation have
an albino child. What is the probability that their next two
children will have the same genotype?
a. 3/8
b. 11/16
c. 7/8
d. 13/16
e. 9/16

A male and a female are each heterozygous for both cystic
fibrosis (CF) and phenylketonuria (PKU). Both conditions are
autosomal recessive and the alleles assort independently.^
Need detail explanation for the question C, and I'm not certain,
if the answer is
correct.<--------------------------------------------
I'm pretty sure that answers for A and B are correct.
A.What proportions of the children of this couple will have
neither condition?
My answer 9/16
B. What proportion of the children will have either PKU or
CF?...

Sex-linked crosses
4. Hemophilia is an X-linked disease, meaning that the gene that
causes the disease is found on the X chromosome. Affected males
only have to have one allele to have the disease
(XhY), while affected females have to have two
alleles (XhXh). Imagine that an
affected male has children with an unaffected carrier female.
a. What are the genotypes of the parents in this cross?
b. Draw a Punnett cross to show the offspring of this
couple.
c....

Frank and Julia have learned that they are both carriers for the
recessive disease PKU and also have the rare AB blood type.
What is the probability that they will have a child with PKU and
the rare AB blood type?
a.
3/16
b.
1/8
c.
1/2
d.
1/16
e.
1/4

If two normal (disease free) parents each carry the recessive
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Select one:
25% (1 out of 4)
100% (all children)
75% (3 out of 4)
50% (2 out of 4)
No children
Clear my choice
If Woody Guthrie had children with a women with Huntingtons
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