Question

About 70% of Americans perceive a bitter taste from the chemical phenylthiocarbamide (PTC). The ability to taste this chemical results from a dominant allele (T) and not being able to taste PTC is the result of having two recessive alleles (t). Albinism is also a single locus trait with normal pigment being dominant (A) and the lack of pigment being recessive (a). A normally pigmented woman who cannot taste PTC has a father who is an albino taster. She marries a homozygous, normally pigmented man who is a taster but who has a mother that does not taste PTC. What are the genotypes of the possible children (choose all that apply)?

Woodrats are medium sized rodents with lots of interesting behaviors. You may know of them as packrats. Let's assume that the trait of bringing home shiny objects (H) is controlled by a single locus gene and is dominant to the trait of carrying home only dull objects (h). Suppose two heterozygous individuals are crossed. How many of each genotype would be expected if only 4 offspring were produced?

Suppose you have two rose plants, both with pink flowers. You cross the two plants and are surprised to find that, while most of the offspring are pink, some are red and some are white. You decide that you like the red flowers and would like to make more. What cross would you perform to produce the most red flowered plants?

A boy, whose parents and grandparents had normal vision, is color-blind. What are the genotypes for his mother and his maternal grandparents? Use XB for the dominant normal condition and Xb for the recessive, color-blind phenotype

In humans, sickle cell anemia is a disease caused by a mutation of a single locus gene which codes for an important blood protein. The allele for the normal protein (S) is dominant to that for the one causing sickle cell anemia. What gametes would be produced by individuals resulting from a SS x ss cross?

A man with group A blood marries a woman with group B blood.

a-What are the genotypes of these individuals?

b-What blood groups and frequencies would you expect in offspring from this marriage?


Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?


a- all three of their children will be of normal phenotype

b- one or more of the three children will have the disease

c. all three children will have the disease

d. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have a child, if the child will have sickle-cell anemia.


A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait.
What fraction of this couple's children would be expected to have extra digits?

Answer 1

1st Question (About 70% of Americans......)

Answer:

The possible genotypes of the children are-

1) Homozygous normal pigmentation (AA), and heterozygous PTC taster (Tt)

2) Homozygous normal pigmentation (AA), and homozygous PTC non-taster (tt)

3) Heterozygous normal pigmentation (Aa), and heterozygous PTC taster (Tt)

4) Heterozygous normal pigmentation (Aa), and homozygous PTC non-taster (tt)

Step-by-step explanation:

From the information given, the woman who had normal pigmentation, but couldn't taste PTC, with an albino father who could taste PTC, would have a hetrozygous allele for pigmentation (Aa), and homozygous for the non-tasting PTC ability (tt). This would likely be because her father is recessive for pigmentation (aa) and heterozygous for the ability to taste PTC (Tt), while her mother would most likely be a normal pigmented woman, without the ability to taste PTC.

The man, however, would likely possess homozygous alleles for pigmentation (AA), with heterozygous alleles for tasting PTC (Tt). This would be as a result of his normal pigmented mother who could not taste PTC.

From the above statement, the parent alleles would be-

Mother- A a t t x Father- A A T t

After crossing, the children's allele would be-

A A T t, A A t t, A a T t, A a t t

Each would possess a percentage possibility of 50%

2nd Question (Wood rats are medium sized.....)

Answer:

HH- 1 (One homozygous allele for liking shiny objects)

Hh- 2 (Two heterozygous alleles for liking shiny objects)

hh- 1 (One homozygous allele for liking dull objects)

This could further be concluded that, there is a good chance, that 3 of the offspring would be attracted to shiny objects, while one would be attracted to dull objects.

Step-by-step explanation:

After crossing the percentage possibilities are-

HH- 25%,

Hh- 50%

hh- 25%

when mutplied by 4 (Number of offsprings) and divided by 100, you get the following-

HH- 1

Hh- 2

hh- 1

3rd Question( Suppose you have two......)

Answer:

The colored flower cross that would produce the most red flowers, would be crossing two red flowers.

4th Question (A boy, whose parents.....)

Answer:

The mother's genotype would be XB Xb (Heterozygous normal vision)

Maternal grandparents genotype would be XB Xb for both grandparents (Heterozygous normal vision), or XB XB, XB Xb, for one each.

Step-by-step explanation:

Since both grandparents and parents had normal vision, the only way for the color blind trait to have been passed across the 2 generation to the child, is for the parents to be heterozygous dominant.

5th Question (In humans, sickle cell.....)

Answer:

The gamete that would be only produced, would be Ss (Heterozygous normal protein)

Step-by-step explanation:

The dominant gene, would most definitely inhibit the expression of the recessive gene, making all offspring just carriers.

6th Question

Answer:

a) Genotype of the man- AA or AO

Genotype of the woman- BB or BO

b) Blood group expected would be AB, with a frequency of 100%

7th Question (Phenylketonuria...)

Answer:

a) 1/3

b)1/3

c)1/3

d) 2/3

8th Question (Karen and Steve.....)

Answer:

The possibility that the next child would have the disease is 25%

Step-by-step explanation:

This is because, from the statement, it can be concluded that both parents are carriers of the disease, thus, providing a percentage possibility of a sickle cell child at 25%, a normal child at 25%, and carriers at 50%.

9th Question (A man has six fingers.....)

Answer:

Around 50% or 1/2 (half of the couples children would be expected to have extra digits.

Step-by-step explanation:

Since, the man had normal digit children, and considering the dominance of the extra digit trait, we can conclude that the man is heterozygous dominant. Thus, if crossed with a homozygous recessive woman, would provide a 50% possibility of having hetrozygous dominant children, and 50% possibility of having homozygous recessive children.