Question 1

What is the probability that two lizards heterozygous for stripes on their tails (Ss) will produce offspring homozygous for no stripes (ss)?

Accepted Answer

When two heterozygous lizards (Ss) mate, the possible genotypes for their offspring are SS, Ss, Ss, and ss. The probability of producing an offspring with the genotype ss is 1 out of 4, or 1/4.

Question 2

What are the possible genotypes of a female child from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics?

Accepted Answer

Hemophilia is a sex-linked recessive trait. A heterozygous woman (X^HX^h) and a normal man (X^HY) can produce a female child with either the genotype X^HX^H (normal) or X^HX^h (carrier).

Question 3

In genetics, map units express the distance between:

Accepted Answer

Map units, also known as centimorgans, express the distance between two loci on a chromosome. This unit of measurement is used in genetic mapping to determine the relative location of genes on a chromosome.

Question 4

A brown-eyed couple already has a child with blue eyes. What is the probability that their next child will have blue eyes, assuming that the brown eye allele is dominant and the blue eye allele is recessive?

Accepted Answer

If both parents are heterozygous (Bb) for brown eyes, there's a 1/4 chance for a child to inherit two recessive blue eye alleles (bb).

Question 5

The sex of a human is determined by the:

Accepted Answer

The answer of The sex of a human is determined...

Question 6

Color-blindness is more common in males than in females because:

Accepted Answer

Males have one X and one Y chromosome, so inheriting one recessive color blindness allele on the X chromosome from their mother will result in color blindness. Females have two X chromosomes, so they would need two copies of the recessive allele to express the trait.

Question 7

A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). Out of 800 total F₁ progeny, 296 are normal gray, 328 are vestigial black, 99 are normal black, and 77 are vestigial gray. What is the best explanation for this result?

Accepted Answer

The fact that some of the progeny have combinations of both wing length and body color traits (i.e. normal gray and vestigial black) suggests that the loci for these traits are on the same chromosome and undergo recombination. The specific ratios of the different phenotypes observed suggest that the genes are linked, but not completely linked (i.e. there is some degree of crossing-over between them). This means that Mendel's principles of segregation and independent assortment do not fully account for the observed results.

Question 8

In humans, assume that the allele for brown eyes is dominant and the allele for blue eyes is recessive. If two brown-eyed individuals have a child with blue eyes, that means:

Accepted Answer

For a child to have blue eyes, both parents must carry the recessive blue eye allele, meaning they are heterozygous (Bb).

Question 9

All calico cats are female because:

Accepted Answer

The calico phenotype is caused by random X chromosome inactivation, where one X chromosome in each cell of a female cat is randomly inactivated and becomes a Barr body. This results in some cells expressing one X chromosome while others express the other X chromosome, leading to the black and orange patches characteristic of calico cats. Since males only have one X chromosome, they cannot have the calico phenotype. Thus, all calico cats are female.

Question 10

If a couple is planning on having two children, what is the probability that both will be male?

Accepted Answer

The probability of having a male or female child is 1/2 each. The probability of having two males would be (1/2) x (1/2) = 1/4.

Question 11

A couple who does not have cystic fibrosis but are both heterozygous has already had three girls with cystic fibrosis, and are hoping to have a normal child for their fourth. What are the chances that the fourth child will be a normal male?

Accepted Answer

The parents are both heterozygous for cystic fibrosis, so each has a 50% chance of passing on the defective gene. The child will be normal if it inherits two normal genes, one from each parent. The probability of inheriting a normal gene from the mother is 1/2, and the probability of inheriting a normal gene from the father is also 1/2. The probability of inheriting two normal genes is therefore (1/2) * (1/2) = 1/4. The child will be male if it inherits a Y chromosome from the father. The probability of inheriting a Y chromosome is 1/2. The probability of the child being a normal male is therefore (1/4) * (1/2) = 1/8.

Question 12

A Barr body in a mammalian female cell represents a(n):

Accepted Answer

A Barr body in a mammalian female cell represents an inactivated X chromosome. During development, one of the X chromosomes in female cells is randomly inactivated, forming a condensed, inactive Barr body. This process ensures that there is an equal amount of X-linked gene expression in both males and females.

Question 13

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that they will have a child with purple eyes?

Accepted Answer

The homozygous Martian has two alleles for red eyes, and the heterozygous Martian has one allele for red eyes and one for purple. Since red is dominant, all offspring will have at least one red allele, resulting in red eyes.

Question 14

A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). If there are 800 F₁ progeny, the expected phenotypes would be:

Accepted Answer

The heterozygous fruit fly (VvBb) will produce four types of gametes in equal proportions: VB, Vb, vB, and vb. The homozygous fruit fly (vvbb) will produce only one type of gamete: vb. The cross between these two flies will produce the following F1 progeny:- 25% VVBb: normal gray- 25% Vvbb: normal black- 25% vvBb: vestigial gray- 25% vvbb: vestigial blackTherefore, the expected phenotypes would be 200 normal gray, 200 normal black, 200 vestigial gray, and 200 vestigial black.

Question 15

What are the predicted phenotypes of the female children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

Accepted Answer

The woman is heterozygous (X^HX^h) and the man is normal (X^HY). The possible female offspring are X^HX^H (homozygous normal) and X^HX^h (carrier), resulting in a 1:1 ratio of homozygous normal to carrier.

Question 16

Mendel's principle of independent assortment is explained by which metaphase I event?

Accepted Answer

The principle of independent assortment is explained by the random orientation of homologous chromosomes on the metaphase plate during metaphase I, which leads to the independent segregation of different pairs of alleles.

Question 17

What are the predicted phenotypes of the male children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

Accepted Answer

Hemophilia is an X-linked recessive disorder. A heterozygous woman (X^HX^h) and a normal man (X^HY) will produce male children with a 50% chance of having hemophilia (X^hY) and a 50% chance of being normal (X^HY).

Question 18

A brown-eyed couple heterozygous for eye color are planning on having two children. What is the probability that both children will have blue eyes, assuming brown eyes is dominant and blue eyes is recessive?

Accepted Answer

Each child has a 1/4 chance of having blue eyes (bb) if both parents are heterozygous (Bb). The probability of both children having blue eyes is (1/4) * (1/4) = 1/16.

Question 19

If a couple is planning on having three children, what is the probability that only one will be male?

Accepted Answer

To solve this problem, we can use the binomial probability formula: P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials (in this case, 3 children), k is the number of successful outcomes (in this case, 1 male), p is the probability of a successful outcome (in this case, 1/2 for a male), and (n choose k) is the number of ways to choose k successes out of n trials, given by the formula (n choose k) = n!/(k!(n-k)!). Putting all these values in, we get:

P(X=1) = (3 choose 1) * (1/2)^1 * (1-1/2)^(3-1)
        = 3 * 1/2 * 1/4
        = 3/8

Therefore, the probability that only one child will be male is 3/8, so the answer is choice C.

Question 20

The sex of most mammals, birds, and insects is determined by:

Accepted Answer

In most mammals, birds, and insects, sex is determined by specific sex chromosomes (e.g., XX for females and XY for males in mammals).

Quiz 11: The Basic Principles of Heredity

What is the probability that two lizards heterozygous for stripes on their tails (Ss) will produce offspring homozygous for no stripes (ss) ?

What are the possible genotypes of a female child from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics?

In genetics, map units express the distance between:

A brown-eyed couple already has a child with blue eyes. What is the probability that their next child will have blue eyes, assuming that the brown eye allele is dominant and the blue eye allele is recessive?

The sex of a human is determined by the:

Color-blindness is more common in males than in females because:

In humans, assume that the allele for brown eyes is dominant and the allele for blue eyes is recessive. If two brown-eyed individuals have a child with blue eyes, that means:

All calico cats are female because:

If a couple is planning on having two children, what is the probability that both will be male?

A couple who does not have cystic fibrosis but are both heterozygous has already had three girls with cystic fibrosis, and are hoping to have a normal child for their fourth. What are the chances that the fourth child will be a normal male?

A Barr body in a mammalian female cell represents a(n) :

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that they will have a child with purple eyes?

A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb) . If there are 800 F₁ progeny, the expected phenotypes would be:

What are the predicted phenotypes of the female children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

Mendel's principle of independent assortment is explained by which metaphase I event?

What are the predicted phenotypes of the male children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

A brown-eyed couple heterozygous for eye color are planning on having two children. What is the probability that both children will have blue eyes, assuming brown eyes is dominant and blue eyes is recessive?

If a couple is planning on having three children, what is the probability that only one will be male?

The sex of most mammals, birds, and insects is determined by:

Quiz 11: The Basic Principles of Heredity

What is the probability that two lizards heterozygous for stripes on their tails (Ss) will produce offspring homozygous for no stripes (ss) ?

What are the possible genotypes of a female child from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics?

In genetics, map units express the distance between:

A brown-eyed couple already has a child with blue eyes. What is the probability that their next child will have blue eyes, assuming that the brown eye allele is dominant and the blue eye allele is recessive?

The sex of a human is determined by the:

Color-blindness is more common in males than in females because:

In humans, assume that the allele for brown eyes is dominant and the allele for blue eyes is recessive. If two brown-eyed individuals have a child with blue eyes, that means:

All calico cats are female because:

If a couple is planning on having two children, what is the probability that both will be male?

A couple who does not have cystic fibrosis but are both heterozygous has already had three girls with cystic fibrosis, and are hoping to have a normal child for their fourth. What are the chances that the fourth child will be a normal male?

A Barr body in a mammalian female cell represents a(n) :

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that they will have a child with purple eyes?

A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb) . If there are 800 F1 progeny, the expected phenotypes would be:

What are the predicted phenotypes of the female children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

Mendel's principle of independent assortment is explained by which metaphase I event?

What are the predicted phenotypes of the male children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

A brown-eyed couple heterozygous for eye color are planning on having two children. What is the probability that both children will have blue eyes, assuming brown eyes is dominant and blue eyes is recessive?

If a couple is planning on having three children, what is the probability that only one will be male?

The sex of most mammals, birds, and insects is determined by:

A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb) . If there are 800 F₁ progeny, the expected phenotypes would be: