Question 1

In humans, assume straight hairline is recessive to widow's peak. A woman with a straight hairline is married to a man with a widow's peak whose mother had a straight hairline. What are the chances of the couple having a child with a straight hairline?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

The man must be heterozygous (Ww) because he has a widow's peak (dominant trait) but his mother had a straight hairline (recessive trait). The woman is homozygous recessive (ww) for a straight hairline. The possible genotypes for their children are Ww (widow's peak) and ww (straight hairline), giving a 50% chance for a straight hairline.

Question 2

In guinea pigs, B = black, b = brown, S = short hair, s = long hair. A guinea pig heterozygous for both traits is mated with a brown, short-haired guinea pig whose mother had long hair. What phenotypic ratio is expected in the offspring?
A) 9 black long hair, 3 black short hair, 3 brown long hair, 1 brown short hair.
B) 9 black short hair, 6 black long hair, 3 brown long hair, 1 black short hair.
C) 6 black short hair, 1 black long hair, 6 brown short hair, 2 brown long hair.
D) 6 brown short hair, 6 black long hair, 2 brown long hair, 2 black short hair.
E) 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair.

Accepted Answer

The answer of In guinea pigs, B = black, b...

Question 3

If the probability in any pregnancy of having a boy is 50%, then what is the probability of having 3 boys in a row?
A) 3/4
B) 1/4
C) 1/8
D) 1/2
E) 3/16

Accepted Answer

The probability of having a boy or a girl is independent of the previous child's gender. Therefore, the chance of having three boys in a row would be (1/2) x (1/2) x (1/2) = 1/8, or 12.5%. Thus, the correct answer is option C.

Question 4

A blue-eyed, left-handed woman marries a brown-eyed, right-handed man who is heterozygous for both of his traits. If blue eyes and left-handedness are recessive, how many different phenotypes are possible in their children?
A) one
B) two
C) three
D) four
E) five

Accepted Answer

Given that blue eyes and left-handedness are recessive traits, and considering the man is heterozygous for both of his dominant traits (brown eyes and right-handedness), their children could exhibit four different phenotypes: 1) brown-eyed, right-handed; 2) brown-eyed, left-handed; 3) blue-eyed, right-handed; 4) blue-eyed, left-handed.

Question 5

Two parents heterozygous for cystic fibrosis have a child with the disease. The parents are expecting a second child. What is the probability that the second child will also have cystic fibrosis?
A) 100% probability
B) 75% probability
C) 50% probability
D) 25% probability
E) 0% probability

Accepted Answer

Cystic fibrosis is an autosomal recessive disorder. When both parents are heterozygous (carriers), each child has a 25% chance of inheriting two recessive alleles and having the disease.

Question 6

When pairs of factors separate independently of other pairs of factors, this is known as
A) the law of segregation.
B) the law of independent assortment.
C) the law of dominance.
D) multiple alleles.
E) codominance.

Accepted Answer

The law of independent assortment states that pairs of factors separate independently of other pairs of factors during gamete formation. This means that the inheritance of one trait is not dependent on the inheritance of another trait, and they are randomly distributed during meiosis.

Question 7

Albinism is inherited as a recessive trait. An albino woman is married to a man with normal coloring and their first child is an albino. What was the genotype of the man?
A) homozygous dominant
B) heterozygous
C) homozygous recessive
D) incompletely dominant
E) sex-linked recessive

Accepted Answer

Since albinism is a recessive trait, it is possible for the albino woman to be homozygous recessive (aa) and the man to be heterozygous (Aa) for normal coloration. The Punnett square for a cross between an aa and Aa individual shows that there is a 50% chance of their child being albino (aa) and a 50% chance of their child having normal coloring (Aa). The fact that their first child is albino confirms that the man must be carrying one copy of the albino allele. Therefore, his genotype is heterozygous (Aa).

Question 8

Both parents are heterozygous for Tay-Sachs disease (an autosomal recessive disorder). Three children in a row were born with Tay-Sachs disease. What is the chance that a fourth child will have Tay-Sachs disease?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

The chance of a child inheriting Tay-Sachs disease when both parents are heterozygous is 25%, regardless of the outcomes of previous pregnancies.

Question 9

If the genotype of an organism is YySsTt, then yST would represent
A) the genotype of the offspring.
B) a possible phenotype of the offspring.
C) a gamete of the parent.
D) a possible zygote.
E) the P generation.

Accepted Answer

yST represents a gamete of the parent because it is a combination of one allele from each of the three genes (Yy, Ss, Tt) in the parent's genotype. It is not necessarily the genotype or phenotype of any specific offspring or zygote, and it is not a reference to the P generation.

Question 10

Which of the following crosses will yield only homozygous recessive offspring?
A) RRYY × RRYY
B) RRYY × rryy
C) RrYy × rryy
D) rryy × rryy
E) RrYy × RrYy

Accepted Answer

In a cross between two homozygous recessive individuals (rryy × rryy), all of the offspring will be homozygous recessive (rryy).

Question 11

Two parents with widow's peak and short fingers have a child with a continuous hairline and long fingers. What are the genotypes of the two parents? (W = widow's peak; S = short fingers)
A) WWSS × wwss
B) wwss × WWSS
C) WWss × WWss
D) WwSs × WwSs
E) WwSs × wwss

Accepted Answer

- Widow's peak is dominant, so one parent must be heterozygous for the trait and the other parent can either be homozygous recessive or heterozygous. Therefore, the possible genotypes for the parents are: WwSs or WWss.
- Short fingers is also dominant, so one parent must be heterozygous for the trait and the other parent can either be homozygous recessive or heterozygous. Therefore, the possible genotypes for the parents are: WwSs or wwSs.
- The child has a continuous hairline and long fingers, which are both recessive traits. This means that the child must be homozygous recessive for both traits (wwss).
- Combining the possible genotypes of the parents (WwSs and WWss for widow's peak, and WwSs and wwSs for short fingers), the only option that can result in a homozygous recessive child is D: WwSs × WwSs.

Question 12

A cross that involves two traits is called a
A) monohybrid cross.
B) dihybrid cross.
C) Punnett square.
D) testcross.
E) trihybrid cross.

Accepted Answer

A cross that involves two traits is called a dihybrid cross. The term "di-" means two, indicating that two traits are being studied in the cross.

Question 13

In guinea pigs, B = black, b = brown, S = short hair, s = long hair. A black, short-haired animal who is homozygous for both traits, reproduces with a brown, long-haired animal. What is the expected phenotypic ratio of the offspring?
A) 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair
B) 3 black short hair, 1 brown long hair, 3 black long hair, 1 brown short hair
C) 9 black short hair, 3 black long hair, 3 brown long hair, 1 brown short hair
D) 100% black short hair
E) 1 black short hair, 1 brown short hair

Accepted Answer

Since the black, short-haired guinea pig is homozygous for both traits (BBSS), and it is crossed with a brown, long-haired guinea pig (bbss), all offspring will inherit a B and an S from one parent and a b and an s from the other, resulting in all offspring being heterozygous (BbSs) but phenotypically black and short-haired.

Question 14

An individual with a dominant phenotype is crossed with an individual with a recessive phenotype. Five out of nine offspring show the dominant phenotype. What is the genotype of the dominant parent?
A) homozygous dominant
B) heterozygous
C) homozygous recessive
D) either homozyous dominant or heterozygous
E) There is not enough information to determine the genotype of the dominant parent.

Accepted Answer

If five out of nine offspring have the dominant phenotype, then the ratio of dominant to recessive offspring is 5:4. This is close to a 1:1 ratio, which suggests that the dominant parent is likely heterozygous (i.e. has one dominant allele and one recessive allele). If the dominant parent was homozygous dominant, then we would expect a 100% dominant phenotype in the offspring. If the dominant parent was homozygous recessive, then all of the offspring would be heterozygous and show the dominant phenotype, which is not the case. Therefore, the best choice is B, heterozygous.

Question 15

Unattached earlobes (EE or Ee) are dominant over attached earlobes (ee). A couple has unattached earlobes. Both notice that one of their parents on both sides have attached earlobes (ee). Therefore, they correctly assume that they are carriers for attached earlobes (Ee). If the couple proceeds to have four children, then
A) they can be certain that three will be heterozygous and one homozygous recessive.
B) if the first three are heterozygous, the fourth must be homozygous recessive.
C) the children must repeat the grandparents' genotype (Ee).
D) all children must have unattached earlobes since both parents possess the dominant gene for it.
E) two heterozygous, one homozygous dominant, and one homozygous recessive is a likely outcome, but all genotype ratios are possible.

Accepted Answer

Since both parents are carriers (Ee) for attached earlobes, there is a 25% chance that each child will inherit the recessive allele (ee) and have attached earlobes, and a 75% chance that they will inherit at least one dominant allele (E or e) and have unattached earlobes. The possible genotypes for the children are therefore: EE (25%), Ee (50%), and ee (25%). This translates to two heterozygous (Ee), one homozygous dominant (EE), and one homozygous recessive (ee) as likely outcomes, but all genotype ratios are technically possible.

Question 16

In cattle, consider that the lack of horns is dominant to the presence of horns. A bull without horns is crossed with a horned cow. All eight offspring lack horns. The bull is most likely
A) homozygous dominant.
B) homozygous recessive.
C) heterozygous.
D) sex-lined recessive.
E) heterozygous recessive.

Accepted Answer

Since all offspring lack horns and the trait for lacking horns is dominant, the bull must be homozygous dominant to ensure all offspring inherit the dominant allele for no horns, regardless of the cow's genotype.

Question 17

Which of the following crosses will yield four phenotypes in a ratio of 9:3:3:1?
A) RRYY × RRYY
B) RRYY × rryy
C) RrYy × RrYy
D) RrYy × rryy
E) rryy × rryy

Accepted Answer

The cross between RrYy and RrYy will yield four phenotypes in a 9:3:3:1 ratio due to the presence of two different genes that exhibit complete dominance and independent assortment. The capital "R" and "Y" alleles will result in a dominant phenotype, whereas the lowercase "r" and "y" alleles will result in a recessive phenotype. When two heterozygous individuals (RrYy) are crossed, the resulting offspring will have the following genotypes and phenotypes:

- 9/16 will be RY and display the dominant phenotype
- 3/16 will be RrYy and display the dominant phenotype
- 3/16 will be RY and display the recessive phenotype
- 1/16 will be rryy and display the recessive phenotype

Thus, the ratio of dominant to recessive phenotypes is 12:4 or 3:1, while the ratio of the two dominant phenotypes to each other is 9:3 or 3:1. None of the other crosses will yield a 9:3:3:1 phenotypic ratio.

Question 18

Which of the following crosses will yield four phenotypes in the 1:1:1:1 ratio?
A) RRYY × RRYY
B) RRYY × rryy
C) RrYy × RrYy
D) RrYy × rryy
E) rryy × rryy

Accepted Answer

Cross D (RrYy × rryy) will yield four phenotypes in a 1:1:1:1 ratio because it involves a dihybrid individual heterozygous for two traits being crossed with a homozygous recessive individual for those same traits. This results in equal probabilities of producing offspring with each of the four possible combinations of the two traits.

Question 19

In 1940, two researchers named Weiner and Landsteiner discovered that about 85% of the human population sampled possessed a blood cell protein called the Rh factor. The presence of Rh factor was labeled Rh positive (Rh⁺) and was found to be dominant over the absence of the blood factor (Rh^-). Rh factor is inherited in a typical Mendelian pattern. Based on this information, one could predict that two Rh⁺ parents could never have an Rh^- child.

Accepted Answer

While it is less likely for two Rh positive parents to have an Rh negative child, it is still possible. If both parents are heterozygous (Rh⁺), there is a 25% chance that their child will inherit Rh^- from both parents and thus be Rh negative.

Question 20

Which of the following would be considered a heterozygous dihybrid genotype?
A) Ww
B) sS
C) WWSs
D) WWss
E) WwSs

Accepted Answer

A heterozygous dihybrid genotype means that the individual carries two different alleles at each of two different gene loci. Option E (WwSs) presents two different alleles (W and w) at one gene locus and two different alleles (S and s) at another gene locus, which makes it a heterozygous dihybrid genotype.

Quiz 23: Patterns of Gene Inheritance

In humans, assume straight hairline is recessive to widow's peak. A woman with a straight hairline is married to a man with a widow's peak whose mother had a straight hairline. What are the chances of the couple having a child with a straight hairline?

In guinea pigs, B = black, b = brown, S = short hair, s = long hair. A guinea pig heterozygous for both traits is mated with a brown, short-haired guinea pig whose mother had long hair. What phenotypic ratio is expected in the offspring?

If the probability in any pregnancy of having a boy is 50%, then what is the probability of having 3 boys in a row?

A blue-eyed, left-handed woman marries a brown-eyed, right-handed man who is heterozygous for both of his traits. If blue eyes and left-handedness are recessive, how many different phenotypes are possible in their children?

Two parents heterozygous for cystic fibrosis have a child with the disease. The parents are expecting a second child. What is the probability that the second child will also have cystic fibrosis?

When pairs of factors separate independently of other pairs of factors, this is known as

Albinism is inherited as a recessive trait. An albino woman is married to a man with normal coloring and their first child is an albino. What was the genotype of the man?

Both parents are heterozygous for Tay-Sachs disease (an autosomal recessive disorder) . Three children in a row were born with Tay-Sachs disease. What is the chance that a fourth child will have Tay-Sachs disease?

If the genotype of an organism is YySsTt, then yST would represent

Which of the following crosses will yield only homozygous recessive offspring?

Two parents with widow's peak and short fingers have a child with a continuous hairline and long fingers. What are the genotypes of the two parents? (W = widow's peak; S = short fingers)

A cross that involves two traits is called a

In guinea pigs, B = black, b = brown, S = short hair, s = long hair. A black, short-haired animal who is homozygous for both traits, reproduces with a brown, long-haired animal. What is the expected phenotypic ratio of the offspring?

An individual with a dominant phenotype is crossed with an individual with a recessive phenotype. Five out of nine offspring show the dominant phenotype. What is the genotype of the dominant parent?

In cattle, consider that the lack of horns is dominant to the presence of horns. A bull without horns is crossed with a horned cow. All eight offspring lack horns. The bull is most likely

Which of the following crosses will yield four phenotypes in a ratio of 9:3:3:1?

Which of the following crosses will yield four phenotypes in the 1:1:1:1 ratio?

Which of the following would be considered a heterozygous dihybrid genotype?