Question 1

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the probability that individual III-1 is Ww?
A) 3/4
B) 1/4
C) 2/4
D) 1

Accepted Answer

The answer of The following question refers to the pedigree...

Question 2

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the likelihood that the offspring of IV-3 and IV-4 will have the trait?
A) 0%
B) 50%
C) 75%
D) 100%

Accepted Answer

IV-3 and IV-4 are both carriers of the trait, as indicated by their half-filled shapes. This means that there is a 50% chance that they will pass the trait to their offspring. However, it is also possible that they could both pass on the dominant normal allele (not shaded in the diagram) to their offspring, in which case the offspring would not have the trait. Therefore, the likelihood of their offspring having the trait is 50%.

Question 3

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the genotype of individual II-5?
A) WW
B) Ww
C) ww
D) ww or Ww

Accepted Answer

The answer of The following question refers to the pedigree...

Question 4

In some parts of Africa, the frequency of heterozygosity for the sickle-cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?
A) Mendel's law of independent assortment
B) Mendel's law of segregation
C) Darwin's explanation of natural selection
D) the malarial parasite changing the allele

Accepted Answer

This relationship is related to Darwin's explanation of natural selection, as the allele that causes sickle-cell anemia confers some resistance to malaria, thereby increasing the fitness of individuals carrying the allele in regions where malaria is prevalent. This results in higher frequency of the allele in populations. This relationship does not involve Mendel's laws or changes in the allele due to the malarial parasite.

Question 5

Feather color in budgies is determined by two different genes: Y for pigment on the outside of the feather, and B for pigment on the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. A blue budgie is crossed with a white budgie. Which of the following results in the offspring is most possible?
A) green offspring only
B) white offspring only
C) blue offspring only
D) blue and white offspring

Accepted Answer

The blue budgie is likely homozygous recessive (yy) for the Y gene and homozygous dominant (BB) for the B gene. The white budgie is likely homozygous recessive (yybb) for both genes. When crossed, all the offspring will be heterozygous for the Y gene and have one copy of the B gene. This means that each offspring will have a 50% chance of inheriting the dominant B gene from the blue budgie and being blue, and a 50% chance of inheriting the recessive bb gene from the white budgie and being white. So, the most likely outcome is blue and white offspring.

Question 6

The figure shows the pedigree for a family. Dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. Males are represented by squares, females by circles.

From this pedigree, this trait seems to be inherited ________.
A) from mothers
B) as an autosomal recessive
C) as a result of epistasis
D) as an autosomal dominant

Accepted Answer

The pedigree shows that affected individuals (dark-shaded symbols) appear in every generation, indicating an autosomal dominant inheritance pattern. Additionally, both males and females are affected, ruling out a X-linked dominant or recessive inheritance.

Question 7

Gene S controls the sharpness of spines in a type of cactus. Cacti with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cacti have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all. A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce ________.
A) all sharp-spined progeny
B) 50% sharp-spined, 50% dull-spined progeny
C) 25% sharp-spined, 50% dull-spined, 25% spineless progeny
D) It is impossible to determine the phenotypes of the progeny.

Accepted Answer

A true-breeding sharp-spined cactus would have the genotype SSNN, and a spineless cactus would have the genotype ssnn. The offspring would all be SsNn, resulting in all sharp-spined progeny.

Question 8

Skin color in a certain species of fish is inherited via a single gene with four different alleles. One fish of this type has alleles 1 and 3 (S₁S₃), and its mate has alleles 2 and 4 (S₂S₄). If each allele confers a unit of color darkness such that S₁ has one unit, S₂ has two units, and so on, then what proportion of their offspring would be expected to have five units of color?
A) 1/4
B) 1/8
C) 1/2
D) 0

Accepted Answer

The answer of Skin color in a certain species of...

Question 9

A gene for the MN blood group has codominant alleles M and N. If both children in a family are of blood type M, which of the following situations is possible?
A) Each parent is either M or MN.
B) Each parent must be type M.
C) Both children are heterozygous for this gene.
D) Neither parent can have the N allele.

Accepted Answer

If both children are of blood type M, each parent can be either M or MN, as the M allele is codominant and can be passed from either parent.

Question 10

Gene S controls the sharpness of spines in a type of cactus. Cacti with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cacti have dull spines. At the same time, a second gene, N, determines whether or not cacti have spines. Homozygous recessive nn cacti have no spines at all. The relationship between genes S and N is an example of which of the following inheritance patterns?
A) incomplete dominance
B) epistasis
C) pleiotropy
D) codominance

Accepted Answer

The relationship between genes S and N is an example of epistasis, where one gene (N) masks the expression of the other gene (S). In this case, the presence of at least one dominant allele of gene N is required for the expression of the S gene, which determines the sharpness of spines. Homozygous recessive nn cacti lack the ability to have spines, regardless of their genotype for gene S.

Question 11

In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: Iᴬ, Iᴮ, and i. The Iᴬ allele codes for the A glycoprotein, The Iᴮ allele codes for the B glycoprotein, and the i allele doesn't code for any membrane glycoprotein. Iᴬ and Iᴮ are codominant, and i is recessive to both Iᴬ and Iᴮ. People with type A blood have the genotypes IᴬIᴬ or Iᴬi, people with type B blood are IᴮIᴮ or Iᴮi, people with type AB blood are IᴬIᴮ, and people with type O blood are ii. If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have?
A) A and B
B) AB and O
C) A, B, and O
D) A, B, AB, and O

Accepted Answer

The answer of In humans, ABO blood types refer to...

Question 12

Gene S controls the sharpness of spines in a type of cactus. Cacti with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cacti have dull spines. At the same time, a second gene, N, determines whether or not cacti have spines. Homozygous recessive nn cacti have no spines at all. If cacti heterozygous for both traits, SsNn, were allowed to self-pollinate, the offspring would segregate into which of the following phenotype ratios?
A) 3 sharp-spined:1 spineless
B) 1 sharp-spined:2 dull-spined:1 spineless
C) 1 sharp-spined:1 dull-spined:1 spineless
D) 9 sharp-spined:3 dull-spined:4 spineless

Accepted Answer

The SsNn x SsNn cross results in a 9:3:3:1 phenotypic ratio, but since nn results in no spines, the 3 sharp-spined:1 dull-spined becomes 9 sharp-spined:3 dull-spined:4 spineless.

Question 13

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Which of the following treatments would be most effective for people with PKU?
A) Feed them the substrate that can be metabolized into phenylalanine.
B) Regulate the diet of the affected persons to severely limit the uptake of phenylalanine.
C) Feed the patients the missing enzymes in a regular cycle, such as twice per week.
D) Feed the patients an excess of the missing product.

Accepted Answer

The most effective treatment for people with PKU is to regulate their diet to limit the intake of phenylalanine. This can be achieved by avoiding foods that are high in phenylalanine such as meat, fish, eggs, and dairy products. Individuals with PKU are also advised to consume a special low-protein diet that is supplemented with synthetic amino acids, vitamins, and minerals to ensure that they have adequate nutrition. This helps to prevent the buildup of toxic byproducts of phenylalanine metabolism, which can cause brain damage and other serious health problems.

Question 14

Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of 13 years. Patients, who look very old even as children, do not live to reproduce. Which of the following statements represents the most likely assumption regarding this disorder?
A) The disease is autosomal dominant.
B) The disorder will increase in frequency in successive generations within a family.
C) The disorder may be due to mutation in a single protein-coding gene.
D) Each patient will have had at least one affected grandparent or parent.

Accepted Answer

The fact that the disorder is exceedingly rare suggests that it is due to a mutation in a single gene rather than being inherited through dominant or recessive traits or increasing in frequency in successive generations within a family. Additionally, the fact that patients with the disorder do not live to reproduce suggests that it is not inherited in a simple Mendelian manner. Thus, C is the most likely assumption.

Question 15

An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient?
A) karyotyping of the woman's somatic cells
B) X-ray
C) amniocentesis or CVS
D) blood transfusion

Accepted Answer

Amniocentesis or CVS (chorionic villus sampling) can both detect biochemical abnormalities in fetal cells and allow for diagnosis of serious disorders in utero. Karyotyping of the woman's somatic cells and X-rays are not relevant for detecting fetal disorders, and blood transfusions would not be helpful in this situation.

Question 16

In rabbits, the homozygous genotype LᶜLᶜ has normal legs, LᶜLᶜ results in deformed legs, and LᶜLᶜ results in very short legs. The genotype FᴮFᴮ produces black fur, FᴮFᵇ brown fur, and FᵇFᵇ white fur. If a cross is made between brown rabbits with deformed legs and white rabbits with deformed legs, what percentage of the offspring would be expected to have deformed legs and white fur?
A) 25%
B) 33%
C) 100%
D) 50%

Accepted Answer

The cross between brown rabbits with deformed legs (FᴮFᵇLᶜLᶜ) and white rabbits with deformed legs (FᵇFᵇLᶜLᶜ) will produce offspring with the following genotypes:FᴮFᵇLᶜLᶜ (brown, deformed legs)FᵇFᵇLᶜLᶜ (white, deformed legs)FᴮFᵇLᶜLᶜ (brown, normal legs)FᵇFᵇLᶜLᶜ (white, normal legs)Therefore, 50% of the offspring would be expected to have deformed legs and white fur (FᵇFᵇLᶜLᶜ).

Question 17

One of two major forms of a human condition called neurofibromatosis (NF1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. Which of the following is the best explanation for why a young, affected child is the first in her family to be diagnosed?
A) The mother carries the gene but does not express it.
B) One of the parents has a mild expression of the gene.
C) The condition skipped a generation in the family.
D) The child has one more chromosome than either of the parents.

Accepted Answer

Neurofibromatosis (NF1) is inherited as a dominant gene, which means that only one copy of the gene is enough to show the condition. If one of the parents has a mild expression of the gene, it is possible that they may not have recognized their symptoms or been diagnosed with the condition. In this case, their child may be the first in the family to be diagnosed as the condition became more severe. The other options (A, C, and D) do not provide a plausible explanation for why the child is the first in the family to be diagnosed.

Question 18

Feather color in budgies is determined by two different genes, Y for pigment on the outside of the feather, and B for pigment on the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. Two blue budgies were crossed. Over the years, they produced 22 offspring, five of which were white. What are the most likely genotypes for the two blue budgies?
A) yyBB and yyBB
B) yyBB and yyBb
C) yyBb and yyBb
D) yyBb and yybb

Accepted Answer

Since two blue budgies (yy) produced white offspring, they must have both been homozygous recessive for the B gene (bb). This rules out options A and B since both suggest at least one of the budgies had a dominant B allele. This leaves options C and D, but since all of the offspring were blue, option D can be ruled out since it suggests at least one of the budgies had a dominant Y allele (yellow). Therefore, the most likely genotypes for the two blue budgies are yyBb and yyBb (option C).

Question 19

In human blood types, Rh positive is a trait that shows simple dominance over Rh negative. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Which of the following phenotypes is possible for the father?
A) A negative
B) O negative
C) B positive
D) AB negative

Accepted Answer

Since the mother is type A positive, she must have at least one dominant A allele and one positive Rh allele. The daughter is type O positive, which means she has two recessive O alleles and one positive Rh allele inherited from the mother. The son is type B negative, which means he has one B allele and two recessive Rh negative alleles. The only way for the father to have children with both B and O blood types is if he has at least one B allele and one O allele. Therefore, he could be type B positive, B negative, or O positive. Of these options, only option C (B positive) is listed.

Question 20

Marfan syndrome in humans is caused by an abnormality of the connective tissue protein fibrillin. Patients are usually very tall and thin, with long spindly fingers, curvature of the spine, sometimes weakened arterial walls, and sometimes eye problems, such as lens dislocation. Which of the following would you conclude about Marfan syndrome from this information?
A) It is recessive.
B) It is dominant.
C) It is pleiotropic.
D) It is epistatic.

Accepted Answer

Marfan syndrome shows pleiotropy, as it affects various parts of the body, such as skeletal system, cardiovascular system, and ocular system, through a single gene defect.

Quiz 14: Mendel and the Gene Idea

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the probability that individual III-1 is Ww?

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the likelihood that the offspring of IV-3 and IV-4 will have the trait?

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the genotype of individual II-5?

In some parts of Africa, the frequency of heterozygosity for the sickle-cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?

A gene for the MN blood group has codominant alleles M and N. If both children in a family are of blood type M, which of the following situations is possible?

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Which of the following treatments would be most effective for people with PKU?

An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient?

One of two major forms of a human condition called neurofibromatosis (NF1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. Which of the following is the best explanation for why a young, affected child is the first in her family to be diagnosed?

In human blood types, Rh positive is a trait that shows simple dominance over Rh negative. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Which of the following phenotypes is possible for the father?