Question 1

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?

Accepted Answer

IV-3 and IV-4 are both heterozygous carriers of the dominant W allele, as indicated by their shaded circles, meaning they each have a 50% chance of passing the allele on to their offspring. Therefore, their offspring have a 50% chance of inheriting the W allele and exhibiting the trait.

Question 2

Feather colour 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?

Accepted Answer

When a blue budgie (yybb) is crossed with a white budgie (yybb), all their offspring will inherit a copy of the recessive alleles for both Y and B, which means their genotype will be yybb. As per the given information, the offspring with the genotype yybb is white. Therefore, the most possible result among the given choices is blue and white offspring. Green offspring are not possible because at least one of the parents must carry the dominant allele for yellow or blue to produce green. Similarly, white offspring are possible, but not the most possible, as the parents can carry the recessive alleles for any colour (Y or B), which make the white offspring possible, but it is less likely to get all white offspring from this cross.

Question 3

Feather colour 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?

Accepted Answer

If two blue budgies produced white offspring, it means that both parents must have carried the recessive alleles for both genes (yybb). This rules out options A and B since both have at least one dominant allele. Therefore, the only option left is C. The parents must have been yyBb and yyBb, which means that each parent passed on a y and a b allele to their offspring. The probability of producing white offspring from this cross is 1/16, which matches with the observed frequency of five out of 22 offspring.

Question 4

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?

Accepted Answer

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

Question 5

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 ________.

Accepted Answer

The answer of Gene S controls the sharpness of spines...

Question 6

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolise 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?

Accepted Answer

The most effective treatment for PKU is to regulate the diet of the affected person to severely limit the uptake of phenylalanine. This is because the accumulation of phenylalanine can lead to severe neurological damage. While feeding them the substrate that can be metabolised into phenylalanine may be a short-term solution, it is not sustainable in the long-term as it could still result in the accumulation of phenylalanine. Providing the missing enzymes or an excess of the missing product is not effective as the underlying issue is the inability to metabolize phenylalanine, not the absence of enzymes or products.

Question 7

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?

Accepted Answer

Marfan syndrome is pleiotropic, as it affects multiple organ systems and has a variety of symptoms that result from the same underlying genetic abnormality. It is neither dominant nor recessive, as it is caused by a mutation in a single gene (FBN1) and can manifest in both heterozygous and homozygous individuals. It is also not epistatic, as it does not involve the suppression or interaction of genes at different loci.

Question 8

W. Affected individuals are indicated by a dark square or circle. 
- What is the genotype of individual II-5?

Accepted Answer

The answer of  W. Affected individuals are indicated by...

Question 9

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 ________.

Accepted Answer

The pedigree suggests that the trait is inherited as an autosomal dominant, as affected individuals (dark-shaded symbols) are present in most generations and both males and females are affected. Autosomal recessive inheritance would require unaffected parents to produce affected offspring, which is not the case here. There is no evidence of epistasis, which refers to the interaction of multiple genes in determining a trait. There is also no evidence to suggest maternal inheritance, as affected individuals have both affected mothers and fathers.

Question 10

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?

Accepted Answer

The genotype of the brown rabbit with deformed legs is LCLc FBFb, while the white rabbit with deformed legs is LcLc FbFb. 
The Punnett Square for this cross would be:

|   | LC c | LC c |
| - | -   | -   |
| F B | LCLC FBFb | LCLc FBFb |
| F b | LCLC FbFb | LCLc FbFb |

From the Punnett Square, we can see that 25% of the offspring would have the genotype LcLc FbFb, which corresponds to having both deformed legs and white fur.

Question 11

W. Affected individuals are indicated by a dark square or circle. 
- What is the probability that individual III-1 is Ww?

Accepted Answer

The answer of  W. Affected individuals are indicated by...

Question 12

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?

Accepted Answer

The answer of A gene for the MN blood group...

Question 13

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?

Accepted Answer

Hutchinson-Gilford progeria is caused by a mutation in the LMNA gene, which codes for the lamin A protein. This mutation causes abnormal processing of the prelamin A protein, leading to premature aging. It is not an autosomal dominant disorder, as it is usually caused by a spontaneous mutation rather than being inherited from a parent. The disorder does not necessarily increase in frequency in successive generations within a family, and there may be cases where there are no affected family members.

Question 14

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?

Accepted Answer

The relationship between genes S and N is an example of epistasis, where one gene masks or affects the expression of another gene. Specifically, the expression of gene S (sharp vs. dull spines) depends on whether or not gene N allows for spine development in the first place. Therefore, the expression of gene S is dependent on the presence or absence of gene N, making this an example of epistasis.

Question 15

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?

Accepted Answer

Inheritance of NF1 is dominant, meaning that if a parent has the gene, there is a 50% chance of passing it on to each offspring. However, the severity of expression can vary widely within families. It is possible that one of the parents has a mild expression of the gene and was not previously diagnosed, leading to their affected child being the first in the family to be diagnosed.

Question 16

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?

Accepted Answer

The answer of Gene S controls the sharpness of spines...

Question 17

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

Accepted Answer

Amniocentesis or CVS (Chorionic Villus Sampling) are procedures that can detect genetic disorders in a foetus by analyzing its DNA. Karyotyping of the woman's somatic cells and X-ray are not relevant to detecting genetic disorders in a foetus. Blood transfusion is not relevant to this situation.

Question 18

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

Accepted Answer

The relationship between sickle-cell anaemia and a high frequency of heterozygosity in certain parts of Africa is related to Darwin's explanation of natural selection. The prevalence of the sickle-cell anaemia allele is thought to have increased in these areas because individuals who carry one copy of the allele (i.e. are heterozygous) have a selective advantage in regions where malaria is common. This is an example of natural selection favoring a particular genotype in response to environmental pressures.

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?

Accepted Answer

The mother must be heterozygous for blood type A (AO genotype) because she has a daughter with type O blood. The son's B negative blood type means he must have inherited a B allele from the father and a negative allele from either parent. Therefore, the father must be either BB or BO for the son to have a B allele.

Since the daughter is A positive, she must have received an A allele from the mother and a positive allele from either parent. If the father were A negative, the daughter would have been A negative. If the father were O negative, the daughter would have been O negative. If the father were AB negative, the daughter would have been AB negative. Therefore, the only possibility for the father's blood type is B positive.

Question 20

Skin colour 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 colour 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 colour?

Accepted Answer

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

Quiz 14: Mendel and the Gene Idea

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?

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolise 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?

W. Affected individuals are indicated by a dark square or circle. - What is the genotype of individual II-5?

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 ________.

W. Affected individuals are indicated by a dark square or circle. - What is the probability that individual III-1 is Ww?

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?

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?

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

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

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?