Chapter 14 Mendelian Genetics Quiz

1. The genetic makeup of an organism

Genotype

Phenotype

Homozygous

Heterozygous

The term "genotype" refers to the genetic makeup of an organism, which includes all the genes and alleles present in its DNA. It determines the characteristics and traits that an organism possesses. The genotype is not directly observable but can influence the observable traits or characteristics, known as the phenotype. The phenotype is the physical expression of the genotype, such as the organism's appearance, behavior, and other observable traits. Homozygous refers to having two identical alleles for a particular gene, while heterozygous refers to having two different alleles for the same gene.

Explanation

The term "genotype" refers to the genetic makeup of an organism, which includes all the genes and alleles present in its DNA. It determines the characteristics and traits that an organism possesses. The genotype is not directly observable but can influence the observable traits or characteristics, known as the phenotype. The phenotype is the physical expression of the genotype, such as the organism's appearance, behavior, and other observable traits. Homozygous refers to having two identical alleles for a particular gene, while heterozygous refers to having two different alleles for the same gene.

2. Two pea plants heterozygous for pod color and shape (dihybrid) are crossed. Draw a punnet square to determine phenotypic ratio of offspring. The ratio will be:

3:1

9:3:3:1

15:1

12:3:1

When two pea plants heterozygous for pod color and shape are crossed, the resulting phenotypic ratio of the offspring is 9:3:3:1. This means that for every 9 offspring, 3 will have both dominant traits (color and shape), 3 will have one dominant trait and one recessive trait, 3 will have the other dominant trait and one recessive trait, and 1 will have both recessive traits. This ratio is a result of the independent assortment of alleles for the two traits during meiosis.

Explanation

When two pea plants heterozygous for pod color and shape are crossed, the resulting phenotypic ratio of the offspring is 9:3:3:1. This means that for every 9 offspring, 3 will have both dominant traits (color and shape), 3 will have one dominant trait and one recessive trait, 3 will have the other dominant trait and one recessive trait, and 1 will have both recessive traits. This ratio is a result of the independent assortment of alleles for the two traits during meiosis.

3. A man with A blood type marries a woman with B blood type. Their child has type O. What is the genotype of the woman?

Bi

BB

B

BO

The genotype of the woman is Bi. This is because the child has type O blood, which is only possible if both parents have at least one recessive allele for O blood type. The man has blood type A, which means he has genotype AA or AO. Since the child has type O, the woman must have genotype BO or OO. Since OO is not an option, the woman's genotype must be BO, making her a carrier of the O allele. Therefore, the correct answer is Bi.

Explanation

The genotype of the woman is Bi. This is because the child has type O blood, which is only possible if both parents have at least one recessive allele for O blood type. The man has blood type A, which means he has genotype AA or AO. Since the child has type O, the woman must have genotype BO or OO. Since OO is not an option, the woman's genotype must be BO, making her a carrier of the O allele. Therefore, the correct answer is Bi.

4. In tigers a recessive gene causes an absence of fur pigmentation and a cross eyed condition. If two phenotypically normal tigers, that are heterozygous for this loci mate, what percentage of their offspring will be cross eyed and white?

O

25

50

100

When two phenotypically normal tigers that are heterozygous for the recessive gene mate, there is a 25% chance that their offspring will be cross-eyed and white. This is because both tigers are carriers of the recessive gene, and there is a 25% chance that each offspring will inherit two copies of the recessive gene, resulting in the cross-eyed and white phenotype. The other 75% of the offspring will either inherit one copy of the recessive gene and be phenotypically normal, or inherit two copies of the dominant gene and also be phenotypically normal.

Explanation

When two phenotypically normal tigers that are heterozygous for the recessive gene mate, there is a 25% chance that their offspring will be cross-eyed and white. This is because both tigers are carriers of the recessive gene, and there is a 25% chance that each offspring will inherit two copies of the recessive gene, resulting in the cross-eyed and white phenotype. The other 75% of the offspring will either inherit one copy of the recessive gene and be phenotypically normal, or inherit two copies of the dominant gene and also be phenotypically normal.

5. A man with A blood type marries a woman with B blood type. Their child has type O. What is the genotype of the man?

AA

AB

AO

Ai

The genotype of the man is Ai. In the ABO blood typing system, A and B are dominant alleles, while O is recessive. The man has blood type A, which means he can have either genotype AA or Ai. Since the child has blood type O, which is recessive, it means that both parents must carry the recessive O allele. Therefore, the man's genotype must be Ai.

Explanation

The genotype of the man is Ai. In the ABO blood typing system, A and B are dominant alleles, while O is recessive. The man has blood type A, which means he can have either genotype AA or Ai. Since the child has blood type O, which is recessive, it means that both parents must carry the recessive O allele. Therefore, the man's genotype must be Ai.

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

1/4

1/8

1/2

1/6

Since extra digits is a dominant trait, it means that if one parent has the trait, there is a 100% chance that their child will also have the trait. In this case, the man has the trait, so all of their children will have the trait. Therefore, the fraction of the couple's children expected to show this condition is 1/2.

Explanation

Since extra digits is a dominant trait, it means that if one parent has the trait, there is a 100% chance that their child will also have the trait. In this case, the man has the trait, so all of their children will have the trait. Therefore, the fraction of the couple's children expected to show this condition is 1/2.

7. A man with A blood type marries a woman with B blood type. Their child has type O. What is the genotype of the child?

ABO

OO

Ii

AB

The genotype of the child is ii. This is because both the father and mother have a different blood type, which means they each have two different alleles for the ABO blood group gene. Since the child has blood type O, which is the absence of both A and B antigens, it indicates that the child inherited two recessive O alleles from both parents. Therefore, the genotype of the child is ii.

Explanation

The genotype of the child is ii. This is because both the father and mother have a different blood type, which means they each have two different alleles for the ABO blood group gene. Since the child has blood type O, which is the absence of both A and B antigens, it indicates that the child inherited two recessive O alleles from both parents. Therefore, the genotype of the child is ii.

8. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what is the probability that the F2 offspring will have the following phenotype (among others, of course)? AABBCCDD

1/256

1/16

1/64

1/128

The genotype of the F1 individuals in a tetrahybrid cross is AaBbCcDd. This means that each individual has one copy of each gene, with one allele coming from each parent. Assuming independent assortment of these four genes, the probability of any particular combination of alleles in the F2 offspring is 1/2 for each gene. Since there are four genes, the probability of the F2 offspring having the AABBCCDD phenotype (all dominant alleles) is (1/2)^8, which simplifies to 1/256.

Explanation

The genotype of the F1 individuals in a tetrahybrid cross is AaBbCcDd. This means that each individual has one copy of each gene, with one allele coming from each parent. Assuming independent assortment of these four genes, the probability of any particular combination of alleles in the F2 offspring is 1/2 for each gene. Since there are four genes, the probability of the F2 offspring having the AABBCCDD phenotype (all dominant alleles) is (1/2)^8, which simplifies to 1/256.

9. Karen and Steve both have sibling with SIckle Cell. Neither of them nor their parents have the disease, but none of them have been tested to reveal the trait. Calculate the probability that Karen and Steve's child may be born with Sickle Cell?

1/16

1/9

1/17

1/1

Since Karen and Steve both have siblings with Sickle Cell, it is possible that they both carry the trait for the disease. The probability of a person carrying the trait for Sickle Cell is 1/4. Therefore, the probability of both Karen and Steve carrying the trait is (1/4) * (1/4) = 1/16. If both Karen and Steve carry the trait, there is a 1/4 chance that their child will inherit the disease. Therefore, the probability that Karen and Steve's child may be born with Sickle Cell is (1/16) * (1/4) = 1/64. However, the question only asks for the probability that their child may be born with Sickle Cell, which means it includes the possibility of the child not inheriting the disease. Therefore, the correct answer is 1 - 1/64 = 63/64. Simplifying this fraction gives us 1/9, which is the answer.

Explanation

Since Karen and Steve both have siblings with Sickle Cell, it is possible that they both carry the trait for the disease. The probability of a person carrying the trait for Sickle Cell is 1/4. Therefore, the probability of both Karen and Steve carrying the trait is (1/4) * (1/4) = 1/16. If both Karen and Steve carry the trait, there is a 1/4 chance that their child will inherit the disease. Therefore, the probability that Karen and Steve's child may be born with Sickle Cell is (1/16) * (1/4) = 1/64. However, the question only asks for the probability that their child may be born with Sickle Cell, which means it includes the possibility of the child not inheriting the disease. Therefore, the correct answer is 1 - 1/64 = 63/64. Simplifying this fraction gives us 1/9, which is the answer.

10. Charles was married once before- he and his first wife had a child with Cystic Fibrosis. His current wife's brother died of Cystic Fibrosis. What is the probability that he and his current wife will have a baby with CF? Neither he nor his wife has CF.

1/2

1/4

1/6

1/8

The probability that Charles and his current wife will have a baby with CF can be determined by considering the inheritance pattern of Cystic Fibrosis. CF is an autosomal recessive disorder, which means that both parents must be carriers of the CF gene in order for their child to have CF. Since Charles has a child with CF from his previous marriage, it is likely that he is a carrier of the CF gene. Additionally, his current wife's brother died of CF, indicating that she is also a carrier. The probability of two carriers having a child with CF is 1/4. However, since Charles and his current wife do not have CF themselves, the probability is reduced to 1/6.

Explanation

The probability that Charles and his current wife will have a baby with CF can be determined by considering the inheritance pattern of Cystic Fibrosis. CF is an autosomal recessive disorder, which means that both parents must be carriers of the CF gene in order for their child to have CF. Since Charles has a child with CF from his previous marriage, it is likely that he is a carrier of the CF gene. Additionally, his current wife's brother died of CF, indicating that she is also a carrier. The probability of two carriers having a child with CF is 1/4. However, since Charles and his current wife do not have CF themselves, the probability is reduced to 1/6.