Question 1

Glucose- 6- phosphate dehydrogenase deficiency G6PD) is inherited as a recessive allele of an X- linked gene in humans. A woman whose father suffered from G6PD marries a normal man. a) What proportion of their sons is expected to be G6PD? b) If the husband was not normal but was G6PD deficient, would you change your answer in part a)?

Accepted Answer

Since the woman is a carrier of the G6PD deficiency gene, she can pass the gene to half of her sons. Thus, the proportion of their sons who are expected to be G6PD deficient is 1/2. If the husband was also G6PD deficient, then all their sons would inherit the gene, so the proportion would be 100%. However, this scenario is not mentioned in the question.

Question 2

In humans, blue eyes are inherited as a recessive autosomal trait and colour blindness is an X- linked recessive trait. A woman with blue eyes and normal colour vision whose father was colour blind marries a man who also has normal colour vision. He has brown eyes but his mother had blue eyes. Which of the following do you expect to be true for their daughters?

Accepted Answer

Blue eyes are recessive, so the woman with blue eyes must have two copies of the blue eye allele. Her father was color blind, so he must have had one copy of the X-linked color blindness allele, which he passed on to his daughter. The man with brown eyes could either have two copies of the brown eye allele or one copy of the brown eye allele and one copy of the blue eye allele. His mother had blue eyes, so he must have received a blue eye allele from her. 
Therefore, the possible genotypes for the parents are:
Woman: bb XBXb
Man: BB or Bb XbB or XbXb
When you combine these genotypes to predict the genotypes and phenotypes of their daughters, you get the following:
50% chance of BB or Bb and XBXb → Normal color vision, brown eyes
50% chance of Bb and XbXb → Normal color vision, blue eyes

Question 3

Regarding an allelic pair for flower colour in snapdragons, heterozygotes have pink flowers, whereas the two homozygotes have red flowers or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring is expected to have pink flowers?

Accepted Answer

This is an example of incomplete dominance, where the heterozygote (RW) expresses a phenotype (pink flowers) that is intermediate between the phenotypes of the two homozygotes (RR for red flowers and WW for white flowers). Crossing red (RR) with white (WW) will produce all RW offspring, which will have pink flowers.

Question 4

The second child of the couple above is unaffected does not display the condition). What is the probability this child is a carrier for the ataxia condition?

Accepted Answer

Since we know that the father must be a carrier (because he has one affected X chromosome and one normal X chromosome) and the mother is unaffected but could be a carrier (because she has one normal X chromosome and one unknown X chromosome), there are four possible outcomes for the second child's genotype: XA (affected female), Xa (carrier female), XA (unaffected male), or Xa (unaffected male). However, we know that the second child is unaffected, so we can eliminate XA from the possible genotypes. This leaves us with three possible genotypes: Xa (carrier female), XA (unaffected male), and Xa (unaffected male). Since each of these three genotypes has an equal chance of occurring, and two out of three of them result in the second child being a carrier, the probability that the second child is a carrier is 2/3. Therefore, the correct answer is C.

Question 5

Whenever the dominant allele of either or both) gene A or gene B is present, red colour is produced. A cross of
AaBb × AaBb can be expected to yield how many red offspring out of 16? Assume the genes are not linked.

Accepted Answer

For red color to be produced, at least one dominant allele of gene A or gene B must be present. The only genotype that does not produce red color is aabb. Using a Punnett square for AaBb × AaBb, we find that 15 out of 16 combinations have at least one dominant allele (A or B), resulting in red offspring. The 16th combination is aabb, which does not produce red color.

Question 6

A couple has a child with Harman's ataxia a recessive disorder). If this couple has second child, what is the probability that child will have ataxia?

Accepted Answer

Since Harman's ataxia is a recessive disorder, both parents must be carriers. The possible genotypes for the children are: AA (normal), Aa (carrier, normal), aa (affected). The probability of having an affected child (aa) is 1/4.

Question 7

$\begin{array} { c c c } \text { Sex } & \text { Phenotype } & \text { Number } \ \text { male } & \text { wild } & 123 \ \text { male } & \text { yellow } & 116 \ \text { female } & \text { wild } & 240 \end{array}$ Figure 14.1 -In a Drosophila experiment, a cross was made between homozygous wild- type females and yellow- bodied males. All of the resulting F₁s were phenotypically wild type. However, adult flies of the F₂generation resulting from matings of the F₁s) had the characteristics shown in the figure above: A. Is the mutant allele for yellow body recessive or dominant? B. Is the yellow locus autosomal not X- linked) or X- linked?

Accepted Answer

The fact that all F₁ flies were phenotypically wild type indicates that they were homozygous for the wild-type allele. However, in the F₂ generation resulting from matings of these flies, some flies had yellow bodies, which indicates that they must have inherited a mutant allele from one of their F₁ parents. This would only be possible if the mutant allele is recessive, as the F₁ parent must have been homozygous for the wild-type allele. Additionally, the fact that the yellow-bodied flies were only found in the F₂ generation and not in the F₁ generation suggests that the yellow locus is not on the X chromosome, as X-linked traits usually show different inheritance patterns between males and females. Thus, the best choice is D) a) recessive; b) X-linked.

Question 8

Radish flowers may be red, purple, or white. A cross between a red- flowered plant and a white- flowered plant yields all- purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait. If true- breeding red long radishes are crossed with true- breeding white oval radishes, the F₁will be expected to be which of the following?

Accepted Answer

The cross between a red-flowered and white-flowered plant yields all-purple offspring, which indicates that purple is the dominant trait in flower color. This means that the offspring from the cross between red long radishes and white oval radishes will all have purple flowers, regardless of whether they are long or oval. However, since long is the dominant trait in radish shape, the offspring will all be long. Therefore, the expected phenotype of the offspring is purple and long (Choice C).

Question 9

Snapdragons are flowers that come in a variety of colours, including red, pink, and white. A series of crosses wit snapdragons having flowers of different colours produced the following results:
Pink × pink: 27 pink, 13 red, 14 white red × red: all red
White × white: all white
Pink × white: 29 pink, 26 white pink × red: 28 pink, 27 red white × red: all pink
Based on the results, what is the most reasonable explanation for the inheritance of these flower colours?

Accepted Answer

The results suggest incomplete dominance because the pink flowers result from the cross between red and white flowers, indicating that neither the red nor the white allele is completely dominant over the other, leading to a blend (pink) in the heterozygous condition.

Question 10

What is the probability that a gamete from an individual with the genotype E₁E₂X₁X₁A₁A₂M₂M₂will contain all paternally derived homologues?

Accepted Answer

The probability of a gamete containing all paternally derived homologues is calculated by multiplying the probabilities of inheriting each paternal allele. For each gene pair (E, X, A, M), the probability is 1/2. Since there are four gene pairs, the probability is (1/2)^4 = 1/16.

Question 11

Hemophilia is caused by several genetic factors; one, a recessive allele of an X- linked gene, is the subject of this problem. Assume that a man with hemophilia marries a normal woman whose father had hemophilia. What is the probability that their first son will have hemophilia?

Accepted Answer

Since hemophilia is a recessive X-linked condition, a man with hemophilia has the genotype X^hY (where X^h represents the hemophilia allele and Y is the normal Y chromosome). A normal woman whose father had hemophilia must be a carrier, as she received one normal X chromosome from her mother and one X^h (hemophilia) chromosome from her father, making her genotype X^hX. When they have a son, the son gets his Y chromosome from his father and has a 50% chance of inheriting the X^h chromosome from his mother, leading to hemophilia.

Question 12

In humans, blue eyes are inherited as a recessive autosomal trait and colour blindness is an X- linked recessive trait. A woman with blue eyes and normal colour vision whose father was colour blind marries a man who also has normal colour vision. He has brown eyes, but his mother had blue eyes. Which of the following do you expect to be true for their sons?

Accepted Answer

The woman is a carrier for color blindness (X^NX^c) and has blue eyes (bb). The man has normal vision (X^NY) and is heterozygous for brown eyes (Bb). Their sons can inherit X^c from the mother and Y from the father, resulting in color blindness, or X^N from the mother and Y from the father, resulting in normal vision. For eye color, sons can inherit either B or b from the father, resulting in brown or blue eyes, respectively. Thus, all combinations are possible.

Question 13

The B- globin gene has many alleles. What is the maximum number of different alleles of the B- globin gene that could be carried by 6 children with the same mom and dad given that the parents are heterozygous?

Accepted Answer

Each parent can pass on one of two alleles, resulting in a maximum of 4 different alleles among the children.

Question 14

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses 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

A true-breeding sharp-spined cactus would have the genotype SSNN (since it is dominant for sharp spines and must have spines), and a spineless cactus would have the genotype ssnn (since it is recessive for no spines). Their offspring would all be SsNn, having sharp spines due to the dominant S allele and the presence of spines due to at least one N allele.

Question 15

Two individuals of genotype AABBCCDD and aabbccdd are mated. The F1 generation is interbred self- fertilized) to yield 5000 progeny. What proportion of the F2 is expected to be AabbCcDD?

Accepted Answer

When the two individuals are crossed, they produce heterozygous (AaBbCcDd) offspring. Therefore, the F1 generation is AaBbCcDd x AaBbCcDd. 
When these F1 individuals are self-fertilized, the resulting genotype proportions can be found using a Punnett square. The possible gametes that can be produced by the F1 generation are AB, Ab, aB, and ab for the A and B loci (and similarly for the C and D loci). 
Therefore, the expected genotype frequencies in the F2 generation are:
- AABBCCDD: (1/4) x (1/4) x (1/4) x (1/4) = 1/256
- AABBCcDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AABBCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AABbCCDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AABbCcDD: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AABbCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AAbbCCDD: (1/4) x (1/2) x (1/4) x (1/2) = 1/64
- AAbbCcDD: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- AAbbCcDd: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- AabbCCDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AabbCcDD: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AabbCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AabbccDD: (1/4) x (1/2) x (1/4) x (1/2) = 1/64
- AabbccDd: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- aabbCCDD: (1/2) x (1/4) x (1/4) x (1/2) = 1/64
- aabbCcDD: (1/2) x (1/4) x (1/2) x (1/2) = 1/32
- aabbCcDd: (1/2) x (1/4) x (1/2) x (1/2) = 1/32
- aabbccDD: (1/2) x (1/2) x (1/4) x (1/2) = 1/32
- aabbccDd: (1/2) x (1/2) x (1/2) x (1/2) = 1/16

Therefore, the proportion of the F2 that is AabbCcDD is the sum of the frequencies of AabbCcDD, AabbCcDd, aabbCcDD, and aabbCcDd:

1/64 + 1/64 + 1/64 + 1/32 = 3/64

This is closest to choice C, 1/64.

Question 16

A man who carries an allele of an X- linked gene will pass it on to

Accepted Answer

A man has one X chromosome and one Y chromosome. The X-linked allele will be passed to all of his daughters because they receive his only X chromosome, while his sons receive his Y chromosome.

Question 17

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses 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. If doubly heterozygous SsNn cactuses were allowed to self- pollinate, the F₂would segregate in which of the following ratios?

Accepted Answer

The SsNn cactuses have two different genes segregating independently. For the S gene, the cross between Ss x Ss will result in a classic 1:2:1 ratio. Therefore, 1/4 of the offspring will be SS (sharp spined), 1/2 will be Ss (sharp spined), and 1/4 will be ss (dull spined). 
For the N gene, the cross between Nn x Nn will result in a classic 1:2:1 ratio. Therefore, 1/4 of the offspring will be NN (spineless), 1/2 will be Nn (spineless), and 1/4 will be nn (have spines).
To determine the combined ratio, we multiply the individual ratios: 
(1/4 SS) x (1/4 nn) = 1/16 SSnn 
(1/4 SS) x (1/2 Nn) = 1/8 SSNn 
(1/4 Ss) x (1/4 nn) = 1/16 Ssnn 
(1/4 Ss) x (1/2 Nn) = 1/8 SsNn 
(1/4 ss) x (1/4 nn) = 1/16 ssnn 
(1/4 ss) x (1/2 Nn) = 1/8 ssNn 
So the final ratio is: 
1 SSnn : 2 SSNn : 1 Ssnn : 2 SsNn : 1 ssnn : 2 ssNn

We can simplify this ratio by combining like terms: 
3 sharp-spined: 3 dull-spined: 4 spineless. 
Therefore, the correct answer is C.

Question 18

Figure 14.3
-The distance between two genes is a determinant of their recombination frequency. Based on the image below, which of the following sets of genes will show the greatest recombination frequency?

Accepted Answer

The greater the distance between two genes on a chromosome, the more opportunity there is for crossing over to occur between them. In this case, genes A and E are the furthest apart, with three units between them, while the others have only two units between them. Therefore, the A and E genes will have the greatest recombination frequency.

Question 19

A homozygous tomato plant with red fruit and yellow flowers was crossed with a homozygous tomato plant with golden fruit and white flowers. The F₁all had red fruit and yellow flowers. The following phenotypes were obtained in the F₂: Red fruit and yellow flowers-41 Red fruit and white flowers-7 Golden fruit and yellow flowers-8 Golden fruit and white flowers-44 How many map units separate these genes?

Accepted Answer

The answer of A homozygous tomato plant with red fruit...

Question 20

There are 40 different alleles for a gene involved in immune function. What is the maximum number of alleles that a human can have?

Accepted Answer

Humans have two copies of each gene, one from each parent. Therefore, the maximum number of alleles that a human can have for a given gene is 2.

Quiz 14: Mendel and the Gene

Regarding an allelic pair for flower colour in snapdragons, heterozygotes have pink flowers, whereas the two homozygotes have red flowers or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring is expected to have pink flowers?

The second child of the couple above is unaffected does not display the condition) . What is the probability this child is a carrier for the ataxia condition?

Whenever the dominant allele of either or both) gene A or gene B is present, red colour is produced. A cross of AaBb × AaBb can be expected to yield how many red offspring out of 16? Assume the genes are not linked.

A couple has a child with Harman's ataxia a recessive disorder) . If this couple has second child, what is the probability that child will have ataxia?

What is the probability that a gamete from an individual with the genotype E₁E₂X₁X₁A₁A₂M₂M₂will contain all paternally derived homologues?

Hemophilia is caused by several genetic factors; one, a recessive allele of an X- linked gene, is the subject of this problem. Assume that a man with hemophilia marries a normal woman whose father had hemophilia. What is the probability that their first son will have hemophilia?

The B- globin gene has many alleles. What is the maximum number of different alleles of the B- globin gene that could be carried by 6 children with the same mom and dad given that the parents are heterozygous?

Two individuals of genotype AABBCCDD and aabbccdd are mated. The F1 generation is interbred self- fertilized) to yield 5000 progeny. What proportion of the F2 is expected to be AabbCcDD?

A man who carries an allele of an X- linked gene will pass it on to

Figure 14.3 -The distance between two genes is a determinant of their recombination frequency. Based on the image below, which of the following sets of genes will show the greatest recombination frequency?

There are 40 different alleles for a gene involved in immune function. What is the maximum number of alleles that a human can have?

Quiz 14: Mendel and the Gene

Regarding an allelic pair for flower colour in snapdragons, heterozygotes have pink flowers, whereas the two homozygotes have red flowers or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring is expected to have pink flowers?

The second child of the couple above is unaffected does not display the condition) . What is the probability this child is a carrier for the ataxia condition?

Whenever the dominant allele of either or both) gene A or gene B is present, red colour is produced. A cross of AaBb × AaBb can be expected to yield how many red offspring out of 16? Assume the genes are not linked.

A couple has a child with Harman's ataxia a recessive disorder) . If this couple has second child, what is the probability that child will have ataxia?

What is the probability that a gamete from an individual with the genotype E1E2X1X1A1A2M2M2 will contain all paternally derived homologues?

Hemophilia is caused by several genetic factors; one, a recessive allele of an X- linked gene, is the subject of this problem. Assume that a man with hemophilia marries a normal woman whose father had hemophilia. What is the probability that their first son will have hemophilia?

The B- globin gene has many alleles. What is the maximum number of different alleles of the B- globin gene that could be carried by 6 children with the same mom and dad given that the parents are heterozygous?

Two individuals of genotype AABBCCDD and aabbccdd are mated. The F1 generation is interbred self- fertilized) to yield 5000 progeny. What proportion of the F2 is expected to be AabbCcDD?

A man who carries an allele of an X- linked gene will pass it on to

Figure 14.3 -The distance between two genes is a determinant of their recombination frequency. Based on the image below, which of the following sets of genes will show the greatest recombination frequency?

There are 40 different alleles for a gene involved in immune function. What is the maximum number of alleles that a human can have?

What is the probability that a gamete from an individual with the genotype E₁E₂X₁X₁A₁A₂M₂M₂will contain all paternally derived homologues?