Question 1

A recessive allele on the X chromosome is responsible for red-green color blindness in humans.A woman with normal vision whose father is color blind marries a color-blind male.What is the probability that this couple's first son will be color blind?

Accepted Answer

The woman is a carrier of the color blindness allele, as she inherited one recessive allele from her father but has a normal phenotype. The color-blind male can only pass on the recessive allele for color blindness to his offspring. Therefore, the probability that their first son will be color blind is 50%, as there is a 50% chance that he will inherit the recessive allele from his mother and a 100% chance that he will inherit the recessive allele from his father.

Question 2

Suppose two AaBbCc individuals are mated.Assuming that the genes are not linked,what fraction of the offspring are expected to be homozygous recessive for the three traits?

Accepted Answer

Each individual can make four types of gametes: AB, Ab, aB, and ab. The probability of two individuals producing a homozygous recessive offspring at one locus is 1/4. This probability can be multiplied by the probability of the same outcome occurring at the other two loci to get (1/4) x (1/4) x (1/4) = 1/64. Therefore, 1/64 of the offspring are expected to be homozygous recessive for all three traits.

Question 3

Assuming independent assortment for all gene pairs,what is the probability that the following parents,AABbCc × AaBbCc,will produce an AaBbCc offspring?

Accepted Answer

First, we need to figure out the probability of each parent passing on the desired allele to form AaBbCc. The AaBbCc genotype requires the parent with the dominant trait A and the recessive trait b to pass on the recessive b, and the parent with the dominant trait B and the recessive trait c to pass on the dominant B.

For the AABbCc parent, there are three gene pairs involved and each of them has a 50% probability of passing on the desired allele: 
A: 1/2 
B: 1/2 
C: 1/2

For the AaBbCc parent, there are also three gene pairs involved but the probability of passing on the desired allele is dependent on which allele each parent actually passes on. 
A: 1/2 (either A or a can be passed on) 
B: 1/2 (either B or b can be passed on) 
C: 1/2 (either C or c can be passed on)

To find the probability of an AaBbCc offspring, we need to multiply the probabilities of each gene pair for each parent and then multiply the two resulting probabilities together:

AABbCc: (1/2) x (1/2) x (1/2) = 1/8 
AaBbCc: (1/2) x (1/2) x (1/2) x (1/2) x (1/2) x (1/2) = 1/64

Multiplying the two probabilities together: 
1/8 x 1/64 = 1/512

So the probability of producing an AaBbCc offspring is 1/512, which is equivalent to approximately 0.002, or 0.2%. Therefore, the best choice is C) 1/8.

Question 4

Different ratios occur in crosses with single gene pairs or two gene pairs.What types of ratios are likely to occur in crosses dealing with a single gene pair?

Accepted Answer

The ratios 3:1 and 1:2:1 are common in crosses with a single gene pair.

Question 5

When Mendel crossed yellow-seeded and green-seeded pea plants,all the offspring were yellow seeded.When he took these F₁ yellow-seeded plants and crossed them to green-seeded plants,what genotypic ratio is expected?

Accepted Answer

The F₁ yellow-seeded plants are heterozygous (Yy). When crossed with green-seeded plants (yy), the expected genotypic ratio is 1:1 (Yy:yy).

Question 6

In rabbits,the homozygous CC is normal,Cc results in deformed legs,and cc results in very short legs.The genotype BB produces black fur,Bb brown fur,and bb 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 cross is BbCc x bbCc. The offspring will be 25% BbCc (brown fur, deformed legs), 25% Bbcc (brown fur, very short legs), 25% bbCc (white fur, deformed legs), and 25% bbcc (white fur, very short legs). Therefore, the percentage of offspring with deformed legs and white fur is 25%.

Question 7

Pseudohypertrophic muscular dystrophy is a human disorder that causes gradual deterioration of the muscles.Only boys are affected,and they are always born to phenotypically normal parents.Due to the severity of the disease,the boys die in their teens.Is this disorder likely to be caused by a dominant or recessive allele? Is its inheritance sex-linked or autosomal?

Accepted Answer

The disorder is likely caused by a recessive, sex-linked allele because it affects only boys and is passed from phenotypically normal parents, indicating it is X-linked recessive.

Question 8

A black guinea pig crossed with an albino guinea pig produced 12 black offspring.When the albino was crossed with a second black animal,6 blacks and 6 albinos were obtained.What is the best explanation for this genetic situation?

Accepted Answer

The fact that the black guinea pig crossed with an albino guinea pig produced 12 black offspring suggests that black is the dominant trait and albino is recessive. This is further supported by the fact that when the albino is crossed with a second black animal, a 1:1 ratio of black to albino offspring is obtained, which is consistent with Mendelian inheritance patterns for a single gene with two alleles.

Question 9

A man and woman are both of normal pigmentation,but both have one parent who is albino (without melanin pigmentation).Albinism is an autosomal (not sex-linked)recessive trait.What is the probability that their first female child will have albinism?

Accepted Answer

Possible gametes of the parents are:

Man: N (normal) x a (albino)
Woman: N (normal) x a (albino)

Thus, their Punnett square would be:

|   | N | a |
|---|---|---|
| N | NN | Na |
| a | Na | aa |

The probability of having an albino child would be 1/4 or 25%, which corresponds to option D.

Question 10

Phenylketonuria is an inherited disease caused by a recessive autosomal allele.If a woman and her husband are both carriers,what is the probability that their first child will be a phenotypically normal girl?

Accepted Answer

The probability of having a phenotypically normal child is 3/4 (since the child can be either homozygous dominant or heterozygous). The probability of having a girl is 1/2. Therefore, the probability of having a phenotypically normal girl is 3/4 * 1/2 = 3/8.

Question 11

Gray seed color in peas is dominant to white.Assume that Mendel conducted a series of experiments where plants with gray seeds were crossed among themselves,and the following progeny were produced:
302 gray and 98 white.(a)What is the most probable genotype of each parent? (b)Based on your answer in (a)above,what genotypic and phenotypic ratios are expected in these progeny? (Assume the following symbols:
G = gray and g = white. )

Accepted Answer

If gray seed color is dominant to white, then the only way to produce white progeny is if both parents contribute a recessive white allele. The fact that 98 out of 400 progeny (or 24.5%) are white suggests a 3:1 genotypic ratio, which is consistent with a heterozygous x heterozygous cross (i.e. Gg x Gg). Therefore, the most probable genotype of each parent is Gg. The expected genotypic ratio in the progeny is 1 GG : 2 Gg : 1 gg, which simplifies to a 1:2:1 genotypic ratio. The expected phenotypic ratio is 3 gray : 1 white, which is also consistent with a 3:1 ratio.

Question 12

A man and woman are both of normal pigmentation and have a one child out of three who is albino (without melanin pigmentation).Albinism is an autosomal (not sex-linked)recessive trait.What are the genotypes of the albino's parents?

Accepted Answer

Both parents must be heterozygous carriers of the recessive allele for albinism, as the trait is recessive and the child is albino.

Question 13

A heterozygous,but phenotypically wild-type fruit fly (gray body color and normal wings)was mated to a black fly with vestigial wings.The offspring had the following phenotypic distribution:
Wild type,720;black-vestigial,780;black-normal,280;gray-vestigial,220.What conclusion is likely from these results?

Accepted Answer

The presence of non-parental phenotypic classes (black-normal and gray-vestigial) and their relatively low frequency suggests that the black and vestigial loci are physically linked on the same chromosome. This linkage causes these alleles to be inherited together more frequently but not in complete linkage as some recombination occurs, leading to the production of the non-parental phenotypes. Therefore, option D is the most likely conclusion.

Question 14

In Drosophila melanogaster,vestigial (short)wings (vg)are caused by a recessive allele of a gene that independently assorts with a gene pair that influences body hair.Hairy (h)results in a hairy body.A cross is made between a fly with normal (wild)wings and a hairy body and a fly with vestigial wings and a normal (wild)body.The phenotypically normal F₁ flies were crossed among each other,and 1024 F₂ flies were reared.What phenotypes would you expect in the F₂,and in what actual numbers (not ratios)would you expect to find them?

Accepted Answer

The F1 flies are heterozygous for both traits (Vgvg Hh). In the F2 generation, a 9:3:3:1 phenotypic ratio is expected for a dihybrid cross with independent assortment. This translates to 576 wild, 192 vestigial, 192 hairy, and 64 vestigial hairy out of 1024 flies.

Question 15

A man and woman are both of normal pigmentation,but both have one parent who is albino (without melanin pigmentation).Albinism is an autosomal (not sex-linked)recessive trait.What is the probability that their first child will be an albino?

Accepted Answer

Both the man and woman have one parent who is albino, which means they both must be carriers of the recessive albinism allele. Therefore, their genotypes must be Aa (where A represents the dominant allele for normal pigmentation, and a represents the recessive allele for albinism).

When these two individuals have a child, there are four possible combinations of their gametes:

- The man can contribute an A allele and the woman can contribute an A allele: the child will be AA (normal pigmentation). 
- The man can contribute an A allele and the woman can contribute an a allele: the child will be Aa (normal pigmentation, but a carrier of albinism). 
- The man can contribute an a allele and the woman can contribute an A allele: the child will be Aa (normal pigmentation, but a carrier of albinism). 
- The man can contribute an a allele and the woman can contribute an a allele: the child will be aa (albino).

Of these four possibilities, there is only one where the child will be an albino (aa genotype). Therefore, the probability is 1/4 or 0.25.

Question 16

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 they will have a daughter with hemophilia?

Accepted Answer

A man with hemophilia must have the affected X allele, which he inherited from his mother since it is X-linked. A normal woman whose father had hemophilia must be heterozygous for the affected allele since she is a carrier. When they have children, their daughters will inherit one X chromosome from each parent. The daughter has a 50% chance of inheriting the affected X allele from her father and a 50% chance of inheriting the normal X allele from her mother. Therefore, the probability of their daughter having hemophilia is (1/2)x(1/2) = 1/4.

Question 17

A man has extra digits (six fingers on each hand and six toes on each foot).His wife and their daughter have a normal number of digits.Having extra digits is a dominant trait.The couple's second child has extra digits.What is the probability that their next (third)child will have extra digits?

Accepted Answer

Since having extra digits is a dominant trait and the man has extra digits while his wife does not, the man must be heterozygous (Hh) to have produced a child with a normal number of digits with his wife, who must be homozygous recessive (hh). Therefore, each child has a 1/2 chance of inheriting the dominant allele (H) from the father and thus having extra digits.

Question 18

Mendel crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F₂ generation.The results were as follows:
6022 yellow and 2001 green (8023 total).The allele for green seeds has what relationship to the allele for yellow seeds?

Accepted Answer

The fact that the F₁ generation produced all yellow seeds indicates that the allele for yellow seeds is dominant over the allele for green seeds. The fact that the F₂ generation produced both yellow and green seeds, with a ratio of approximately 3:1, indicates that green seed color is a recessive trait. Therefore, the allele for green seeds is recessive to the allele for yellow seeds.

Question 19

In tigers,a recessive allele causes a white tiger (absence of fur pigmentation).If one phenotypically normal tiger that is heterozygous is mated to another that is phenotypically white,what percentage of their offspring is expected to be white?

Accepted Answer

A heterozygous normal tiger (genotype Tt) mated with a white tiger (genotype tt) will produce offspring with a 50% chance of being white (tt) and a 50% chance of being heterozygous normal (Tt).

Question 20

In birds,sex is determined by a ZW chromosome scheme that is much like the typical XY scheme seen in humans and many other organisms,except that the system is reversed:
Males are ZZ (similar to XX in humans)and females are ZW (similar to XY in humans).A lethal recessive allele that causes death of the embryo occurs on the Z chromosome in pigeons.What would be the sex ratio in the offspring of a cross between a male heterozygous for the lethal allele and a normal female?

Accepted Answer

The male is heterozygous (ZlZ) and the female is normal (ZW). The possible offspring are ZlZ (male, lethal), ZZ (male, normal), ZlW (female, lethal), and ZW (female, normal). The lethal combinations (ZlZ and ZlW) die, leaving a 2:1 ratio of normal males (ZZ) to normal females (ZW).

Quiz 13: Mendel and the Gene

A recessive allele on the X chromosome is responsible for red-green color blindness in humans.A woman with normal vision whose father is color blind marries a color-blind male.What is the probability that this couple's first son will be color blind?

Suppose two AaBbCc individuals are mated.Assuming that the genes are not linked,what fraction of the offspring are expected to be homozygous recessive for the three traits?

Assuming independent assortment for all gene pairs,what is the probability that the following parents,AABbCc × AaBbCc,will produce an AaBbCc offspring?

Different ratios occur in crosses with single gene pairs or two gene pairs.What types of ratios are likely to occur in crosses dealing with a single gene pair?

When Mendel crossed yellow-seeded and green-seeded pea plants,all the offspring were yellow seeded.When he took these F₁ yellow-seeded plants and crossed them to green-seeded plants,what genotypic ratio is expected?

A black guinea pig crossed with an albino guinea pig produced 12 black offspring.When the albino was crossed with a second black animal,6 blacks and 6 albinos were obtained.What is the best explanation for this genetic situation?

A man and woman are both of normal pigmentation,but both have one parent who is albino (without melanin pigmentation) .Albinism is an autosomal (not sex-linked) recessive trait.What is the probability that their first female child will have albinism?

Phenylketonuria is an inherited disease caused by a recessive autosomal allele.If a woman and her husband are both carriers,what is the probability that their first child will be a phenotypically normal girl?

A man and woman are both of normal pigmentation and have a one child out of three who is albino (without melanin pigmentation) .Albinism is an autosomal (not sex-linked) recessive trait.What are the genotypes of the albino's parents?

A man and woman are both of normal pigmentation,but both have one parent who is albino (without melanin pigmentation) .Albinism is an autosomal (not sex-linked) recessive trait.What is the probability that their first child will be an albino?

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 they will have a daughter with hemophilia?

A man has extra digits (six fingers on each hand and six toes on each foot) .His wife and their daughter have a normal number of digits.Having extra digits is a dominant trait.The couple's second child has extra digits.What is the probability that their next (third) child will have extra digits?

Mendel crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F₂ generation.The results were as follows: 6022 yellow and 2001 green (8023 total) .The allele for green seeds has what relationship to the allele for yellow seeds?

In tigers,a recessive allele causes a white tiger (absence of fur pigmentation) .If one phenotypically normal tiger that is heterozygous is mated to another that is phenotypically white,what percentage of their offspring is expected to be white?