Question 1

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

D

Question 2

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

E

Question 3

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 offspring did not follow a 9:3:3:1 phenotypic ratio, indicating that the black and vestigial loci are not assorting independently, and the fact that the black and vestigial traits are found together in some offspring shows that they are not allelic. Therefore, the only possible explanation is that the two loci are linked.

Question 4

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 allele for green seeds is recessive to the allele for yellow seeds, as it only appears in the F₂ generation when both alleles are present in a homozygous recessive genotype. This is supported by the ratio of 3:1 for yellow to green offspring in the F₂ generation, which is characteristic of a monohybrid cross with a dominant/recessive relationship between the alleles.

Question 5

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 determine the possible alleles from each parent:

AABbCc 
Possible alleles: ABc, Abc

AaBbCc 
Possible alleles: ABc, Abc, aBc, abc

To get an AaBbCc offspring, we need to get at least one A from the first parent, one B from either parent, and one C from either parent.

Probability of getting an A from the first parent: 
1 (AA is homozygous dominant)

Probability of getting a B from either parent: 
3/4 (Bb or bB)

Probability of getting a C from either parent: 
3/4 (Cc or cC)

Multiplying these probabilities together gives: 
1 × (3/4) × (3/4) = 9/16

Therefore, the probability of producing an AaBbCc offspring is 9/16, which is answer choice C.

Question 6

A man and woman are both of normal pigmentation and have a child 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) for the recessive allele in order to have a child with albinism, which is a recessive trait.

Question 7

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 offspring can be: 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 females (ZW).

Question 8

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

The man and woman are both carriers of the albino gene, since they each have an unaffected parent and an albino parent. This means that each of them has a 1/2 chance of passing on the recessive allele to their offspring. The probability of their child inheriting the recessive allele from each parent is 1/2 x 1/2 = 1/4. Therefore, the probability of their first child inheriting the recessive allele from both parents and being an albino is 1/4 or 25%.

Question 9

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 10

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 parent has one copy of each gene, with one dominant allele and one recessive allele. To have a homozygous recessive individual, both alleles must be recessive. 
For one trait, the probability of an offspring being homozygous recessive is 1/4 (1/2 x 1/2). For three traits, the probability is (1/4)^3 = 1/64, which is option D.

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

The most probable genotype of each parent is Gg, as this is the only genotype combination that can produce the observed phenotypic ratio of 3:1 (gray:white) in the offspring. 
When two heterozygous parents are crossed (Gg x Gg), the expected genotypic ratio is 1:2:1 (GG:Gg:gg) and the expected phenotypic ratio is 3:1 (gray:white). Therefore, option B is the best choice.

Question 12

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 wings and a hairy body and a fly with vestigial wings and a normal 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 (vg+ vg, h+ h). In the F2 generation, the expected phenotypic ratio for two independently assorting genes is 9:3:3:1. Therefore, out of 1024 flies, you would expect approximately 576 wild, 192 vestigial, 192 hairy, and 64 vestigial hairy.

Question 13

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 between brown rabbits with deformed legs (BbCc) and white rabbits with deformed legs (bbCc) will produce offspring with the following genotypes:BbCc (brown fur, deformed legs)Bbcc (brown fur, very short legs)bbCc (white fur, deformed legs)bbcc (white fur, very short legs)Therefore, 25% of the offspring would be expected to have deformed legs and white fur.

Question 14

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 the albino guinea pig produced all black offspring suggests that black is dominant and albino is recessive. In the second cross, when the albino was crossed with a second black animal, 6 blacks and 6 albinos were obtained, indicating that the black animal in this cross was heterozygous (Bb). The cross can be represented as follows:

Bb (black) x aa (albino) = B_ (black offspring)

B_ (heterozygous black) x aa (albino) = 1 BB (black), 2 Bb (black), 1 bb (albino)

This is consistent with the inheritance pattern of a dominant/recessive trait.

Question 15

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

Accepted Answer

If both tigers are heterozygous (Ww), they have a 25% chance of producing offspring with the homozygous recessive genotype (ww), which results in white fur. Therefore, 25% of their offspring is expected to be white.

Question 16

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 because she has a normal vision. Her father, who is color blind, can only pass on the recessive allele for color blindness to his daughter. The male is color-blind, which means he has two copies of the recessive allele. The Punnett square for the cross is as follows: | | X^N | X^n | | - | - | - | | Y | XY | X_nY | | Y | XY | X_nY | where X^N and X^n represent the normal and color blindness alleles, respectively. 50% of the male offspring will inherit the X^n allele from the mother and the Y chromosome from the father, resulting in color blindness. Therefore, the answer is B) 50%.

Question 17

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

Crosses dealing with a single gene pair typically follow a 3:1 or 1:2:1 ratio, depending on whether the gene is dominant or recessive, respectively.

Question 18

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

The man and woman are both carriers of the recessive gene for albinism, since they are of normal pigmentation despite having one albino parent each. When they have a child, each of them will pass on one of two possible alleles for the albinism gene, either "A" for normal pigmentation or "a" for albinism. The chances of passing on an "a" allele are 1/2 for each parent. To determine the probability of their first female child having albinism, we can use a Punnett square:

|  A      |  a
    ---|--------|--------
     A |  AA     |  Aa
    ---|--------|--------
     a |  Aa     |  aa
     
The possible genotypes for the child are AA (normal pigmentation), Aa (carrier of albinism gene), and aa (albino). The chance of a child being albino (having the aa genotype) is 1/4 or 25%. Therefore, the probability of their first female child having albinism is 1/4 or 0.25, which is closest to answer choice D (1/4).

Question 19

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 20

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.

Quiz 13: Mendel and the Gene

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?

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?

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?

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

A man and woman are both of normal pigmentation and have a child 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?

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?

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?

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?

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

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?

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?

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?

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?