Chapter 4: Extensions of Mendelian Genetics

Question 1

True or False?

In a cross between two strains that are true breeding for purple and white flowers, the F2 phenotypic ratio would be ¼ purple, ¼ white, and ½ lavender if the flower color phenotype exhibits incomplete dominance.

Answer 1

True

The heterozygous offspring exhibit a phenotype that is intermediate between the parents.

Question 2

Which of the following statements about an individual with the Bombay phenotype is correct?

Answer 2

The individual lacks the enzyme required to produce the H substance.

Individuals with the Bombay phenotype are homozygous hh and do not make the enzyme required to produce the H substance. Thus, they cannot modify the H substance to produce the antigens on which type A and B blood phenotypes are based.

Question 3

What is the expected phenotypic ratio of a cross between a disc-shaped squash that is heterozygous at both loci and a long squash?

Answer 3

4⁄16 disc, 8⁄16 sphere, 4⁄16 long

The genotypic ratio for this cross is 4⁄16 AB, 4⁄16 Ab, 4⁄16 aB, and 4⁄16 ab, making the phenotypic ratio 4⁄16 disc, 8⁄16 sphere, 4⁄16 long.

Question 4

How many different phenotypes are possible in a one gene/three allele system that displays codominance to each other?

Answer 4

6

With three alleles, a1, a2, a3, each unique combination of two alleles results in a distinct phenotype. Possible combinations are:
a1/a1
a1/a2
a1/a3
a2/a2
a2/a3
a3/a3

Question 5

Which of the following is the most likely example of codominance?

    A pure-breeding tall plant is crossed to a pure-breeding short plant. All their progeny are of medium height.
A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have pink flowers.
A plant with red flowers is crossed to a plant with white flowers. Half their progeny have red flowers, the other half have white flowers.
A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have flowers with some red patches and some white patches.

Answer 5

A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have flowers with some red patches and some white patches.

Heterozygotes express both alleles.

Question 6

The color dilution gene in horses is an example of incomplete dominance.

Answer 6

True

Question 7

In mice, agouti fur is a dominant trait. A mouse with solid color fur is the recessive phenotype (A = agouti; a = solid color).

A separate gene, which is not linked to the agouti gene, can result in either a dominant black pigment or a recessive brown pigment (B = black; b = brown).

A litter of mice from the mating of two agouti black parents includes offspring with the following fur colors:
solid color, black
solid color, brown
agouti black
agouti brown

What would be the expected frequency of agouti brown offspring in the litter?

Answer 7

3/16

Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/4, and the probability of bb offspring is 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.

Question 8

In addition to A and a, the agouti gene has a third allele, AY.
The AY allele is dominant to both A and a.
The homozygous genotype (AYAY ) results in lethality before birth.
The heterozygous genotypes (AYA or AYa) result in yellow fur color, regardless of which alleles are present for the B/b gene. (This effect exhibited by the AY allele is known as epistasis–when the expression of one gene masks the expression of a second gene.)

In a mating of mice with the genotypes AYaBb x AYaBb , what is the probability that a Live-Born offspring will have yellow fur?

Answer 8

2/3

Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.

Question 9

In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse.

Answer 9

Because the C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally.

AaBbcc-Albino,
AaBBCC-Agouti Black
Aabbcc-Albono
AAbbCc Agouti Brown
aaBbCc-Solid color, Black
AABBcc-Albino

Question 10

piebaldism

Answer 10

Piebaldism is a rare autosomal dominant disorder of melanocyte development. Common characteristics include a congenital white forelock, scattered normal pigmented and hypopigmented macules and a triangular shaped depigmented patch on the forehead.

Although piebaldism may visually appear to be partial albinism, it is a fundamentally different condition. The vision problems associated with albinism are not usually present as eye pigmentation is normal. Piebaldism differs from albinism in that the affected cells maintain the ability to produce pigment but have that specific function turned off. In albinism the cells lack the ability to produce pigment altogether.

Question 11

In the same mouse species, a 4th unlinked gene (gene P/p) also affects fur color.
For mice that are either homozygous dominant (PP) or heterozygous (Pp), the organism’s fur color is dictated by the other three genes (A/a, B/b, and C/c).
For mice that are homozygous recessive (pp), large patches of the organism’s fur are white. This condition is called piebaldism.

In a cross between 2 mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur?

Answer 11

9/256

Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BBor Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.

Question 12

White eye color is an X-linked trait in one line of fruit flies. White eyes is recessive to red eyes. If a red-eyed female and a white-eyed male are crossed, _______.

Answer 12

some of their male progeny may have white eyes

If the female is heterozygous, approximately half of the male progeny will have white eyes.

Question 13

TRUE or FALSE?

For X-linked traits in Drosophila, the male phenotype is determined by the maternally inherited allele.

Answer 13

True

Males inherit only one X chromosome. That chromosome is contributed by the female parent.

Question 14

Eye color in Drosophila is an X-linked trait. White eyes is recessive to red eyes. If a Drosophila male has white eyes, which of the following must also be true?

Answer 14

His mother had at least one white allele.

Because this male had white eyes, he must have inherited a white allele from his mother.

Question 15

The inheritance of eye color in Drosophila is controlled by genes on each of the fly’s four chromosome pairs. One eye-color gene is on the fly’s X chromosome, so the trait is inherited in a sex-linked manner. For this sex-linked trait, the wild-type (brick red) allele is dominant over the mutant vermilion (bright red) allele.
A homozygous wild-type female fly is mated with a vermilion male fly.

Predict the eye colors of F1 and F2 generations. (Assume that the F1 flies interbreed to produce the F2 generation.)

Answer 15

Both F1 females and F1 males will have the wild-type eye color, but because the trait is sex-linked, F1 females will be heterozygous for the trait.
As a result, F2 males have a 1/2 chance of inheriting each allele, and thus of having that eye color. F2 females, on the other hand, will all have the wild-type eye color because they inherit the dominant allele on the X chromosome from their fathers (the F1 males).

Question 16

In cases involving complete dominance, homozygous dominant individuals cannot be distinguished from heterozygous individuals by phenotype alone. Therefore, this phenotypic category contains two genotypes.

Answer 16

In cases involving incomplete dominance, each genotype has a distinct phenotype.

Question 17

Like incomplete dominance, codominance produces a distinct phenotype for each of the three genotypes in a monohybrid cross.

Answer 17

Unlike incomplete dominance, however, codominance results in both alleles fully expressing their phenotype in the heterozygote.

Question 18

Is the condition recessive or dominant?

Answer 18

Dominant conditions require that an affected individual have at least one affected parent.
An affected offspring of two unaffected parents indicates a recessive condition.

Question 19

Is the condition autosomal or X-linked?

Answer 19

An affected female offspring of two unaffected parents indicates an autosomal recessive condition. If the condition were X-linked, the male parent would be affected.
To distinguish between rare X-linked and autosomal recessive conditions, choose the simpler mode for the observed pattern—that is, the mode that requires fewer unrelated individuals to carry a rare allele. In this example, the autosomal recessive mode would require two unrelated horses to carry a rare recessive allele for condition B, whereas the X-linked recessive mode requires only one. Therefore, the X-linked recessive mode is a better choice.

Question 20

A pedigree provides a pictorial representation of the genetic relationships in a family and shows each person’s sex and phenotype with respect to the trait in question. In this family, Leah is the only family member (on both sides) that has galactosemia.

Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia?

Answer 20

autosomal recessive

Because neither Jane nor John has the same condition as their daughter, and there is no evidence of sex-linkage, galactosemia must be an autosomal recessive trait.

Question 21

If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactose- and galactose-free diet as soon as possible after birth. Even on this diet, affected individuals may still suffer from learning disabilities, ovarian failure (in young women), late-onset cataracts, and early death.
Which of the following tests would be most useful for Jane and John to have?

Answer 21

newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn’s blood)

Chromosome analysis will not identify this abnormality because it is caused by a mutation in a single gene.

Newborn screening by either of these methods would be effective in identifying the condition in a newborn. It would provide definitive information early enough to begin dietary restrictions, if needed.

Question 22

The study of modern genetics began with Thomas Hunt Morgan in the early 1900s. Morgan’s experiments with the fruit fly Drosophila were the first to demonstrate that genes are located on chromosomes and are the basis of heredity.

One of Morgan’s first challenges was to find fruit flies with mutant phenotypes. He discovered a single male fly with white eyes, instead of the usual (wild-type) red eye color. Through his genetic experiments with the white-eyed fly, Morgan deduced that a fruit fly’s eye color was somehow linked to its sex.

Reciprocal crosses
When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait’s inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds, and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds.

Unlike Mendel, however, Morgan obtained very different results when he carried out reciprocal crosses involving eye color in his fruit flies. The diagram below shows Morgan’s reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male.

Answer 22

When Morgan crossed a homozygous red-eyed female with a white-eyed male, all of the offspring had red eyes. However, in the reciprocal cross (homozygous white-eyed female with a red-eyed male), all of the females had red eyes while all of the males had white eyes.

Question 23

When a homozygous red-eyed female was crossed with the white-eyed male (w+w+ × wY), the resulting F1 females were w+w and the F1 males were w+ Y. Crossing the F1 males and F1 females would yield these results:

Answer 23

All the F2 females would have red eyes, although some would be homozygous (w+w+ ) and others would be heterozygous (w+w).

Half the F2 males would have red eyes (w+ Y), and half would have white eyes (wY).

Question 24

Use these rules when assigning genotypes for X-linked recessive conditions:

Answer 24

Males have only one allele for every X-linked gene. Affected males have the recessive allele, and non-affected males have the dominant (wild-type) allele.
Unaffected females with affected sons are heterozygous (carriers).
Unaffected females with affected fathers are heterozygous.
All sons of affected females will be affected.
All daughters of affected females will be heterozygous

Question 25

Use these rules when assigning genotypes for autosomal recessive conditions:

Answer 25

Unaffected parents with an affected offspring are both heterozygous.
Unaffected offspring of one affected parent will be heterozygous.
Unaffected offspring of two heterozygous parents may be homozygous dominant or heterozygous.

Question 26

Alterations of expected Mendelian frequencies

“Exceptions” to the rule…

Answer 26

Multiple alleles

Partial (incomplete) dominance

Codominance

X-linked genes

Question 27

ABO Blood Types

Answer 27

Codominant

Blood types: A, B, AB, O

Phenotype determined by Isoagglutanin alleles

IA produces A antigen on cell surface
IB produces B antigen on cell surface
IO produces no antigen

IA and IB behave dominant to IO, but codominant to each other

Transfusion with incompatible blood type:
Immune system antibodies will react to non-self antigens, coagulate the transfused blood: potentially lethal!!

Question 28

The biochemical basis of the ABO blood groups.

Answer 28

The H allele, present in almost all humans, directs the conversion of a precursor molecule to the H substance by adding a fucose molecule to it. The IA and IB alleles are then able to direct the addition of terminal sugar residues to the H substance.

The IO allele is unable to direct either of these terminal additions. Gal: galactose; AcGluNH: N-acetyl glucosamine; AcGalNH: N-acetylgalactosamine.

Failure to produce the H substance results in the Bombay phenotype where individuals are type O, regardless of the presence of an IA or IB allele.

Question 29

moiety

Answer 29

The definition of moiety in chemistry is a part of a functional group of a molecule or a whole functional group of a molecule. Moiety refers to specific groups of atoms or atomic bonds within a molecule.

Moiety is used synonymously with the term “functional group.” It refers to the structure of a molecule. A molecule is made up of a number of atoms and bonds. Each grouping of atoms within the molecule is a moiety or functional group. Each moiety is responsible for the chemical and physical characteristics of the molecule. Each moiety or functional group also has an impact on the characteristics and reactivity of neighboring moieties.

Moiety is an extremely important concept in chemistry. Because atoms within a molecule can be broken down into moieties, it is possible to predict the impact each moiety has on surrounding moieties. Breaking molecules down into moieties and studying the reactivity and chemical reactions of these functional groups also allows scientists to create new compounds with beneficial moieties to produce the desired outcome. In drug research, for example, altering one or more moieties within a molecule may result in a reduction of side effects or a more effective drug therapy.

Question 30

Isoagglutanin genotypes

Answer 30

IAIO----> Blood type A
IAIA----> Blood type A
IBIO----> Blood type B
IBIB----> Blood type B
IAIB----> Blood type AB
IOIO----> Blood type O

Question 31

Universal Donor Blood Type:

Answer 31

O is the Universal Donor

O has no antigens, will not cause immune reaction in A, B, or AB

Question 32

Universal Acceptor

Answer 32

AB is the Universal Acceptor

AB blood type does not have anitbodies to A or B

Question 33

Incomplete dominance

Answer 33

Neither allele is dominant, both alleles expressed

Eg., phenotype of flowers are of intermediate color

Examples of incomplete dominance are rather rare

Question 34

Other genes can affect ABO phenotype:

A case of epistasis

Answer 34

ABO blood types= codominance

Other genes influence blood antigen expression, like the Bombay phenotype, genetically B, phenotypically O.

Question 35

“H” substance

Answer 35

A & B antigens are carbohydrate groups added to “H” substance

hh mutants do not have H substance, cannot add A or B antigen, makes genetically A, B, or AB individuals look O phenotypically

Cross of IAIB Hh x IAIB Hh reveals altered phenotypic ratios

Example: Functionally, her ABO blood group behaves as type O. Genetically, she is type B.

Question 36

Bombay

Answer 36

Mutant hh individuals

Produces NO H substance

Bombay individuals have antibodies to O (= the H substance)

Can only receive blood from Bombay individuals

Question 37

Follow 2 characters altered ratios

Answer 37

Albinism in mice (typical monohybrid inheritance pattern, 3:1 ratio)

Blood type (codominant)

What is the phenotypic result?
Not the expected 9:3:3:1…

Question 38

Coat color in mice

Answer 38

Wild type color is agouti: A allele

Mutant allele AY produces yellow mice in heterozygotes, therefore AY is dominant

However, when AY is homozygous, it is lethal (recessive lethality)

Results in altered phenotypic ratios

the normal wild-type agouti allele (A); the mutant yellow allele (AY) . The mutant allele behaves dominantly to the normal allele in controlling coat color, but it also behaves as a homozygous lethal allele. The genotype AYAY does not survive.

Is AY a dominant or a recessive allele?
It depends on the phenotype that we are looking at:
AY is dominant for yellow fur color, but AY recessive for lethality!
This is called Pleiotropy

Question 39

pleiotropy

Answer 39

Condition in which a single mutation causes multiple phenotypic effects.

Pleiotropy occurs when one gene influences multiple, seemingly unrelated phenotypic traits

Example: Phenylketonuria is a human disease that affects multiple systems but is caused by one gene defect.

Consequently, a mutation in a pleiotropic gene may have an effect on some or all traits simultaneously. Pleiotropic gene action can limit the rate of multivariate evolution when natural selection, sexual selection or artificial selection on one trait favours one specific version of the gene (allele), while selection on other traits favors a different allele.

The underlying mechanism of pleiotropy in most cases is the effect of a gene on metabolic pathways that contribute to different phenotypes. Genetic correlations and hence correlated responses to selection are most often caused by pleiotropy.

Question 40

Labrador coat color genetics: Epistasis

Answer 40

Coat colors: Black, chocolate brown, golden yellow

Color depends on assortment of 2 independently assorting coat color genes

B is dominant, determines black if present

bb recessive is brown

Second gene, dominant E allele has no effect on black or brown

However, recessive homozygote ee in black or brown yields golden yellow

Phenotypic Results:

9: B- E- (Black)
3: bb E- (Brown)
4: -- ee (Yellow)

Question 41

epistasis

Answer 41

Nonreciprocal interaction between nonallelic genes such that one gene influences or interferes with the expression of another gene, leading to a specific phenotype.

Question 42

Recessive epistasis

Answer 42

Phenotypic Results: 9:3:4

the allele causing the epistasis is recessive. The recessive ee homozygote is considered epistatic to any allelic combination at the first gene

Take home point: Altered Mendelian phenotypic ratios because of epistatic effects

Question 43

Penetrance

Answer 43

percentage of individuals that show some degree of expression of the mutant genotype

The frequency, expressed as a percentage, with which individuals of a given genotype manifest at least some degree of a specific mutant phenotype associated with a trait.

Example: In Huntingtion’s disease, all with the dominant mutant gene get the disease: Penetrance is 100%

Question 44

Expressivity

Answer 44

The degree to which a phenotype for a given trait is expressed.

Observed range of the mutant phenotype

Example: trisomy 21 (Down Syndrome) has variable expressivity in observed phenotype

Question 45

Sex-limited and Sex-Influenced Inheritance

Answer 45

Autosomal genes, but expression dependent on hormonal constitution of the individual

Question 46

Sex-limited expression

Answer 46

phenotype is absolutely limited to one sex

Example: Tail and neck plumage in domestic fowl

Question 47

Sex-Influenced expression:

Answer 47

sex of an individual influences the expression of a phenotype

Example: Pattern baldness in humans