Chapter 4: Extensions Mendelian Genetics

A loss-of-function mutation __________.

results in an increased quantity of the normal gene product

is the wild-type allele

results in reduced or no production of the gene product

is a neutral mutation

Results in reduced or no production of the gene product.

In incomplete dominance, heterozygotes have a phenotype that is __________.

intermediate between the two homozygous phenotypes

more environmentally influenced than the phenotypes of the homozygotes

indistinguishable from one of the homozygous phenotypes

characteristic of both homozygous phenotypes

Intermediate between the two homozygous phenotypes

A loss-of-function mutation __________.

results in an increased quantity of the normal gene product

is the wild-type allele

results in reduced or no production of the gene product

is a neutral mutation

Results in reduced or no production of the gene product.

In incomplete dominance, heterozygotes have a phenotype that is __________.

intermediate between the two homozygous phenotypes

more environmentally influenced than the phenotypes of the homozygotes

indistinguishable from one of the homozygous phenotypes

characteristic of both homozygous phenotypes

Intermediate between the two homozygous phenotypes

In the ABO blood types, all of the following offspring are commonly possible EXCEPT__________.

an individual with type B blood from a mother with type A and a father with type AB

an individual with type AB blood from a mother with type B and a father with type AB

an individual with type A blood from a mother with type O and a father with type AB

an individual with type O blood from a mother with type A and a father with type AB

an individual with type O blood from a mother with type A and a father with type AB

In a cross between two strains that are true breeding for red and white flowers, the flower color trait exhibits incomplete dominance. If these strains are crossed, what F₂ phenotypic ratio would be produced?

1/4 red; 1/4 white; and 1/2 pink

100 percent red; 0 percent white

3/4 red;1/4 white

1/2 red; 1/2 white

Phenotypic ratio:

• RR: Red

• RW: Pink

• WW: White

• So the ratio is 1 RR (red) : 2 RW (pink) : 1 WW (white).

• This translates to 1/4 red, 1/2 pink, and 1/4 white.

T/F: Mendelian crosses involving alleles at two segregating loci will always produce a 9:3:3:1 phenotypic ratio.

False; The 9:3:3:1 ratio is conditional:

• This ratio is specifically observed in dihybrid crosses (involving two genes) where:

  • Both genes exhibit complete dominance.

  • The genes assort independently (meaning they are not linked).

A mother who is blood type AB has a child who is AB also. A potential father is blood type O. What would a well-informed geneticist conclude?

Hint: There are three alleles that directly determine blood type; the alleles A and B allow production of cell-surface receptor proteins (A and B) or the allele O results in no production of surface proteins. A rare Bombay phenotype is an exception.

He is very likely to be the father.

He cannot be the father.

He might be the father, but it is unlikely.

He or any other man of blood type O could be the father.

He might be the father, but it is unlikely

The child has type AB blood, which has the genotype IAIB. Therefore, the potential father with type O blood could not have contributed either the IA or IB allele and thus cannot be the father.

In a plant species, if the B allele (blue flowers) and the b allele (white flowers) are incompletely dominant (B b is light blue), what offspring ratio is expected in a cross between a blue-flowered plant and a white-flowered plant?

1/4 blue:1/2 light blue:1/4 white

1/2 blue:1/2 white

3/4 blue:1/4 white

1/3 blue:1/3 light blue:1/3 white

all light blue

All light blue

F1 generation:

• BB x bb

• All offspring will have the genotype Bb.

• Since Bb results in light blue flowers, all offspring will be light blue.

If a mother has type A blood and her son has type O blood, what are the possible blood types of her son’s father?

types B or O

types A, B, or O

type O only

any blood type

Type A, B, or O

For the son to have type O blood (ii), he must have received an 'i' allele from each parent. Since the son received an 'i' allele from his mother, we know the mother's genotype must be IAi.

Now, let's consider the father's possible genotypes:

• He must have an 'i' allele to pass to his son.

• Therefore, his genotype could be:

  • ii (type O)

  • IAi (type A)

  • IBi (type B)

FRQ: In a disputed parentage case, the child is blood type O, while the mother is blood type A. What blood type would exclude a male from being the father? Would the other blood types prove that a particular male was the father? 

The father could not have an AB blood type. This is because the mother's blood type indicates that she could have either an AA or Ai genotype. While the child carriers an ii genotype, so they have to inherit one i allele from each parent. This means that the father cannot have blood type AB because that would mean they could only give an A or B allele. Whereas the child requires an O allele from both parents. 

Yes, the other blood types would prove that a particular male was the father. For example, if the father has the same blood type as the child (O), this could be a strong sign that he is the father. 

In this pedigree of the Bombay phenotype, the arrow points to the proband and demonstrates that A B O blood type ________.

demonstrates codominance

demonstrates incomplete dominance

demonstrates complete dominance

is due to multiple alleles

is dependent on another gene

Is dependent on another gene

The Bombay phenotype is a classic example of epistasis, where one gene masks the expression of another. In this case, the H gene is epistatic to the ABO gene.

• The H gene is required for the production of the H antigen, which is a precursor to the A and B antigens.

• Individuals with the Bombay phenotype have a homozygous recessive mutation (hh) at the H gene.

• As a result, they cannot produce the H antigen, even if they have the IA or IB alleles.

• This means that their ABO genotype is masked, and they phenotypically appear to have type O blood, even if their genotype indicates otherwise.

The color of sunflower pollen can be orange, yellow, or white-cream. Crosses reveal that true-breeding orange is dominant to both yellow and white-cream. However, a cross between true-breeding white-cream and yellow plants produced F1 plants with orange pollen. The F2 populations of this cross resulted in a ratio of nine orange: four white-cream: three yellow. What type of inheritance do these results suggest?

dependent assortment

codominance

recessive epistasis

dominant epistasis

incomplete dominance

Recessive epistasis

• The 9:3:4 ratio is a classic indicator of recessive epistasis.

• Orange is dominant to both yellow and white-cream.

• White-cream x yellow = orange F1.

• F2 ratio: 9 orange, 4 white-cream, 3 yellow.

• In recessive epistasis, a homozygous recessive genotype at one locus masks the expression of alleles at another locus.

You have read that in guinea pigs, black fur is dominant to white fur. You are surprised then when your two white guinea pigs have all black F1. When you cross the F1, you observe 9:7 black:white. How do you best explain this?

Lethal alleles

Two genes are responsible for the trait

Codominance

Sex-limited expression

Incomplete dominance

Two genes are responsible for the trait

• A 9:7 ratio is a classic deviation from the standard 9:3:3:1 dihybrid ratio.

• This indicates that two genes are involved in the fur color trait, and they exhibit epistasis.

• Why the other options are incorrect:

  • Lethal alleles: Lethal alleles would result in a different ratio, often with a missing class of offspring.

  • Codominance: Codominance would result in a blended or combined phenotype, not a 9:7 ratio.

  • Sex-limited expression: Sex-limited expression would mean the trait is only expressed in one sex, not a specific phenotypic ratio.

  • Incomplete dominance: Incomplete dominance would result in a blended phenotype, not a 9:7 ratio.

The following interactions are found in mice. The albino allele, b, eliminates all color in homozygotes and is recessive to the normal pigment allele, B, which yields a black color. A second allele, A, allows the black pigment to be converted to agouti, whereas this does not occur in the homozygous recessive condition in which the mouse is homozygous for the null, a allele. A mouse of genotype Aabb will __________.

be an albino

have a spotted pattern

have a black coat

have an agouti coat

Be an albino

• bb: This homozygous recessive genotype means the mouse is albino. The "b" allele eliminates all color expression.

• The "A" allele is irrelevant due to the homozygous recessive "bb" combination.

Therefore, a mouse with genotype Aabb will:

• be an albino

A ratio of 12:3:1 is most characteristic of a cross involving __________.

four gene pairs

three gene pairs

two gene pairs

one gene pair

Two gene pairs

• Let's say gene 1 has alleles A and a, and gene 2 has alleles B and b.

• If the 'A' allele is dominant epistatic to the 'B' gene, then any genotype with 'A' will have the same phenotype, regardless of the 'B' alleles.

• This leads to a modified dihybrid ratio, not the standard 9:3:3:1.

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

 

Hint: Disc-shaped squash have dominant alleles (one or two copies) at both loci, whereas long squash have all recessive alleles at both loci.

9/16 disc, 6/16 long, 1/16 sphere

4/16 disc, 8/16 sphere, 4/16 long

4/16 disc, 8/16 long, 4/16 sphere

9/16 disc, 6/16 sphere, 1/16 long

4/16 disc, 8/16 sphere, 4/16 long

A pleiotropic mutation __________.

is also known as a conditional mutation

is a mutation that causes multiple phenotypic effects on the organism

is expressed only in the homozygous recessive condition

is expressed more frequently in males than in females

is a mutation that causes multiple phenotypic effects on the organism

A woman is heterozygous for the recessive X-linked gene for Lesch-Nyhan syndrome. What proportion of her daughters will be carriers for the trait if their father is NOT affected?

75 percent

50 percent

0 percent

25 percent

50%

Proportion of Daughters as Carriers:

• The daughters are XX and Xx.

• Only Xx daughters are carriers.

• There is 1 Xx out of 2 daughters (XX and Xx).

• Therefore, 1/2 or 50% of the daughters will be carriers.

Sex-influenced inheritance differs from sex-limited inheritance in which of the following ways?

Sex-influenced inheritance is X-linked, but sex-limited inheritance is not.

These are both terms for the same thing.

Expression of the phenotype occurs only in one sex in sex-limited inheritance, but not in sex-influenced inheritance.

Sex-limited traits occur only in males.

Expression of the phenotype occurs only in one sex in sex-limited inheritance, but not in sex-influenced inheritance.

• Sex-limited inheritance:

  • The expression of the trait is absolutely limited to one sex or the other.

  • The genes responsible for the trait may be autosomal or sex-linked, but the phenotype manifests only in one sex.

  • Example: Milk production in mammals is limited to females.

• Sex-influenced inheritance:

  • The expression of the trait is influenced by the sex of the individual.

  • The trait can occur in both sexes, but the expression may differ between males and females.

  • Often, this involves differences in dominance between males and females.

  • Example: Pattern baldness in humans, where the allele acts dominantly in males and recessively in females.

FRQ:  In the guinea pig, one locus involved in the control of coat color may be occupied by any of four alleles: C (full color), ck (sepia), cd (cream), or ca (albino), with an order of dominance of: C > ck > cd > ca. (C is dominant to all others, ck is dominant to cd and ca, but not C, etc.) In the following crosses, determine the parental genotypes and predict the phenotypic ratios that would result:

 

(a) sepia X cream, where both guinea pigs had one albino parent

 

(b) sepia X cream, where the sepia guinea pig had an albino parent and the cream guinea pig had two sepia parents

 

(c) sepia X cream, where the sepia guinea pig had two full-color parents and the cream guinea pig had two sepia parents

 

(d) sepia X cream, where the sepia guinea pig had  full-color parent and an albino parent and the cream guinea pig had two full-color parents 

FRQ: Below are three pedigrees. For each trait, consider whether it is consistent with X-linked recessive inheritance. Indicate why or why not for each case.

(a). This is consistent with X-linked recessive inheritance. The male would contribute to an X chromosome carrying the a mutation to the aa daughter. The mother would have to be a carrier.

(b). This is consistent with X-linked recessive inheritance. This is because the mother could potentially act as a carrier and pass the a allele to one son and her A allele to the other.  

(c). This is not consistent with X-linked recessive inheritance. The carrier mother cannot have an affected son.

FRQ: Contrast penetrance and expressivity as the terms relate to phenotypic expression.

Penetrance is when a gene is the probability that a gene will be expressed in an individual. For example, if a gene is 100% penetrance, then every individual will show that same trait. 

Expressivity is when a phenotype is expressed by individuals having a particular genotype. This specifically refers to the degree in which a trait is expressed in a specific individual who has that gene. This varies from penetrance because even though individuals have the same gene it can be expressed in various different ways. For example, in terms of genetic disorders some people may have more severe symptoms while others may present milder. 

---

Penetrance refers to the proportion of individuals with a particular genotype who actually express the associated phenotype. It is usually expressed as a percentage.

• Complete Penetrance: When all individuals with a specific genotype express the expected phenotype, the trait is said to have complete penetrance.

• Incomplete Penetrance: When some individuals with a specific genotype do not express the expected phenotype, the trait exhibits incomplete penetrance.

Example: If a genetic disorder has 80% penetrance, it means that 80% of individuals with the genetic mutation will exhibit symptoms of the disorder, while 20% will not.

Expressivity refers to the range of phenotypes expressed by individuals with a particular genotype. It describes the variation in phenotypic expression among individuals with the same genotype.

Example: In a condition with variable expressivity, individuals with the same genetic mutation might exhibit symptoms ranging from mild to severe.

• Penetrance is about whether the phenotype is expressed (and how much in %) at all, while expressivity is about the extent of the phenotype that is expressed.

These concepts help explain why individuals with the same genetic mutation can sometimes show different physical traits or symptoms.

FRQ: Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where the color can be red, pink (the heterozygote), or white. The second pair leads to personate (dominant) or peloric (recessive) flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed to those that are white, peloric, and dwarf. Determine the F1 genotype(s) and phenotype(s). If the F1 plants are interbred, what proportion of the offspring will exhibit the same phenotype as the F1 plants? 

Three independently assorting characteristics are being dealt with: (1) flower color (incomplete dominance), (2) flower shape (dominant/recessive), and (3) plant height (dominant/recessive).

 

Establish appropriate gene symbols:

Flower color: RR = red; Rr = pink; rr = white Flower shape: P = personate; p = peloric Plant height: D = tall; d = dwarf

RRPPDD X rrppdd results in 

RrPpDd (pink, personate, tall) Use components of the forked-line method as follows:

 

2/4 pink X 3/4 personate X 3/4 tall = 18/64