Bombay.html: 04_03-Bombay.jpg
The Bombay phenotype is a rare mutation in the FUT1 gene that prevents the formation of the H substance, resulting in an apparent O phenotype.

The proband (first occurrence of a phenotype in a pedigree) exhibits an apparent O phenotype, even though she is genetically type B.

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Drosophila-X-crossB.html: 04_12-Drosophila-X-crossB.jpg

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In Drosophila, the wild-type red eye color is dominant to white eyes.

Reciprocal crosses between white-eyed and red-eyed flies yielded different results.

The results can best be explained if the white (w) locus is X-linked and is present on the X rather than an autosome. Males carry only one allele for X-linked genes and are called hemizygous.

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In the presence of the wild-type bw⁺ (brown) allele, the scarlet pigment drosopterin is synthesized.
In the presence of the wild-type st⁺ (scarlet) allele, the brown pigment xanthommatin is synthesized.
Heterozygotes carrying both pigments yield red eyes; homozygous recessives lack the pigments and yield white eyes.

H_substance-blood_ABO.html: 04_06-H_substance-blood_ABO.jpg
Even though only a single character was followed, the phenotypic ratio was expressed in sixteenths, indicating that a second gene pair is interacting with that controlling the A and B antigens during the expression of this phenotype.

H_substance.html: 04_06-H_substance.jpg
The Bombay phenotype is an example of epistasis in which the homozygous recessive condition at one locus masks the expression of a second. The hh individual is phenotypically type O regardless of the I genotype.

The Mendelian dominance of the h locus and the multiple alleles inheritance pf ABO blood types yield offspring ratios that also differ from classic 9:3:3:1 dihybrid ratio.

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X-linkage exhibits a crisscross pattern of inheritance. Recessive X-linked alleles such as color blindness are passed from homozygous mothers to all sons. Heterozygous females are carriers and pass the allele to half of their sons, who develop the disorder. Affected fathers pass the allele to his daughters, who are usually carriers.

X-linked.html: 04_T03-X-linked.jpg

agouti-alleles.shtml:
 

	Gene B		Gene A
Precursor	↓	Black	↓	Agouti
Molecule	→	Pigment	→	Pattern
(colorless)	B -		A -

In the presence of a B allele, black pigment can be made from a colorless substance. In the presence of an A allele, the black pigment is deposited during the development of hair in a pattern producing the agouti phenotype.
If the aa genotype occurs, all of the hair remains black.
If the bb genotype occurs, no black pigment is produced, regardless of the presence of the A or a alleles, and the mouse is albino.
Therefore, the homonzygous b genotype masks or suppresses the expression of the A allele; this is referred to as recessive epistasis.

agouti-pigment.shtml:
The wild type agouti (gray-brown) phenotype is caused by yellow pigment deposited in a band on a black hair shaft.

The A^Y mutation is a deletion spanning the regulatory region for the yellow pigment and extending into an adjacent gene (Merc), which is critical to embryonic development.

The "loss of function" in a Merc mutant causes A^Y/A^Y homozygotes to die before birth.

Heterozygotes deposit yellow pignment along the entire length of hair shafts.

agouti.html: 04_04-agouti.jpg
Inheritance patterns involving the normal agouti allele (A) and the mutant yellow allele (A^Y) in the mouse.

The mutation is a homozygous recessive lethal since the genotype A^Y/A^Y does not survive. But it is dominant to the wild type agouti allele (A) in producing yellow coat color in a heterozygote.

agouti.shtml:

Crosses
(A)	agouti	X	agouti	→	all agouti
(B)	yellow	X	yellow	→	2/3 yellow: 1/3 agouti
(C)	agouti	X	yellow	→	1/2 yellow: 1/2 agouti

Inheritance patterns in three crosses involving the normal agouti allele and the mutant yellow allele in the mouse.

albinism-blood_ABO.html: 04_05-albinism-blood_ABO.jpg
Instead of the dihybrid cross yielding the four phenotypes in the Mendelian 9:3:3:1 ratio, six phenotypes occur in a 3:6:3:1:2:1 ratio due to the two modes of inheritance of the two loci.

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Albinism exhibits Mendelian dominance, while ABO blood types are determined by multiple alleles with AB codominance. The forked-line method can be used to calculate the predicted offspring of two humans who are both heterozygous for the recessive albinism allele and who are both of blood type AB.

blood-MN.shtml:

The MN Blood Group
Genotype	M molecules	N molecules	Phenotype
LM LM	+	-	M
LM LN	+	+	MN
LN LN	-	+	N
LM LN   X   LM LN ↓ 1/4 LM LM 1/2 LM LN 1/4 LN LN

The MN blood group exhibits codominance; a heterozygote expresses both allele products.

The phenotype is determined by the production of two specific molecules located on the surface of red blood cells.

Crossing the heterozygote results in a 1:2:1 genotype and phenotype ratio.

blood_ABO-alleles.html: 04_02-blood_ABO-alleles.jpg
The I^A allele produces an enzyme that adds the sugar AcGalNH to the H substance to produce the A antigen.
The I^B allele produces an enzyme that adds a galactose sugar to produce the B antigen.
The O phenotype results from an absence of either sugar, and may be due to a mutation in the FUT1 locus.

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The I^A and I^B alleles are codominant. Blood type is one tool for excluding parenthood of individuals; for example, a type O person should not have a parent who is type AB.

blood_ABO.shtml:

The ABO Blood Groups
Genotype	Antigen	Phenotype
IAIA	A	A
IAIO	A
IBIB	B	B
IBIO	B
IAIB	A, B	AB
IOIO	Neither	O

Each individual is A, B, AB, or O phenotype as a result of dominance of the I^A and I^B alleles to the I^O allele and codominance of the I^A and I^B alleles to each other.

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A pea combed chicken crossed to a rose combed rooster produces F₁ that are all walnut combed.

The F₂ phenotype ratios are 9 walnuts, 3 roses, 3 peas, and 1 single comb, indicating that two pairs of genes control the comb shape of chickens.

complementary_gene_interaction.shtml:

P1: AAbb X aaBB      white  white
↓
F1: All AaBb (purple)
↓
F2 ratio	Genotype	Phenotype	Final Phenotypic ratio
9/16	A- B-	purple	9/16 purple
3/16	A- bb	white	7/16 white
3/16	aa B-	white
1/16	aa bb	white

Complementary Gene Interaction

A cross between two strains of white-flowered sweet peas, the F₁ plants are all purple, and the F₂ occurred in a ratio of 9/16 purple to 7/16 white.

The presence of at least one dominant allele of each of two gene pairs is required for purple flowers.

Sometimes gene interaction can yield novel phenotypes in the F₂ generation.

epistasis-dominant.shtml:

F1: AaBb X AaBb
↓
F2 ratio	Genotype	Phenotype	Final Phenotypic ratio
9/16	A- B-	white	12/16 white    3/16 yellow    1/16 green
3/16	A- bb	white
3/16	aa B-	yellow
1/16	aa bb	green

Dominant Epistasis

The dominant allele A results in white fruit color of the summer squash regardless of the genotype at a second locus, B.

In the absence of a dominant A allele (the aa genotype), BB or Bb results in yellow color, while bb results in green color.

Crossing two white-colored double heterozygotes (AaBb) produces a phenotypic ratio of 12:3:1.

epistasis-recessive.shtml:

F1: AaBb X AaBb
↓
F2 ratio	Genotype	Phenotype	Final Phenotypic ratio
9/16	A- B-	agouti	9/16 agouti   4/16 albino   3/16 black
3/16	A- bb	albino
3/16	aa B-	black
1/16	aa bb	albino

Recessive Epistasis

In a cross between agouti (AABB) and albino (aabb), the F₁ are all AaBb and have agouti coat color.

The F₂ produced from crossing these double heterozygotes are shown at right.

In the presence of a B allele, black pigment is made. In the presence of an A allele, the black pigment is deposited on hair in a pattern producing the agouti phenotype.

In the aa genotype, all of the hair remains black. In the bb genotype, no black pigment is produced; this genotype masks or suppresses the expression of the A allele, and the mouse is black.

epistasis.html: 04_07-epistasis.jpg
Epistasis results in modified dihybrid ratios in a cross between individuals heterozygous in two genes.
Examples of epistasis include recessive and dominant epistasis, and complementary gene interaction.

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Genomic Imprinting

The mouse Igf2 gene produces a growth factor, and homozygous mutants are dwarf mice.

Heterozygotes that receive the normal allele from their father are normal in size.

Heterozygotes that receive the normal allele from their mother, which has been imprinted, are dwarf.

The normal Igf2 gene is imprinted to function poorly during the course of egg production, but functions normally when it has passed through sperm-producing tissue in males. This process may involve DNA methylation.

incomplete_dominance-snapdragon.html: 04_01-incomplete_dominance-snapdragon.jpg
Incomplete Dominance

Neither allele is dominant. The F₁ generation exhibits an intermediate phenotype.

The genotypic ratio (1:2:1) of the F₂ generation is identical to that of Mendel's monohybrid cross, but the phenotypic ratio is identical to the genotypic ratio.

Superscripts are used to denote the red and white alleles as R¹ and R². Other ways of representing these alleles include W¹ and W², or C^W and C^R.

The expression of these phenotypes are due to the production of a red pigment by the R enzyme; the R² allele is unable to produce the pigment, and a heterozygote produces half as much pigments as the homozygous R¹.

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Some examples of epistasis with modified F₂ dihybrid ratios.

novel_phenotype.shtml:

P1: AABB X AABB     disc   long
↓
F1: All AaBb (disc)
↓
F2 ratio	Genotype	Phenotype	Final Phenotypic ratio
9/16	A- B-	disc	9/16 disc   6/16 sphere   1/16 long
3/16	A- bb	sphere
3/16	aa B-	sphere
1/16	aa bb	long

Novel Phenotypes

In this example of gene interaction, both gene pairs influence fruit shape equally. When a summer squash with disc-shaped fruit (AABB) are crossed with plants with long fruit (aabb), the heterozygous F₁ all have disc fruit.

However, in the F₂ progeny, some individuals show a novel shape, sphere, if it possesses a dominant allele at either locus.

In the absence of dominant alleles, the fruit is long. if both dominant alleles are present, the fruit is flattened into a disc shape.

position_effect.html: 04_17-position_effect.jpg
In the Drosophila white locus, the w⁺/w genotype normally results in a wild-type red eyes.

However, if the wild-type w⁺ allele is translocated to a heterochromatic region (where gene expression is often inhibited), the eyes are mottled with red and white patches (variegated), reflecting intermittent expression of the dominant w⁺ allele due to a position effect.

sex-influenced_baldness.shtml:

Genotype	Phenotype
BB	Bald	Bald
Bb	Not bald	Bald
bb	Not bald	Not bald

Sex-influenced Pattern Baldness

Pattern baldness (where the hair is very thin or absent on the top of the head) is much more prevalent in males, since the heterozygous genotype exhibits this phenotype only in males.

Females who do inherit the BB genotype exhibit a much less pronounced phenotype and express it later in life.

sex-limited_feathers.shtml:

Genotype	Phenotype
H H	Hen feathered	Hen feathered
H h	Hen feathered	Hen feathered
h h	Hen feathered	Cock feathered

Sex-limited Feathers

In domestic fowl, tail and neck plumage is controlled by a single autosomal locus whose expression is modified by the individual's sex hormones: cock feathering is longer and pointed than hen feathering.

Leghorn hens who are hh can be induced to produce cock feathering (at the next molt) by removing the ovaries, indicating that female sex hormones have an inhibitory effect on cock feathering.

temperature.html: 04_18-temperature.jpg
Many chemical reactions are affected by the temperature, which can then influence phenotype.

Siamese cats and Himalayan rabbits exhibit dark fur in regions where the body temperature is cooler; probably because the pigment-producing enzyme is temperature-sensitive, being more active at the lower temperatures in the extremities.

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Variable expressivity in the eyeless mutation in Drosophila.

Environmental factor often affect gene expression. This can be quantified by two measures.

Expressivity of a genotype measures the range of expression of the mutant genotype. Flies homozygous for the recessive eyeless allele show phenotypes that range from the normal eyes to a partial reduction in size to the complete absence of one or both eyes, so the expressivity of eyeless ranges from complete loss of both eyes to completely normal eyes.

Penetrance measures the percentage of individuals that show at least some expression of the mutant genotype. For example, if 15% of mutant flies show the wild-type appearance, the mutant allele has a penetrance of 85%.

white_locus.html: 04_T02-white_locus.jpg
Eye color in Drosophila is controlled by over 100 alleles. In this locus, eye color ranges from complete absence of pigment in the white (w) allele to a buff color in the white-buff (Wbf) allele, in which the amount of pigment in the eyes is reduced to less than 20 percent of that found in the red wild-type eye.