Extensions of and Deviations from Mendelian Genetic Principles II

Copyright © 2010 Pearson Education Inc.

Relationship between phenotype and gene can be studied through mutants identified by phenotype distinct from wild type.

Complementation test (cis-trans test) determines whether independently isolated mutations for the same phenotype are in the same or different genes by crossing two mutants.◦ a. If mutations are in different

genes, phenotype will be wild type (complementation).

◦ b. If mutations are in the same gene, phenotype will be mutant (no complementation).

Wild-type body color is grayish yellow.

If two true-breeding mutant black-bodied strains are crossed, all F1 are wild type.◦ a. Genes are e (ebony) and b

(black). Black parents are homozygous mutant but in different genes (e/e b+/b+) and (e+/e+ b/b).

◦ b. F1 are heterozygous at both loci (e+/e b+/b) and therefore wild type, showing complementation.

Phenotypes result from complex interactions of molecules under genetic control. Genetic analysis can often detect the patterns of these reactions. For example:◦ a. In the dihybrid cross of independently

assorted genes A/a B/b x A/a B/b, nine genotypes will result.

◦ b. If each allelic pair controls a distinct trait and exhibits complete dominance, a 9:3:3:1 phenotypic ratio results.

◦ c. Deviation from this ratio indicates that interaction of two or more genes is involved in producing the phenotype

Two types of interactions occur:◦ a. Different genes control the same general trait,

collectively producing a phenotype.◦ b. One gene masks the expression of others

(epistasis) and alters the phenotype. Examples here are dihybrid, but in the “real

world” larger numbers of genes are often involved in forming traits.

The molecular explanations offered here are currently hypothetical models and await rigorous analysis using the tools of molecular biology.

Nonallelic genes that affect the same characteristic may interact to give novel phenotypes, and often modified phenotypic ratios. Examples include:◦ a. Eye color in Drosophila, in

which 2 loci (bw and st) are involved. i. At least one wild-type allele for each

locus must be present to produce wild-type red eyes. (1) The bw locus encodes a red

pigment. (2) The st locus encodes a brown

pigment. ii. Flies mutant for both bw and st have

white eyes.

In epistasis, one gene masks the expression of another, but no new phenotype is produced.◦ a. A gene that masks another is epistatic.◦ b. A gene that gets masked is hypostatic.

Several possibilities for interaction exist, all producing modifications in the 9:3:3:1 dihybrid ratio:◦ a. Epistasis may be caused by recessive alleles, so that

a/a masks the effect of B (recessive epistasis).◦ b. Epistasis may be caused by a dominant allele, so that

A masks the effect of B.◦ c. Epistasis may occur in both directions between genes,

requiring both A and B to produce a particular phenotype (duplicate recessive epistasis).

In recessive epistasis, A/_ b/b and a/a b/b have the same phenotype, producing an F2 ratio of 9:3:4. Examples include:◦ a. Coat color determination

in rodents: i. Wild mice have individual hairs with

an agouti pattern, bands of black (or brown) and yellow pigment. Agouti hairs are produced by a dominant allele, A (agouti signal protein). Mice with genotype a/a do not produce yellow bands and have solid-colored hairs.

Ii. The B allele (encoding tyrosinase-related protein 1) produces black pigment, while b/b mice produce brown pigment. The A allele is epistatic over B and b, in that it will insert bands of yellow color between either black or brown.

Iii. The C allele encodes a tyrosinase, required for development of any color at all, and so it is epistatic over both the agouti (A) and the pigment (B) gene loci. A mouse with genotype c/c will be albino, regardless of its genotype at the A and B loci.

The A locus encodes agouti signaling protein, expressed in only some tissues. Its product is secreted and involved in regulating melanin production.

The B locus encodes tyrosine-related protein, involved in coloration of skin, hair, and irises of the eyes.

The C locus encodes tyrosinase, and inactive forms of the gene result in oculocutaneous albinism.

Coat color determination in labrador retriever dogs◦ i. Gene B/_ makes black pigment, while b/b makes

brown.◦ ii. Another gene, E/_, allows expression of the B

gene, while e/e does not. The E locus encodes the melanocortin 1 receptor, a regulator of hair and skin color.

◦ Iii. Genotypes and their corresponding phenotypes:

(1) B/_ E/_ is black. (2) b/b E/_ is brown

(chocolate). (3) _/_ e/e produces yellow with

nose and lips either dark (B/_ e/e) or pale (b/b e/e).

In dominant epistasis A/_ B/_ and A/_ b/b have the same phenotype, producing an F2 ratio of 12:3:1. Examples:a. Summer squash fruit have three common colors: white, yellow, and green.i. Yellow is recessive to white but dominant to green.ii. Gene pairs are W/w and Y/y.(1)W/_ are white no matter the genotype of the other locus.(2) w/w are yellow in Y/_ and green in y/y.

The postulate is that a white molecule is converted to a green intermediate and then to a yellow end product.◦ (1) Dominant Y required to

convert green intermediate to yellow.

◦ (2) Dominant W inhibits white-to-green conversion, resulting in white fruit.

Duplicate recessive epistasis (complementary gene action) involves two loci that each produce an identical phenotype, 9:7 ratio. Flower color determination in sweet peas.◦ a. Purple is dominant for flower color,

true-breeding purple plant is crossed with a true-breeding white one, the F2 shows a typical 3:1 ratio.

◦ b. White strains usually breed true, but occasionally the cross of two different white strains (p/p C/C x P/P c/c) will produce an F1 that is entirely purple (P/p C/c).

The cross of two P/p C/c :◦ The F2 9⁄16 purple (P/_ C/_) and

7⁄16 white (3⁄16 P/_ c/c : 3⁄16 p/p C/_: 1⁄16 p/p c/c).

All of the white F2 plants will breed true, as will 1⁄9 of the purple F2 plants (P/P C/C).

The C/c alleles determine whether the flower can have color, and the P/p alleles determine whether purple is produced

A form of nonallelic interaction that is milder than epistasis involves modifier genes, which may either enhance or reduce phenotypic expression of another gene.

A modifier that shifts the phenotype of a mutant allele of another gene toward wild-type is a suppressor gene.

An example is coat color in cats and rodents.◦ a. D is a dominant allele of the dense pigment gene

involved in pigment deposition in hairs (B/_ D/_; black cat). ◦ b. Recessive homozygotes (d/d) have a lightened color

(B/_ d/d; gray cat) Symptoms of human genetic diseases may be

affected by modifier genes. Examples include:◦ a. The variable phenotypes of cystic fibrosis. ◦ b. The severity of hearing loss associated with mutations

in the cadherin 3 gene is affected by the V586M gene.

Extranuclear DNA is found in mitochondria and chloroplasts. Genes include rRNA for ribosomes of the organelles, tRNAs, and a few organelle proteins.

In many organisms, extranuclear genes inheritance is maternal. This differs from maternal effect in two ways:◦ a. Extranuclear inheritance is determined by genes

in an organelle, while maternal effect derives from nuclear genes.

◦ b. Extranuclear genotype matches individual’s phenotype, while in maternal effect the individual’s phenotype results from its mother’s genes.

Mitochondria, found in all aerobic eukaryotic cells, oxidize pyruvate (from glycolysis) to CO2 and water, producing ATP.◦ a. Mitochondrial genomes are usually circular supercoiled

dsDNA molecules; linear genomes are found in some protozoa and fungi.

◦ b. Mitochondrial genome encodes rRNAs, tRNAs, and some mitochondrial proteins. Other mitochondrial proteins are encoded in the nuclear genome.

Chloroplasts, found in green plants and photosynthetic protists, perform photosynthesis, converting water and CO2 into carbohydrate using light energy. ◦ a. All known chloroplast genomes are circular supercoiled

dsDNA.◦ b. Chloroplast genome encodes rRNAs, tRNAs, and some

chloroplast proteins. Other chloroplast proteins are encoded in the nuclear genome.

Extranuclear genes display non-Mendelian inheritance, which has four characteristics:◦ a. Typical Mendelian ratios do not occur, because

meiosis-based segregation is not involved.◦ b. Reciprocal crosses usually show uniparental

inheritance. All progeny have the phenotype of one parent, generally the mother because the zygote receives nearly all of its cytoplasm (including organelles) from the ovum.

◦ c. Extranuclear genes cannot be mapped to chromosomes in the nucleus.

◦ d. If a nucleus with a different genotype is substituted, non-Mendelian inheritance is unaffected.

Leber’s hereditary optic neuropathy (LHON, OMIM 585000). Optic nerve degeneration results in complete or partial blindness in midlife adults.◦ LHON is caused by mutations in mtDNA genes for electron

transport chain proteins. Kearns-Sayre syndrome (OMIM 530000) produces

three types of neuromuscular defects:◦ i. Progressive paralysis of certain eye muscles.◦ ii. Abnormal pigment accumulation on the retina,

causing chronic inflammation and degeneration of the retina.

◦ iii.Heart disease.

Hybrid crops are agriculturally important, exhibiting heterozygote superiority (heterosis), and require controlled crosses by plant breeders.

Corn was the first hybrid crop plant because hybrids can be produced easily by manual emasculation (detasseling) and fertilization. In other species, mutations causing male sterility are used for emasculation.◦ a. Nuclear gene mutations produce genic male

sterility.◦ b. Extranuclear gene mutations produce cytoplasmic

male sterility (CMS).

CMS (defective pollen formation) is produced by a mitochondrial mutation.◦ a. Plant mitochondria are entirely maternal, and so

when a CMS plant is the female parent, hybrid seed produce progeny plants that are male sterile.

◦ b. This poses a problem for fertilization, since the self-fertilization is not possible.

◦ c. The answer is Rf (restorer of fertility) genes, where the dominant Rf allele overrides CMS but the recessive rf cannot.

Hybrid seeds are generated using [CMS] rf/rf females and [CMS] Rf/rf males. Hybrids segregate 1 Rf/rf : 1 rf/rf.◦ a. Rf/rf are male fertile

due to overriding by Rf.◦ b. rf/rf are male sterile but

may be fertilized by pollen from Rf/rf plants in the field.

When the female gamete contributes most of the cytoplasm, maternal inheritance is the usual explanation for extranuclear mutations. However, exceptions occur. ◦ a. PCR analysis shows heteroplasmy in mice, with

paternal mtDNA present at a frequency of 1024 relative to maternal mtDNA. This heteroplasmy may facilitate recombination between the mtDNAs, creating more diversity in mtDNA than previously believed.

◦ b. In plants, the angiosperms show variation in plastid inheritance; most inherit only maternal plastids, but others inherit from both parents, or from the paternal parent. Paternal inheritance is also found in gymnosperms.