Transcript Document

Extensions of and
Deviations from Mendelian
Genetic Principles II
Copyright © 2010 Pearson Education Inc.
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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).
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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.
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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
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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.
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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.
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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 wildtype 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.
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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.
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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).
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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 solidcolored hairs.
 Ii. The B allele (encoding tyrosinaserelated 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.
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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.
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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).
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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.
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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 whiteto-green conversion, resulting
in white fruit.
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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).
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The cross of two P/p C/c :
◦ The F2 9⁄16 purple (P/_ C/_) and
7⁄
3
3
16 white ( ⁄16 P/_ c/c : ⁄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
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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.
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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.
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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.
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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.
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Extranuclear genes display non-Mendelian
inheritance, which has four characteristics:
◦ a. Typical Mendelian ratios do not occur, because meiosisbased 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.
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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.
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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.
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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).
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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 selffertilization is not possible.
◦ c. The answer is Rf (restorer of fertility) genes, where the
dominant Rf allele overrides CMS but the recessive rf
cannot.
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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.
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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.