Transcript CHAPTER 23 Non-Mendelian Inheritance

Peter J. Russell
A molecular Approach 2nd Edition
CHAPTER 23
Non-Mendelian Inheritance
edited by Yue-Wen Wang Ph. D.
Dept. of Agronomy,台大農藝系
NTU
遺傳學 601 20000
Chapter 23 slide 1
Origin of Mitochondria and Chloroplasts
1. Endosymbiosis is believed to account for mitochondria
and chloroplasts.
a. Mitochondria appear to be derived from a photosynthetic purple
nonsulfur bacterium that entered a eukaryotic cell about a billion
(109) years ago. They provide oxidative phosphorylation to the
cell.
b. Chloroplasts appear to derive from entry of a photosynthetic
cyanobacterium.
2. Many proteins of both mitochondria and chloroplasts are
encoded by nuclear genes, indicating that genes have
moved from the organelles to the nuclear DNA.
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Chapter 23 slide 2
Organization of Extranuclear Genomes
Mitochondrial Genome
1. Mitochondria perform cellular respiration after the cytosolic glycolysis step
(Figure 23.1). They contain enzymes that include::
a. Pyruvate dehydrogenase.
b. Electron transport and oxidatiyphosphory1ation enzymes.
c. Citric acid cycle enzymes.
d. Fatty acid oxidation enzymes
2. Mitochondrial genomes (mtDNA) are sequenced for several species.
a. Many are circular, double-stranded and supercoiled (Figure 23.2). Linear genomes
occur in mitochondria of some protozoa and fungi.
b. GC content of mtDNA often differs from nuclear DNA, allowing separation by CsCl
density gradient centrifugation.
c. Mitochondrial DNA lacks histone-like proteins.
d. Multiple genomic copies are in multiple nucleoid regions within the mitochondrion.
e. Gene content is conserved across species, but size of the mtDNA varies widely:
i. Animal mtDNA is less than 20 kb (human is 16,569 bp). Essentially all of this
DNA encodes products.
ii. Yeast mtDNA is about 80 kb, and not all of this DNA encodes products.
iii. Plant mtDNA ranges from 100 kb to 2 million base pairs. Not all encodes
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Chapter 23 slide 3
products.
3. Replication of mtDNA is semi-conservative, uses
mitochondrial DNA polymerases and RNA
primers, and involves no proofreading.
a. Replication of mtDNA occurs throughout the cell
cycle. (Contrast with nuclear DNA, which replicates
only in S phase.)
b. Both strands of mtDNA in most animals replicate in a
continuous manner, with replication of one strand
initiating well before the other (Figure 23.3).
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Chapter 23 slide 4
Fig. 23.3 Model for mitochondrial DNA replication that involves the formation of a D
loop structure
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 23 slide 5
4. Mitochondria are not synthesized de novo, but arise from growth and division of
preexisting mitochondria (Luck, 1963). Some of their genetic information is in
the mitochondrial chromosome, the rest in the nuclear DNA.
a. The human mtDNA map (Figure 23.4) contains information for:
i. tRNAs.
ii. rRNAs.
iii. Some polypeptide subunits of cytochrome oxidase, NADHdehydrogenase, and
ATPase.
b. Nuclear DNA encodes other mitochondrial components, including:
i. DNA polymerase and other replication proteins.
ii. RNA polymerase and other transcription proteins.
iii. Ribosomal proteins, translation factors, aminoacyl tRNA synthetases.
iv. The other subunits of cytochrome oxidase, NADH-dehydrogenase and ATPase.
(Figure 23.5)
c. The mtDNA has genes on both DNA strands. They were identified in two ways:
i. Computer-based search for ORFs (also called URFs, unidentified reading
frames).
ii. Aligning 5’ and 3’ sequences of mitochondrial mRNAs with the corresponding
mtDNA sequence.
iii. All known human mtDNA ORFs have been assigned a function
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Chapter 23 slide 6
Fig. 23.4 Map of the genes of human mitochondrial DNA
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Chapter 23 slide 7
Fig. 23.5 Synthesis of the multisubunit protein cytochrome oxidase takes place on
both cytoplasmic and mitochondrial ribosomes
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 8
5. Mitochondrial ribosomes translate mRNAs from the mitochondrial chromosome
within the organelle. For example, human mitochondrial ribosomes:
a. Have two subunits, 45S and 35S, forming a 60S mitochondrial ribosome.
b. Have only two rRNAs, 16S rRNA in the large subunit, and 12S rRNA in the small
subunit.
c. Usually have only one gene for each rRNA in the mtDNA.
d. Ribosomal proteins are usually encoded by nuclear DNA, and move into the mitochondria
from the cytoplasm.
6. Mammalian mtDNA is transcribed into a single large RNA molecule that is cleaved
to produce mRNAs, tRNAs and rRNAs. These RNAs are then processed:
a. mRNAs receive 3’ poly(A) tails.
b. tRNAs receive 3’ CCA sequences.
c. Mitochondrial mRNAs have no 5’ caps.
7. The much larger mitochondrial genomes of yeast and plants differ from animal
mitochondria:
a. tRNA genes do not separate genes, and other DNA sequences signal transcription
termination.
b. Large gaps occur between genes in the mtDNA.
c. Introns are found in mitochondrial genes from these organisms, but never in animal
mtDNA genes.
d. mtDNA sequences encoding some mRNAs do not include a complete stop codon. Instead,
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Chapter 23 slide 9
the 3’ end is U or UA, and the poly(A) tail completes
stop601
codon
(UAA).
8. Translation of mitochondrial mRNAs is distinct from
cytoplasmic translation:
a. Mitochondrial mRNAs do not have a 5' cap.
i. Yeast and plant mRNAs have a 5' leader sequence, and so
initiation can occur at the first AUG codon.
ii. Animal mitochondria lack the leader sequence, and so initiation
must occur in a unique way.
b. In translation, mitochondria have similarities to bacteria:
i. Both use fMet-tRNA in initiation.
ii. Mitochondrial initiation (IF), elongation (EF) and release (RF)
factors are distinct from those in the eukaryotic cytoplasm.
iii. mt ribosomes are sensitive to the same agents as bacterial
ribosomes (e.g., streptomycin, neomycin, chloramphenicol).
iv. mt ribosomes are insensitive to agents that inactivate eukaryotic
cytoplasmic ribosomes (e.g., cycloheximide).
v. These sensitivities are used in research to determine which
proteins derive from nuclear DNA, and which from mtDNA.
c. Only plant mitochondria use the “universal” genetic code. Others
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(Table
23.1). Chapter 23 slide 10
9. Analysis of mtDNA can reveal genetic relationships. In humans:
a. There is a 400-bp polymorphic region.
b. Mitochondria are maternally inherited, and so maternal-line relations can be
analyzed using PCR.
c. An example involves a woman's claim be Princess Anastasia of Russia, sole
member of the Royal Romanov family to survive the Bolshevik Revolution
(1917).
i. Remains of the executed royal family matched mtDNA of living
members related through maternal lines.
ii. The woman's mtDNA did not match the mtDNA of living members
related through maternal lines. Thus, she was not Anastasia.
d. Conservation biology also uses mtDNA台大農藝系
analysis to
determine
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Chapter 23 slide 11
variability in populations (e.g., grizzly bears in Yellowstone Park).
Chloroplast Genome
1. Chloroplasts have a double membrane, internal lamellar structure containing
chlorophyll, and protein-rich stroma. Chloroplasts divide and grow in the same
way as mitochondria.
2. The chloroplast genome (cpDNA) is not as well characterized as mtDNA, but
some things are known:
a. Structurally, cpDNA is similar to mtDNA. It is dsDNA, a super- coiled circle lacking
structural proteins.
b. The GC content of cpDNA often differs from both nuclear and mtDNA, allowing
separation on CsCl density gradients.
c. The size of cpDNA varies from 80 kb-600 kb. All chloroplast genomes carry
noncoding DNA.
d. Each chloroplast has multiple copies of cpDNA in several nucleoid regions. An
example is Chiamydomonas with 500-1,500 cpDNA molecules per chloroplast.
3. Nuclear genes encode some chloroplast components, while cpDNA genes
(which may include introns) encode the rest, including (Figure 23.7):
a. Two copies of each chloroplast rRNA (16S, 23S, 4.5S and 5S). The copies are
included with other genes in inverted repeats designated IRA and IRB that define
the short (SSC) and long (LSC) single copy regions of the cpDNA.
b. The tRNAs (30 in tobacco and rice, 32 in the liverwort Marchantia).
c. Almost 100 highly conserved ORFs. Approximately 60 are now correlated with
proteins needed for chloroplast transcription,台大農藝系
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and photosynthesis.
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Chapter 23 slide 12
Fig. 23.7 Organizations of the chloroplast genome of rice (Oryza sativa)
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 23 slide 13
4. Chloroplasts have 70S ribosomes consisting of 50S and 30S subunits.
a. The 50S subunit includes a 23S, 5S and 4.5S rRNA.
b. The 30S subunit includes a 16S rRNA.
c. Ribosomal protein number is unclear. Some proteins are encoded in nuclear DNA,
others in cpDNA.
d. Translation is similar to prokaryotes:
i. Initiation uses fMet-tRNA.
ii. Chloroplast-specific initiation (IF), elongation (EF) and release (RF) factors are
used.
iii. The universal genetic code is used.
e. Chloroplast ribosomes have the same antibiotic inhibition profile as mitochondria,
and can be studied in the same way. Ribulose bisphosphate decarboxylase is an
example:
i. This enzyme controls the first step in photosynthetic fixation of carbon. It is the
most prevalent protein on earth.
ii. The enzyme contains four identical small peptides (encoded by nuclear DNA)
and four identical large peptides (encoded by cpDNA).
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Chapter 23 slide 14
Rules of Non-Mendelian Inheritance
1. 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, with all
progeny having 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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Chapter 23 slide 15
Examples of Non-Mendelian Inheritance
Shoot Variegation in the Four O’Clock
Animation: Shoot Variegation in the Four O’Clock
1. Variegated-shoot phenotype in four o’clocks involves non-Mendelian inheritance of
chloroplasts in the shoots (stem, leaves and flowers).
a. Green shoots have normal chloroplasts.
b. White shoots have only leucoplasts, which lack chlorophyll, and are incapable of
photosynthesis.
c. Variegated shoots received both chloroplasts and leucoplasts, which segregated during
cell division. Progeny cells are therefore green or white, in a variegated (mixed) pattern
(Figure 23.9).
2. Results of crosses between plants with shoots that are variegated illustrate this
phenomenon (Table 23.2):
a. When ova are from green plants, only green progeny result, regardless of pollen source.
b. When ova are from white plants, only white progeny result (but soon die from lack of
chlorophyll), regardless of pollen source.
c. When ova are from variegated plants, all three types of progeny result, regardless of
pollen source.
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3. Shoot color in these plants therefore shows a pattern of maternal inheritance.
There are three assumptions in the model:
a. Pollen contributes no chloroplasts or leucoplasts to the zygote.
b. The chloroplast genome replicates autonomously, so that progeny plastids retain the
same color phenotype as the original plastid.
c. Segregation of plastids during eukaryotic cell division is random, providing some
offspring cells with chloroplasts, some with leucoplasts, and some with a mixture.
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Chapter 23 slide 17
Fig. 23.8 Variegation in the four o’clock
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 23 slide 18
Fig. 23.9 Model for the inheritance of shoot color in the four o’clock
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 19
The [poky] Mutant of Neurospora
1.
2.
3.
Neurospora crassa is an aerobe, and so requires mitochondrial functions to grow. The [poky] mutation in mtDNA
has an altered cytochrome complement, leading to slow growth of the fungus.
a.
Normal Neurospora have cytochromes a + a3, b and c.
b.
Neurospora with the [poky] mutation lack a + a3 and b, and have an excess of c.
Experimental crosses in Neurospora involve fusion of nuclei from mating type A and a parents. Crosses can occur
two ways:
a.
Place both parents on medium at the same time.
b.
Inoculate one parent onto medium, and add the second parent several days later. The first parent produces all the protoperithecia
(fruiting bodies containing the ascospores).
Protoperithecia have much more cytoplasm than conidia (asexual spores).
a.
The strain producing protoperithecia is similar to the female parent.
b.
The second strain, which contributes conidia, is analogous to the male parent.
c.
Assigning strains these roles allows reciprocal crosses to be made (Figure 23.10).
i. Protoperithecia from [poky] parent, and wild-type conidia results in all [poky] progeny.
ii. Protoperithecia from wild-type parent, and [poky] conidia results in all wild-type progeny.
iii. Results show maternal inheritance.
d.
Tetrad analysis allows correlation of meiotic events with spore organization within the ascus:
i. Protoperithecia from [poky] parent, and wild-type conidia results in all [poky] progeny (an 8:0 ratio).
ii. Protoperithecia from wild-type parent, and [poky] conidia results in all wild-type progeny (a 0:8 ratio).
iii. In the same experiments, nuclear genes segregate at a 4:4 ratio.
e.
The [poky] allele results from a 4-bp deletion in the promoter for 19S rRNA in the mtDNA, resulting in deficiency
of small ribosomal subunits and greatly decreased mitochondrial protein synthesis.
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Chapter 23 slide 20
Fig. 23.10 Results of reciprocal crosses of [poky] and normal (wild-type) Neurospora
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 21
Yeast petite Mutants
1. Yeast can grow either anaerobically by fermentation (slow growth) or aerobically using
mitochondria (fast growth), forming colonies from single cells on solid media.
2. Yeast petite colonies are much smaller than those formed by wild-type cells, due to
cytochrome deficiencies that prevent aerobic respiration.
a. On a medium that supports only aerobic respiration, petite cells are unable to grow.
b. The spontaneous mutation rate is 0.1–1%, but exposure to an intercalating agent (e.g., ethidium
bromide) raises the rate to 100%.
c. This allows isolation of different petite cell lines, containing different mutations.
3. Yeast crosses between petite and wild-type cells (a X α crosses) determine the
mechanism of inheritance for this phenotype.
a. The zygote formed from mating is grown into a colony to check its phenotype, and when it
sporulates by meiosis, the tetrad of ascospores can also be grown into colonies for phenotype
analysis.
b. Some petite X wild-type crosses give 2:2 segregation (wild-type:petite).
i. This is the same ratio as seen in nuclear genes, so these petite mutants are nuclear
(segregational) petites, written pet- (Figure 23.11).
ii. The cross in this case was pet- X pet+. Diploid was pet-/pet+ (hence wild-type) and the
spore tetrad contained 2 pet- and 2 pet+ spores.
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Chapter 23 slide 22
Fig. 23.11a Inheritance of yeast petite mutants
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 23
c. Another class of petite mutants is the neutral petites ([rho-N]).
i. When crossed with wild-type ([rho-N] X [rho+N]) produce wild-type diploids ([rhoN]/[rho+N]) and spores that segregate 0:4 (no petite : 4 wild-type).
ii. This is an example of uniparental (not maternal, since gametes are same size)
inheritance.
iii. In [rho-N] mutants, nearly 100% of the mtDNA is missing, and so mitochondrial
functions are also missing.
iv. Spores produce only wild-type colonies because normal mitochondria from the wildtype parent provide normal mitochondria for the progeny. The petite trait thus is lost after
one generation.
d. Most petite mutants are of the suppressive ([rho-S]) type. They differ from neutral petites by
having an effect on the wild-type, although both are mutations in mtDNA.
i. A [rho+/rho-S] diploid has a respiratory-deficient phenotype, and if it divides mitotically
the progeny will nearly all be petites.
ii. Sporulation of the petite [rho+/rho-S] diploid produces tetrads with a 4:0 (petite : wildtype) ratio.
iii. Sporulation of the rare wild-type [rho+/rho-S] diploid produces tetrads with a 0:4
(petite : wild-type) ratio.
iv. Suppressive petite mutants start with deletions in mtDNA. The amount of mtDNA is
restored by duplications of existing mtDNA, often creating gene deletions and
rearrangements that cause deficiencies in the enzymes for aerobic respiration.
v. The suppressive effect over normal mitochondria might result from either:
(1)Faster replication of the mutant mitochondria, outcompeting wild-type, or
(2)Fusion with normal mitochondria and recombination between [rho -S] mtDNA and
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Chapter 23 slide 24
wild-type mtDNA.
Fig. 23.11b Inheritance of yeast petite mutants
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 25
Fig. 23.11c Inheritance of yeast petite mutants
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 26
Non-Mendelian Inheritance in Chlamydomonas
1.
The motile haploid algae Chlamydomonas reinhardtii have a single chloroplast with many copies of cpDNA.
a.
Its mating types are mt+ and mt-. Mating is syngamous (zygote is formed by fusion of equal-sized cells).
b. Zygote forms a thick-walled cyst, and then produces four haploid cells by meiosis. Mating type results from a nuclear
gene, and thus segregates 2:2 (mt+: mt-).
2.
A chloroplast-encoded trait in Chlamydomonas is erythromycin resistance ([eryr]), with wild-type cells
sensitive to the antibiotic ([erys]).
a.
The cross mt+[eryr] X mt-[erys] produces about 95% erythromycin resistant progeny (thus uniparental inheritance).
b. The cross mt-[eryr] X mt+[erys] produces about 95% erythromycin sensitive progeny (also uniparental inheritance).
c.
3.
Other Chlamydomonas chloroplast alleles also show uniparental inheritance, with the progeny phenotype
always the same as the mt+ parent’s phenotype (Figure 23.13).
a.
4.
In both crosses, progeny resemble the mt+ parent, reflecting either
Both mt and mt contribute cpDNA equally to the zygote, but the mt
is only lightly methylated
cpDNA becomes highly methylated while mt
b. Highly methylated mt
cpDNA is replicated efficiently, but lightly methylated mt
c.
cpDNA is destroyed within hours after mating.
Lightly methylated mt
cpDNA is not.
While 95% of progeny show uniparental inheritance, the remaining 5% display biparental inheritance, with
both types of cpDNA present and active.
a.
The genetic condition of these zygotes is called cytohet (cytoplasmically heterozygous).
b. Cytohet cells usually produce pure types on successive mitotic divisions. (This is similar to variegation in Mirabilis.)
c.
Occasionally, a biparental zygote retains both traits in subsequent generations. This results from recombination in the
cpDNA.
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Fig. 23.13 Uniparental inheritance in Chlamydomonas
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 28
Human Genetic Diseases and Mitochondrial
DNA Defects
iActivity: Mitrochondrial DNA and Human Disease
1. Mutations in mtDNA can produce human genetic disorders. Examples:
a. Leber’s hereditary optic neuropathy (LHON). Optic nerve degeneration results in
complete or partial blindness in mid-life adults.
i. LHON is caused by mutations in mtDNA genes for electron transport chain
proteins. (These include ND1, ND2, ND4, ND5, ND6, cyt b, COI, COIII, and
ATPase 6.)
ii. LHON results from defects in the enzymes of oxidative phosphorylation.
Without ATP production, the optic nerve dies.
b. Kearns-Sayre syndrome 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.
iv. Kearns-Sayre syndrome results from deletions in mtDNA. A model for the
disorder is that tRNA genes are removed, disrupting mitochondrial translation.
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c. Myoclonic epilepsy and ragged-red fiber disease (MERRF). Symptoms include:
i. Microscopic tissue abnormality, “ragged-red fibers.”
ii. Myoclonic seizures (jerking spasms).
iii. Ataxia (uncoordinated movement).
iv. Accumulation of lactic acid in blood.
v. Additional symptoms are sometimes present, including:
(1)Dementia.
(2)Loss of hearing.
(3)Difficulty speaking.
(4)Optic atrophy.
(5)Involuntary jerking of eyes.
(6)Short stature.
vi. Mitochondria have abnormal appearance.
vii. The cause is a single nucleotide substitution in the lysine tRNA gene. Mitochondrial
protein synthesis is affected, and in some way this phenotype is produced.
2. In most mtDNA disorders, cells of affected individuals have a mix of normal and mutant
mitochondria (heteroplasmy).
a. Proportions of the two mitochondrial types vary between tissues, and between individuals.
b. Severity of disease correlates with the relative amount of mutant mitochondria.
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Cytoplasmic Male Sterility and Hybrid Seed
Production
1. Hybrid crops are agriculturally important, exhibiting heterozygote
superiority (heterosis), and require controlled crosses by plant breeders.
2. 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).
3. 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.
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4. Hybrid seed are generated using [CMS] rf/rf females and [CMS] Rf/rf males (Figure
23.14). 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.
5. Genetic engineering is also used to make transgenic male-sterile plants by transformation.
a. Two genes are needed, both from the soil bacterium Bacillus anyloliquefaciens.
i. The barnase gene encodes an RNase secreted as a defense against other
organisms.
ii. The barnstar gene produces an inhibitor of barnase, protecting the bacterium.
b. A wild-type plant receives the barnase gene fused to the TA29 promoter (expressed
only in the tapetum tissue of the pollen sac). Barnase is made in the tapetum, locally
destroying RNAs and making the plant male sterile
c. The TA29-barnase plant is used as the female in a cross with a TA29-barnstar plant.
i. Seeds produce plants with both the barnase and barnstar proteins in the
tapetum
ii. Barnstar binds barnase, inhibits RNase activity, and allows pollen production.
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Fig. 23.14 Production of hybrid seed using cytoplasmic male sterility (CMS) and a
restorer of fertility gene
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Exceptions to Maternal Inheritance
1. When the female gamete contributes most of the
cytoplasm, maternal inheritance is the usual explanation
for extranuclear mutations. However, exceptions occur.
Examples:
a. PCR analysis shows heteroplasmy in mice, with paternal mtDNA
present at a frequency of 10-4 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,
with most inheriting only maternal plastids, but others inheriting
from both parents, or from the paternal parent. Paternal
inheritance is also found in gymnosperms.
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Infectious Heredity: Killer Yeast
1. Symbiotic bacteria or viruses in eukaryotic cytoplasm may also produce
extranuclear inheritance. An example is the killer phenotype in yeast:
a. Killer cells secrete a toxin that kills sensitive cells, but not killer strains.
b. Killer phenotype results from two cytoplasmic viruses, L and M. Neither virus
harms the host cell (Figure 23.15).
i. L virus is 4.6 kb dsRNA, in a protein capsid. L-dsRNA encodes capsid
proteins used for both viruses, and viral RNA polymerase for replication.
ii. M virus is found only in cells also containing L virus. M contains two
copies of a 1.8-kb dsRNA. M-dsRNA encodes killer toxin, which also
confers immunity on the host cell.
c. There are two types of yeast cells sensitive to killer toxin (both lack the M
virus):
i. Those with the L virus.
ii. Those with neither L nor M virus.
d. Transmission of these viruses between yeast cells occurs in mating. All
progeny of the mating inherit copies of the parental viruses.
2. Killer yeast cells are an example of an infectious mechanism of
cytoplasmic inheritance.
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Fig. 23.15 The killer phenomenon in yeast
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Contrasts to Non-Mendelian Genetics
Maternal Effect
Animation: Maternal Effect
1. Some maternally-derived phenotypes are
produced by the maternal nuclear genome
(maternal effect), rather than inherited as
extranuclear genes (maternal inheritance).
a. Proteins and/or mRNA deposited in the oocyte before
fertilization direct early development in the embryo.
b. The genes encoding these products are on nuclear
chromosomes. No mtDNA is involved.
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2. An example is shell coiling in the snail Limnaea peregra (Figure 23.16).
a. Shell coiling is determined by a pair of nuclear alleles, with the dominant D,
producing a dextral (right) coil, and the recessive d producing sinistral (left) coiling.
b. The shell coiling phenotype is always determined by the mother’s genotype.
c. In all crosses of true-breeding dextral and sinistral snails, the F1s have the same
genotype (D/d) but the reciprocal crosses produce different phenotypes (Figure 7.8).
i. A dextral female (D/D) crossed with a sinistral (d/d) male produces a dextral F1
(D/d) (Figure 23.17).
(1) The F2 genotypes have a 1:2:1 ratio (D/D : D/d : d/d). All F2 snails,
including those with genotype d/d, have dextral shells.
(2) Selfing the F2 produces an F3 that is 3⁄4 dextral and 1⁄4 sinistral. The
sinistral snails are the progeny of F2 d/d mothers (who had dextral shells).
ii. A sinistral female (d/d) crossed with a dextral male (D/D) produces a sinistral
F1 (D/d).
(1) The F2 genotypes also have a 1:2:1 ratio (D/D : D/d : d/d). All F2 snails
have sinistral shells.
(2) Selfing the F2 produces an F3 that is all dextral, due to the D/d genotype
of the F2 mothers (who had sinistral shells).
台大農藝系 遺傳學 601 20000
Chapter 23 slide 38
Fig. 23.17 Maternal effect: Inheritance of the direction of shell coiling in the snail
Limnaea peregra
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 39
3. This inheritance pattern is very different from extranuclear inheritance.
a. In extranuclear inheritance, the mother and progeny share a phenotype and an
extranuclear genotype.
b. In maternal effect, the progeny phenotype is determined by the genotype of the
mother, and not by the alleles the progeny carry.
4. In the snail shell example, direction of coiling is determined by the orientation
of the mitotic spindle in the first mitotic division following fertilization.
Maternal products within the oocyte direct orientation of the mitotic spindle,
and thus shell coiling.
a. This is supported experimentally:
i. When eggs of d/d mothers are injected with cytoplasm from dextral snails,
dextral progeny result.
ii. When eggs of D/_ mothers are injected with cytoplasm from sinistral
mothers, the progeny still have dextral shells.
b. Interpretation is that:
i. The D allele produces a cytoplasmic product that causes dextral coiling.
ii. The d allele does not produce this product, and sinistral coiling is produced
by default.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 40
Genomic Imprinting
1. Most nuclear genes function independently of maternal or paternal origin. The
expression of some nuclear genes, however, is determined by genomic
imprinting (parental imprinting), where one parent’s allele is expressed
preferentially.
2. Human examples of genomic imprinting include:
a. Prader-Willi syndrome (PWS):
i. Affected individuals are small and weak at birth, with retardation and poor feeding.
ii. Feeding difficulty resolves into uncontrollable eating and associated problems that
are typically fatal by age 30.
iii. The cause is a disruption in chromosomal region 15q11-q13.
(1) In 70–80% of cases, the disruption is on the father’s chromosome 15.
(2) In PWS patients, the maternal 15q11-q13 region is normally suppressed by
methylation of the genes. Paternal genes are needed for normal
development, and when they are disrupted PWS results.
b. Angelman syndrome (AS) produces severe motor and intellectual retardation, small
head, jerky movements, hyperactivity, and unprovoked laughter.
i. About 50% of AS patients have a deletion of region 15q11-q13.
ii. In AS, maternal alleles are needed for normal development, because paternal
genes are inactivated by methylation prescribed by genomic imprinting. Defective
maternal genes result in disease.
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Chapter 23 slide 41