Presentation on theme: "台大農藝系 遺傳學 601 20000 Chapter 15 slide 1 CHAPTER 15 Non-Mendelian Inheritance Peter J. Russell edited by Yue-Wen Wang Ph. D. Dept. of Agronomy, NTU."— Presentation transcript:
台大農藝系 遺傳學 Chapter 15 slide 1 CHAPTER 15 Non-Mendelian Inheritance Peter J. Russell edited by Yue-Wen Wang Ph. D. Dept. of Agronomy, NTU
台大農藝系 遺傳學 Chapter 15 slide 2 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 (10 9 ) 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.
台大農藝系 遺傳學 Chapter 15 slide 3 Organization of Extranuclear Genomes Mitochondrial Genome 1. Mitochondrial functions include energy-generating oxidative reactions requiring a variety of enzymes, including: 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 15.1). 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 products.
台大農藝系 遺傳學 Chapter 15 slide 5 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. 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 mtDNA map (Figure 15.3) 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. 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
台大農藝系 遺傳學 Chapter 15 slide 7 5. Mitochondrial ribosornes 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, the 3’ end is U or UA, and the poly(A) tail completes the stop codon (UAA).
台大農藝系 遺傳學 Chapter 15 slide 8 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 have differences that show no single pattern (Table 15.1). d. Mitochondria show extended wobble in base pairing of tRNA with codon, allowing translation with only 22 tRNAs (compared with 32 tRNAs theoretically needed for translation using standard wobble). e.. Structural features that never differ in cytoplasmic tRNAs sometimes do differ in mitochondrial tRNAs.
台大農藝系 遺傳學 Chapter 15 slide 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 analysis to determine genetic variability in populations (e.g., grizzly bears in Yellowstone Park).
台大農藝系 遺傳學 Chapter 15 slide 11 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 15.5): a. Two copies of each chloroplast rRNA (loS, 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 chioroplast transcription, translation aiid photosynthesis.
台大農藝系 遺傳學 Chapter 15 slide 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).
台大農藝系 遺傳學 Chapter 15 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.
台大農藝系 遺傳學 Chapter 15 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 15.7). 2.Results of crosses between plants with shoots that are variegated illustrate this phenomenon (Table 15.2): 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.
台大農藝系 遺傳學 Chapter 15 slide 16 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.
台大農藝系 遺傳學 Chapter 15 slide 19 The [poky] Mutant of Neurospora 1.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 + a 3, b and c. b.Neurospora with the [poky] mutation lack a + a 3 and b, and have an excess of c. 2.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). 3.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. 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.
台大農藝系 遺傳學 Chapter 15 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 15.9). 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.
台大農藝系 遺傳學 Chapter 15 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 ([rho - N]/[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 wild- type 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 : wild- type) 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 wild-type mtDNA.
台大農藝系 遺傳學 Chapter 15 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 ([ery r ]), with wild-type cells sensitive to the antibiotic ([ery s ]). a.The cross mt + [ery r ] 3 mt - [ery s ] produces about 95% erythromycin resistant progeny (thus uniparental inheritance). b.The cross mt - [ery r ] 3 mt + [ery s ] produces about 95% erythromycin sensitive progeny (also uniparental inheritance). c.In both crosses, progeny resemble the mt + parent, reflecting either i. Biased segregation of one chloroplast type, or ii. Inactivation of one chloroplast type. 3.Other Chlamydomonas chloroplast alleles also show uniparental inheritance, with the progeny phenotype always the same as the mt + parent’s phenotype. a.Density labeling experiments with mt + and mt - cpDNA show that cpDNA from the mt - parent disappears in mt + /mt - zygotes. b.The mechanism is unknown. 4.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.
台大農藝系 遺傳學 Chapter 15 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.
台大農藝系 遺傳學 Chapter 15 slide 29 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.
台大農藝系 遺傳學 Chapter 15 slide 30 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 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.
台大農藝系 遺傳學 Chapter 15 slide 31 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 15.12). 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.
台大農藝系 遺傳學 Chapter 15 slide 32
台大農藝系 遺傳學 Chapter 15 slide 33 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.
台大農藝系 遺傳學 Chapter 15 slide 34 2.An example is shell coiling in the snail Limnaea peregra (Figure 15.13). 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 F 1 s 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 F 1 (D/d) (Figure 15.14). (1)The F 2 genotypes have a 1:2:1 ratio (D/D : D/d : d/d). All F 2 snails, including those with genotype d/d, have dextral shells. (2)Selfing the F 2 produces an F 3 that is 3⁄4 dextral and 1⁄4 sinistral. The sinistral snails are the progeny of F 2 d/d mothers (who had dextral shells). ii. A sinistral female (d/d) crossed with a dextral male (D/D) produces a sinistral F 1 (D/d). (1)The F 2 genotypes also have a 1:2:1 ratio (D/D : D/d : d/d). All F 2 snails have sinistral shells. (2)Selfing the F 2 produces an F 3 that is all dextral, due to the D/d genotype of the F 2 mothers (who had sinistral shells).
台大農藝系 遺傳學 Chapter 15 slide 36 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 Chapter 15 slide 37 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.