Transmission (Classical, Mendelian) Genetics Ch 11 Gregor Mendel –Experiments in Plant Hybridization, 1865 Simple, controlled, data collection, mathematical analysis

Pisum sativum, the garden pea What makes this a good model organism? easy to grow hundreds of offspring per cross short generation time can self fertilize or cross –Paint pollen (sperm) from one plant onto the female parts of another (emasculated) parental plants true breeding strains

Genes and alleles of Pisum sativum GeneAlleles Pea color? Flower colorwhite, purple Pod shapeconstricted, inflated Pea surface? Stem heighttall, dwarf

I. Monohybrid cross

Mendel’s results from the monohybrid cross F2787 long277short = PHENOTYPES Ratio tall/dwarf = 4 Conclusions: 1.Genes discrete units passed on from parent to offspring 2.A dominant allele masks expression of recessive allele 3.“Unit factors in pairs”- each gene has 2 alleles why?

Genotypes Homozygous dominant = Heterozygous = Homozygous recessive =

4. Random segregation –gamete receives ONE allele per gene –random segregation of alleles 50/50 –Humans Sperm + eggfertilized egg chromosomes? 1 setalleles + 1 set ? haploid + haploid ?

Summing it up Sperm and egg (gametes) are haploid. –Each contains half the genes, or one allele for each gene When sperm and egg unite, organism formed is diploid –2 alleles per gene

More results of Mendel’s monohybrid crosses Parental StrainsF2 progenyRatio Tall X dwarf787 tall, 277 dwarf Round seeds X wrinkled5474 round, 1850 wrinkled Yellow seeds X green6022 yellow, 2001 green Violet flowers X white705 violet, 224 white Inflated pods X constricted882 inflated, 299 constricted Green pods X yellow428 green, 152 yellow Axial flowers X terminal651 axial, 207 terminal gene = ? alleles = ?

II. Test cross (one gene) mouse “Z” has black fur, what are its 2 possible genotypes? Test cross mouse “Z” to homozygous recessive mouse Z was test crossed and 6 offspring were black 2 were white. What is Z’s genotype?

Autosomal recessive inheritance (bb) unaffected parents can have affected offspring affected progeny male, female May “skip” a generation Two affected parents cannot have an unaffected child

Phenylketonuria pp (Ch. 4 pg 73) PKU (1/12,000) Mutation in gene encoding phenylalanine hydroxylase enzyme needed for phe metabolism

missing phenylalanine hydroxylase enzyme If plasma phe level is too high, phe is converted into a phenylpyruvate toxic to brain tissue

Pleiotropic effects no tyrosine (little melanin) slow growth retardation blue eyes low adrenaline ) No nutrasweet low phe diet ($5K/yr

1902 Archibald Garrod: One gene: one enzyme “Inborn errors of metabolism” PKU Albinism Alkaptonuria Tyrosinemia Black urine arthritis Page 68

One gene/one enzyme Garrod’s work on alkaptonuria “Inborn Errors of Metabolism” 1902 Autosomal recessive metabolic disease

All people have harmful recessive alleles, small chance That 2 people with same rare alleles will mate Consanguinous marriage increases the chance

Fill in genotypes. If II,1 and II, 4 mate, what is the chance of offspring having PKU? II, 1 XII, 4p(aa) p(aa AND a girl)? How do we know this is autosomal recessive?

If III-3 and II-1 mate p (normal child) p (affected boy) ?

Autosomal dominant disorders Aa and AA are affected aa is unaffected Tend to show up in every generation 2 affected parents can have unaffected child 2 unaffected parents cannot have an affected child

Dominant pedigree

Achondroplasia -1/20,000 births Mutation in one allele of FGFR3 gene Chromosome 4 Affects cartilage growth needed for bone lengthening Most affected individuals are Aa why? Most cases are spontaneous aa X aa Pg. 291

P(III, 3 and III, 5 have a child of normal height) P ( II, 3 and III, 7 have a boy with achondroplasia)

Fruit fly nomenclature pg 317 box 12.1 Red eyes is normal phenotype, brown is mutant bw+ = wildtype allele bw = brown allele genotypephenotype red brown

Try it: Wingless is recessive mutant (wg allele) Genotype of wildtype, heterozygote, mutant?

Sex-linked genes Ch 12 pg 314 – 317, Human Female = XX –two alleles for each X-linked gene –normal application of recessive and dominance X H X H X h X h

Human Male XY X H Y X h Y

Sex-linked genes Most on X chromosome Hemophilia (recessive) 1/5000 males –Mutation in gene for clotting factor

Mate III 13 with III 1 Probability of a hemophiliac son? Mate IV 2 with homozygous normal female p (hemophilia)? *Criss cross inheritance of X linked traits

w+ = wildtype allelew = white allele X-linked recessive The mutant fruit fly discovered by Thomas Hunt Morgan

A white-eyed female is crossed with a red- eyed male. An F1 female from this cross is mated with her father and an F1 male is mated with his mother. What will be the eye color of the offspring of these two crosses?

Mendel’s Law of Independent assortment - each allele for a trait is inherited independently of other alleles Seeds: G = yellow alleleg = green allele gene? W = roundallelew = wrinkled allele gene? Dihybrid cross – 2 genes

Parents = GGWW X ggww phenotype? Gametes? F1 genotype ? F1 phenotype ? F1 Gametes?

Forked line method for phenotypes GgWw X GgWw

Product rule- the probability that two outcomes occur simultaneously is product of their individual probabilities assumes independent assortment of genes GgWw X GgWw What is the probability of a yellow AND wrinkled? p(G-ww) ¾ X ¼ = 3/16 Probability

Trihybrid cross AaBbCc X AaBbCc p(A-B-cc) AabbCcDD X AaBbCcDd p(triply recessive)

Modified Mendelian Ratios 1. INCOMPLETE DOMINANCE R = red flower (snapdragon) R’ = white flower * allele symbols do not connote dominance * phenotypic ratio = genotypic ratio = ? PC r C r X C w C w F1 F2

Incomplete dominance

2. Codominance Each allele encodes separate gene product distinct in phenotype of heterozygote L gene for human blood cell surface protein L M = M antigen L N = N antigen

A man with the M bloodtype has a child with a woman of the MN bloodtype Expected ratio of offspring?

3. Multiple alleles (more than 2 alleles for gene in population) Example: Blood Groups Karl Landsteiner 1900s

ABO blood system = polymorphic I gene Blood typegenotype AI A I A or I A i BI B I B or I B i AB? Oii What is the mechanism of inheritance of A, B, AB, O? Autosomal or sex chromosome?

Example A child has type O blood. The mother of the child has Type B blood. What could the blood types of the father be?

4. Dominance series – C series/ rabbits c+ = full color c ch = chinchilla (hypomorphic) c h = himalayan (hypomorphic) c = albino (apomorphic allele = nonfunctional) Chinchill a Himalayan Albino

Genotypephenotype? c ch c ch c h c h c c+ c ch c+ = full color c ch = chinchilla (hypomorphic) c h = himalayan (hypomorphic) c = albino (apomorphic allele = nonfunctional)

5. Lethal alleles MM = normal spine MM’ = manx cat (no tail) M’M’ = lethal Cross two manx, what is ratio of phenotypes in offspring? How do breeders obtain manx cats?

A product of one gene influences, or masks, the expression of another gene(s) Modification of dihybrid cross ratio AaBbXAaBb 9:3:3:1 6. Epistasis- gene product interactions. Table 13.4 page 355 (look at 4 phenotypic classes and fewer than 4)

Epistasis in Cats W = whitew = not white B = blackb = brown Mate 2 heterozygous cats What is the expected ratio?

Epistasis in labrador retrievers B and E color genes (labs) B black b brown E color e no color (yellow) ee is epistatic Cross two double heterozygotes Phenotypes of parents? Phenotypes of offspring? ratio?

7. Penetrance % individuals that exhibit phenotype corresponding to genotype Polydactyly, dominant Pp pp 5,5 6, 5 6, 6

8. Expressivity (ex. Piebald spotting) – the extent to which a trait is exhibited osteogenesis imperfecta pg. 359

Penetrance AND expressivity NF-1 = Neurofibromatosis1 –(1/4000, 17q11.2) –(350 kb gene, 60 exons) Autosomal dominant trait N- (in many it’s a spontaneous mutation) 50 – 80% penetrance Expressivity –Pigmented skin to tumors on nerve CT coverings (neurofibromas) on skin, eyes, organs, face –Speech, blood pressure, spine curvature, headaches

9. Quantitative (multifactorial) traits Vary continuously –Weight, height, IQ

Gene expression also affected by: Sex (baldness) Temperature (melanin in Siamese cats) Chemicals (PKU) Diet (height, cancer) + many other factors!

Chromosome Theory of Inheritance 1902 Sutton and Boveri –A chromosome is a linkage group of Mendelian factors (GENES) 1920s Morgan et al. –Genes are in a linear sequence on the chromosomes, they can be mapped

Chromosomes in most animals pairs of autosomes 1 pair sex chromosomes –XY heterogametic –XX homogametic Human karyotype ->

Do more chromosomes mean more intelligence? Human46 Chimpanzee48 Dog78 Cat72 Alligator32 Goldfish94 Mosquito6 Potato48 Baker’s yeast34

I. Mammalian sex determination = the Y system A. Embryo is neither male nor female Week 7 How does embryo “know to become male?

XY embryo sex chromosomes The Y determines sex…. XY = male XX = female

B. SRY gene encodes TDF (Testes determining factor, 1990) SRY (sex determining region Y) TDF stimulates the growth of testes --> testosterone ---> sperm ducts, male brain “sensitization”

XX males: If SRY crosses over to the X chromosome during meiosis (formation of sperm) Father during meiosis X from father + X from mother

XX males Add SRY DNA to female mouse embryo 3. Experiments with transgenic mice

A 17 year old female presented with “streak” ovaries, no uterus, no menstrual cycle XY female –Embryo has Y chromosome but does not develop as male Mutation in SRY  ?

II. Other sex determination systems A. Drosophila Ratio of X to sets of autosomes –embryo “calculates” ratio X/A = 1 or > > female X/A = 0.5 or male X/A between 0.5 and 1 ---> intersex

What is the sex of an XY fly with 2 sets of autosomes? 1X/2A = 0.5 = male What is the sex of a fly with with 2 sets of autosomes but 1 X chromosome 1X/2A = 0.5 = male What is the sex of a triploid fly with 2 X chromosomes? 2X/3A = 0.66 = intersex

B. ZW system - birds Females are ZW (heterogametic) Males are ZZ (homogametic)

C.Temperature sex determination (TSD) In some reptiles sex is not determined genetically! (Varies widely) majority of endangered reptiles use TSD - sea turtles, Galapagos tortoises, alligators, crocodiles sex determined during mid-trimester of development by T of incubation

IV. Dosage Compensation (mammals) Females have 2 Xs, males have 1 X. Do females have an extra dose of X-linked genes/alleles?

X chromosome inactivation Lyon, 1961 Observe dense “Barr body” at edge of nucleus in female cells –Heterochromatic (stains darker) Male cellfemale cellcell with 2 Barr bodies Number of X chromosomes?

Female mosaics All females heterozygous for X-linked traits are mosaics for those traits. red/green colorblindness X C X c phenotype = ? Look at retina of heterozygous female

Anhydrotic ectodermal dysplasia X A X a females X a Y males Tooth + nail abnormalities, life threatening hyperthermia, sparse hair

Most genes on Y are for development and fertility

76 Chromosomal Abnormalities KARYOTYPE Obtain white blood cells from or fetal cells from amniotic fluid Detects number of chromosomes, sex, chromosomal abnormalities

77 Amniocentesis usually done week 14 Karyotype and analyze fluid for enzyme defects

78 Chorionic villus sample (CVS) usually done week 8 More risk, but earlier results

79 Arrange in pairs according to: decreasing size centromere position banding pattern ACROCEN TRIC METACENT RIC SUBMETACEN TRIC

80 metacentric(1)submetacentric (9) p arm is the upper, shorter arm

81 Normal male 46, XY Which are meta-, submeta- acrocentric?

82 Aneuploidy (versus euploidy) Trisomy (not triploid) 47, may be small, but contains 33,546,361 bp of DNA!

83 Down Syndrome (J. Langdon H. Down, 1866) effects –Developmental delays –Possible heart defects, hearing loss, hypotonia, thyroid problems, obesity –Epicanthic eye folds –Wide tongues –Greater risk of Alzheimer’s

84 Trisomy 13 (Patau) Fatal< 1 year (usually) Deaf, blind, clyclopia, polydactly, cleft palate 1/5000 live births 47, XY, 13+

85 47, XY, 18+ (Edward’s) < few months 1/5000 live births

86 Human trisomies of the sex chromosomes (see pg 293) 47, XXY = Klinefelter’s 47, XXX 47, XYY

87 Monosomy (only 1 viable in humans!) 45 X, –Turner Syndrome (1/2000 live births) Partial monosomy 46, 5p- –Cri du Chat –Arises due to a deletion on the short arm of chromosome 5

88 Cri du Chat 46, 5p- pseudodominance for deleted region Microcephaly, myotonia, “cry of cat”, retardation

89 Somatic mosaics More than one genetically distinct population of cells in an individual (like random X inactivation) Example: 46XX embryo, one cell loses an X ---  46,XX/45X mosaic –Symptoms less severe than the standard Turners syndrome

90 Polyploidy = extra SETS of chromosomes # in humans Triploid, tetraploid Octoploid etc..

91 Many plants are polyploid Some bees and wasps are monoploid

92 can observe large ones by karyotype If centromere is lost, then chromosome will be lost Heat, chemicals, radiation Unequal crossing over during meiosis Deletions (del)

93 46,XX,del(7)(q21.12,q21.2) Leads to problems during meiosis  Pseudodominance  Lethal if both chromosomes

94 Duplications segment of a chromosome doubles May be tandem or reverse problems during meiosis

95 dup(5)(qter->q33.1::p15.3->qter) Sample of cord blood from stillborn male with anencephaly

96 Inversions (inv) –180o turnaround of segment no loss of genetic material may change length ratio of p/q arms –Position effect change in gene position with respect to centromere being near heterochromatic region may influence expression

97 Translocations - segment moves to other chromosome - interstitial or reciprocal exchange t (13;14) Individual has all genetic material, but what about gametes?

98 t(11;13) (q21;q14.3) the parent who has the translocation is phenotypically normal as all genetic info is present The gametes, however, are not as evidence by multiple miscarriages