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)

Mendel’s conclusions Unit factors in pairs - Genes are physical units 2 alleles for each gene 1 allele inherited from each parent

Genes and alleles of Pisum sativum Gene Alleles Pea color ? Flower color white, purple Pod shape constricted, inflated Pea surface ? Stem height tall, dwarf

2. Principle of Dominance One allele is dominant the other is recessive The dominant allele is expressed in the phenotype

Gene for flower color P allele = purple p allele = white GENOTYPES PHENOTYPE Homozygous dominant = Heterozygous = Homozygous recessive =

What is the phenotype of all offspring in F1 generation? 3. Random segregation of alleles into gametes gamete receives ONE allele per gene random segregation of alleles 50/50 P generation PP pp What is the phenotype of all offspring in F1 generation?

Note that the P generation is true breeding P P p Genotype Phenotype

How did Mendel do it? The Monohybrid cross YY yy Which allele is dominant? What is the genotype of the f1 generation?

Cross 2 f1 plants (or let one self-fertilize) What is the ratio of phenotypes?

Results of Mendel’s monohybrid crosses Parental Strains F2 progeny Ratio Tall X dwarf 787 tall, 277 dwarf Round seeds X wrinkled 5474 round, 1850 wrinkled Yellow seeds X green 6022 yellow, 2001 green Violet flowers X white 705 violet, 224 white Inflated pods X constricted 882 inflated, 299 constricted Green pods X yellow 428 green, 152 yellow Axial flowers X terminal 651 axial, 207 terminal gene = ? alleles = ?

Test cross (one gene) A mouse has black fur, what are its 2 possible genotypes? Test cross mouse to homozygous recessive mouse If black mouse is BB  If black mouse is Bb  Mouse was test crossed and 7 offspring were black 2 were white. What is mouse’s genotype?

Autosomal recessive inheritance (bb) unaffected parents can have affected offspring May “skip” a generation Two affected parents cannot have an unaffected child Not sex related

Examples of autosomal recessive traits Sickle cell disease Albinism Cystic fibrosis O blood type

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

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

no tyrosine (little melanin) Why are these babies normal when born? Pleiotropic effects no tyrosine (little melanin) slow growth retardation blue eyes low adrenaline No nutrasweet low phe diet ($5K/yr)

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

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

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

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

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

Autosomal dominant disorders Aa and AA =affected aa =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 (Parents are 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 wildtype phenotype, brown is mutant bw+ = wildtype allele bw = brown allele genotype phenotype red brown

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

Autosomal Sex-linked

Genetics Home Reference page (National Library of Medicine) Collagen Blood clotting factor Red blood cell enzyme Dystrophin muscle protein Color vision gene

Sex-linked genes Ch 12 pg 314 – 317, 326 - 328 Human Female = XX two alleles for each X-linked gene normal application of recessive and dominance XHXH XHXh XhXh

Human Male XY XHY XhY

X-linked genes Hemophilia (recessive) 1/5000 males Mutation in gene for clotting factor Xq28

*Criss cross inheritance of X linked traits 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

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

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

Note that each gene gives the 3:1 ratio of a monohybrid cross Yellow/green ratio = Round/ wrinkled =

Forked line method for phenotypes GgWw X GgWw

Test cross G-W- X ggww If all yellow and round: A pea is round and yellow. What is its genotype? G-W- X ggww Note the cross of the “unknown” to a homozygous recessive If all yellow and round: If all yellow and some wrinkled: If all round and some green: If 1:1:1:1:

Probability 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)

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 = ? P CrCr X CwCw 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 LM = M antigen LN = 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 Chromosome 9 I gene

Blood type genotype ABO blood system = polymorphic I gene A IAIA or IAi B IBIB or IBi AB ? O ii What is the mechanism of inheritance of A, B, AB, O? Autosomal or sex chromosome?

ABO dominance and codominance

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

Genotype phenotype? cch cch cch ch ch c c+ cch c+ = full color cch = chinchilla (hypomorphic) ch = himalayan (hypomorphic) c = albino (apomorphic allele = nonfunctional)

5. Lethal alleles MM’ = manx cat (no tail) M’M’ = lethal 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?

6. Epistasis- gene product interactions. Table 13 6. Epistasis- gene product interactions. Table 13.4 page 355 (look at 4 phenotypic classes and fewer than 4) A product of one gene influences, or masks, the expression of another gene(s) Modification of dihybrid cross ratio AaBb X AaBb 9:3:3:1

Epistasis in Cats W = white w = not white B = black b = 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?

Polydactyly, dominant 7. Penetrance % individuals that exhibit phenotype corresponding to genotype 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 Autosomal dominant trait N- 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!