Transmission (Classical, Mendelian) Genetics Ch 11

• Gregor Mendel– Experiments in Plant Hybridization, 1865

• Controlled experiments, 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 plant)

Mendel’s conclusions

1. 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 dominant, the other 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 =

3. Random segregation of alleles into gametes – gamete receives ONE allele per gene– random segregation of alleles 50/50

PP pp

What is the phenotype of all offspring in F1 generation?

P generation

P P

p

p

Note that the P generation is true breeding

Genotype

Phenotype

How did Mendel do it? The Monohybrid cross

Which allele is dominant?What is the genotype of the f1 generation?

YY yy

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 RatioTall X dwarf 787 tall, 277 dwarfRound seeds X wrinkled 5474 round, 1850 wrinkledYellow seeds X green 6022 yellow, 2001 greenViolet flowers X white 705 violet, 224 whiteInflated pods X constricted 882 inflated, 299 constrictedGreen pods X yellow 428 green, 152 yellowAxial flowers X terminal651 axial, 207 terminalgene = ?alleles = ?

In mice, black fur is dominant over white fur

Cross heterozygous mouse with homozygous recessive mouse B = black b = white

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

A mouse was test crossed and 7 offspring black, 2 white. What is mouse’s genotype?

All possible one gene crosses

Parents ratio offspring

Pedigree Analysis Ch 11

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

Autosomal recessive traits

• Sickle cell disease

• Albinism

• Cystic fibrosis

• O blood type

Phenylketonuria (Ch.4)

• PKU (1/12,000) Mutation in gene encoding phenylalanine hydroxylase enzyme needed for phe metabolism

Chromosome 1212q24.1

missing phenylalanine hydroxylase enzyme

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

Why are these babies normal when born?

Pleiotropic effectsno tyrosine (little melanin)slow growthretardationblue eyeslow adrenaline

No nutrasweetlow phe diet ($5K/yr)

1902 Archibald Garrod: One gene: one enzyme“Inborn errors of metabolism”PKUAlbinismAlkaptonuriaTyrosinemia

Black urinearthritis

Ch 4

Fill in genotypes. If II,1 and II, 4 mate, what is the chance of offspring having PKU?

II, 1 X II, 4 p(aa)

•p(aa AND a girl)?

How do we know this is autosomalrecessive?

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

Product rule:

p (affected boy)?

All people have harmful recessive alleles, small chanceThat 2 people with same rare alleles will mateConsanguinous marriage increases the chance

Bedoin intermarriage

Autosomal dominant disordersAa and AA =affectedaa =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 FGFR3 gene Chromosome 4• Affects cartilage growth needed for bone

lengthening• Affected individuals Aa why not AA?

• Most cases spontaneous (associated with increasing paternal age)

what is the genotype of parents in this case?

P(III, 3 and III, 5 have a child of normal height)

P ( II, 3 and III, 7 have a boy with achondroplasia)

• Mutation in dog FGF4 gene