AP Bio: Thursday 3/17/11Genetic Patterns of InheritanceHappy St. Patrick’s Day / Evacuation Day!Homework: PS 16 #1-7 (be ready to discuss

these on Monday)Do Now: Get some breakfast, get out your

notebooks, get ready to learn.Today’s Goals:

Explain how Mendel’s Laws of Inheritance are based on the events of meiosis

Solve genetics problems involving one gene, two genes, multiple alleles, incomplete dominance, sex-linked traits, and epistasis

Agenda: Lecture mixed w/ practice problems

Genetic Patterns of Inheritance

AP BiologyChapters 14-15

Karyotypes

Autosome = Chromosomes

1-22

Sex Chromosome = X or Y

Trisomy 21 (Down Syndrome)

Mendel’s 1st Law:

Law of SegregationAlleles “segregate” into different gametes

during meiosis

a a A A

a Aa A

Monohybrid CrossInvolves one geneCrossing two heterozygotes yields:

A

a

aA Phenotypes:

Ratios:

Mendel’s 2nd Law:

Law of Independent AssortmentAlleles on one set

of chromosomes segregate independently from alleles on other sets of chromosomes.

Results in different allele combinations in different gametes, and in different combinations of traits

Dihybrid CrossInvolves two genes on different chromosomes

First step is always to arrange the gametesCrossing two heterozygotes yields:

RY

ry

RY

Ry

rYRy

ry

rY

Beyond the Dominant/Recessive Paradigm…Codominance

Both alleles fully expressed1:2:1 phenotypic ratioEx: red, white, red/white

spotsIncomplete Dominance

Dominant allele partially expressed

1:2:1 phenotypic ratioEx: red, white, pink

Multiple AllelesMore than two versions

(alleles) of the gene

ABO Blood Types –Show Multiple Alleles AND Co-dominance

Gene I controls ABO Blood Type.

Three possible alleles:IA – makes A antigensIB – makes B antigensi – makes no antigens

Possible genotypes (sets of two alleles):

Beyond the Dominant/Recessive Paradigm…

Polygenic traitsControlled by many genesOften show continuous

variation in phenotypes

Beyond the Dominant/Recessive Paradigm…

EpistasisThe effects of one

gene hide or alter the effects of another gene

9:3:4 phenotypic ratio

Hemophilia in the Royal Family

Sex-Linked Traits

XR – normal visionXr – red-green colorblind (recessive)

XR Xr = carrier female (normal vision)XrY = colorblind male (only needs one recessive allele to be

colorblind)

Controlled by genes on sex chromosomes

X-linked traits – on X chromosomeHemophiliaRed-green

colorblindness

More on X-Linked TraitsTry these sample

problems…Carrier female x normal

male

Normal female x colorblind male

Hairy Ears

Why do only males have them?

Why do only males produce the SRY (testis-determining) protein?

They are Y-linked!

Hairy Ears

If a man with hairy ears marries a woman with non-hairy ears, what is the chance that (a) their daughters will have hairy ears?(b) their sons will have hairy ears?

The end of this lecture.

Pedigree AnalysisGet a worksheet…

Evaluating the Validity of Genetics Experiments… Using Chi-SquaredStatistics! Woot woot!Get a worksheet… enough of this powerpoint

business.

More Chi-Squared PracticeTwo Drosophila genesBody color: gray (G) or black (g)Wing shape: normal (N) or vestigial (n)Cross a double heterozygote with a double

recessiveGgNn x ggnn

Possible phenotypes

Expected Outcomes

Observed Outcomes

Gray/Normal 250 410

Gray/Vestigial 250 90

Black/Normal 250 80

Black/Vestigial

250 420

Linked GenesChi-Squared Test REJECTS our null hypothesis.Independent Assortment of the two genes

doesn’t seem to apply Why?

Conclusion: Two genes on same chromosome (linked)• Recombinant genotypes created by

crossing over between the two genes

• Frequency of recombinant genotypes based on the distance between genes on the chromosome

Finding Distances Between Linked GenesCalculate the recombination frequency –

how frequent are the genotype combinations that AREN’T the same as the parents?

Percentage of recombinants = distance in map units

Practice this with the fruit fly example…

Linkage MapsShow the relative location of genes on a

chromosomeFirst, determine recombination

frequenciesBody color & wing shape = 17% recombinantsBody color & eye color = 7% recombinantsWing shape & eye color = 23% recombinants

1 % recombination frequency = 1 map unit(Can’t be over 50%... why not?)

Create a linkage map (aka genetic map)