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Probability, Mendel, and Extensions to Mendel

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<strong>Probability</strong>, <strong>Mendel</strong>, <strong>and</strong><br />

<strong>Extensions</strong> <strong>to</strong> <strong>Mendel</strong><br />

Set up <strong>and</strong> work probability problems<br />

Medelian <strong>and</strong> non-mendelian modes of<br />

inheritance <strong>and</strong> their significane<br />

The F 1 <strong>and</strong> F 2 Offspring of a Cross between Pure Lines<br />

Introduction <strong>to</strong> Rules of <strong>Probability</strong><br />

Genotypic ration VS Phenotypic Ratio<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

monohybrid<br />

cross<br />

1


Practice Problem<br />

• A homozygous yellow corn is crossed with a homozygous purple<br />

corn. The F1 offspring are crossed <strong>and</strong> produce 139 purple kernels<br />

<strong>and</strong> 61 yellow kernels (F2 offspring).<br />

• What is the genotype of the purple kernels in the P generation<br />

<strong>and</strong> the F1 generation?<br />

• Give a genetic explanation (mode of inheritance) for the<br />

differences in kernel color.<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

Chi-square Test<br />

• Are the observed data consistent with the data expected under the<br />

hypothesis being tested?<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

• χ 2 = Σ [(o-e) 2 / e] Values are significantly<br />

different from expected<br />

2


A Cross between Pure Lines for Two Traits<br />

F2 offspring genotypes:<br />

9/16 R–Y– : 3/16 R–yy : 3/16 rrY– : 1/16 rryy<br />

F2 offspring phenotypes: 9/16 : 3/16 : 3/16 : 1/16<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

Hypothesis of independent assortment:<br />

Alleles of different genes assort independently when gametes form.<br />

A Cross between Pure Lines for Two Traits<br />

Are these data are consistent with the<br />

predictions of independent assortment?<br />

<strong>Mendel</strong>’s results<br />

Conduct a X 2 Test.<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

3


Chi-square Test<br />

• Are the observed data consistent with the data expected under the<br />

hypothesis being tested?<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

• χ 2 = Σ [(o-e) 2 / e] Values are significantly<br />

different from expected<br />

Dihybrid cross: Phenotypic ratio 9:3:3:1<br />

• 9/16 dominant phenotypes at both genes<br />

(A_B_)<br />

• 3/16 dominant phenotype at first gene;<br />

recessive phenotype at second gene<br />

( A_bb)<br />

• 3/16 recessive phenotype at first gene;<br />

dominant phenotype at second gene<br />

(aaB_)<br />

• 1/16 recessive phenotypes at both genes<br />

(aabb)<br />

4


Dihybrid Cross problem solving<br />

• Memorize 9:3:3:1 or be able <strong>to</strong> reproduce it.<br />

• Not all genetics problems deal with dihybrids!<br />

• The genotypes may change by the principles<br />

are the same.<br />

Practice problems<br />

• P (dominant phenotype for both traits)<br />

• AaBBccDd X AAbbCcDd<br />

give P (AABbccDD offspring)<br />

•Forkline method for dihybrid<br />

Variations on <strong>Mendel</strong>’s Laws: Incomplete Dominance<br />

Flower color is variable in four-o’clocks.<br />

origin of ‘blending’ hypothesis?<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

5


Variations on <strong>Mendel</strong>’s Laws: Incomplete Dominance<br />

Incomplete dominance in flower color<br />

Parental<br />

generation<br />

F1 generation<br />

F2 generation<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

Self-fertilization<br />

Purple Lavender White<br />

Incomplete Dominance<br />

Heterozygote’s phenotype between those of<br />

the two homozygotes<br />

Example: Familial hypercholesterolemia<br />

(FH): normal alleles encode liver recep<strong>to</strong>rs<br />

for LDL cholesterol<br />

mutant alleles do not produce any recep<strong>to</strong>rs<br />

Phenotype measured by serum cholesterol<br />

levels.<br />

Heterozygote has ~ 1/2 the normal number<br />

of recep<strong>to</strong>rs<br />

Homozygote the mutant allele <strong>to</strong>tally lacks<br />

the recep<strong>to</strong>r (serum cholesterol level is very<br />

high)<br />

Monohybrid cross results in 1:2:1<br />

genotypic ratio <strong>and</strong> 1:2:1 phenotypic<br />

ratio.<br />

6


Variations on <strong>Mendel</strong>’s Laws: Co-Dominance<br />

• Human Blood type<br />

– I A I A or I A i<br />

– I B I B or I B i<br />

– I A I B<br />

– ii<br />

Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings/Addison-Wesley<br />

Multiple Alleles: the Rule<br />

• An individual carries two alleles for each<br />

au<strong>to</strong>somal gene<br />

• However, a gene can have multiple alleles in<br />

the population because its sequence can<br />

deviate in many ways<br />

• Different allele combinations can produce<br />

variations in the phenotype<br />

• Examples:<br />

– ABO (3 alleles; four phenotypes);<br />

– PKU gene (hundreds of alleles; four basic<br />

phenotypes),<br />

– CF gene (>1500 alleles!; many phenotypes)<br />

7


• Multiple Alleles: I A ,<br />

I B , I 0<br />

• Each person still has<br />

only 2 alleles.<br />

• Co‐Dominance:<br />

Both alleles are<br />

expressed<br />

I A I B = type AB<br />

• Simple Dominant<br />

<strong>and</strong> Recessive: Both<br />

I A <strong>and</strong> I B are<br />

dominant <strong>to</strong> I 0<br />

ABO Blood Type<br />

Lethal Alleles<br />

• Alleles that are incompatible with life.<br />

• Result in missing or dead genotypic classes.<br />

• Recessive lethal = 2 copies required for death<br />

– Cystic fibrosis, sickle cell anemia<br />

– Homozygous recessive is missing or dies.<br />

• Dominant lethal = 1 copy required for death<br />

– Hunting<strong>to</strong>n’s disease, achondroplastic dwarfism<br />

– Heterozygotes <strong>and</strong> homozygous dominant are<br />

missing or die (homozygous dominants are typically<br />

very severely affected <strong>and</strong> often die early in<br />

development.<br />

8


PENETRANCE<br />

• PENETRANCE = proportion of a genotype<br />

showing the expected phenotype<br />

• Non‐penetrant = none show phenotype<br />

• Complete or 100% penetrance = all relevant<br />

genotypes express the trait<br />

• Incompletely penetrant = some show the trait,<br />

others do not<br />

• ALL OR NONE EXPRESSION OF A SINGLE GENE<br />

EXPRESSIVITY<br />

• EXPRESSIVITY = degree <strong>to</strong> which a phenotype<br />

is expressed<br />

• Variable Expressivity ~ range of phenotypes<br />

from slight <strong>to</strong> extreme<br />

• Variability in expression in different<br />

individuals<br />

• Variability in different parts of the same<br />

individual<br />

• Example: Polydactyly<br />

• SEVERITY OR EXTENT OF PHENOTYPIC<br />

EXPRESSION OF A SINGLE GENE<br />

9


Hunting<strong>to</strong>n’s Disease<br />

• Dominant lethal<br />

• Incomplete penetrance: disease is never<br />

manifested in some carriers.<br />

• Variable expressivity: appears at variable<br />

ages from infancy <strong>to</strong> old age.<br />

• Remains in populations because death<br />

typically occurs after reproductive age.<br />

Gene Interactions<br />

• > 1 gene pair contributes <strong>to</strong> the<br />

phenotype<br />

• Many, if not most, traits are the<br />

result of biochemical pathways.<br />

Each step is controlled by a<br />

different gene product.<br />

• Segregation <strong>and</strong> Independent<br />

Assortment hold<br />

• Phenotypic ratios are changed<br />

from 9:3:3:1 for 2 gene pairs<br />

• EPISTASIS = phenotypic<br />

expression of one gene pair hides<br />

the effect of another, different<br />

gene pair<br />

• Examples:<br />

– albinism (hides the effect of genes for eye color,<br />

skin color, hair color)<br />

– Bombay phenotype (hides the effect of ABO<br />

blood type genes)<br />

10


Bombay Phenotype<br />

AB AO<br />

AB O A B<br />

• A, B antigens are<br />

sugars + precursor<br />

• H gene controls<br />

synthesis of precursor<br />

• H = normal, dominant<br />

allele (can add sugars)<br />

• h = abnormal,<br />

recessive allele<br />

(cannot add A or B<br />

regardless of<br />

genotype)<br />

11

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