CHAPTER 14 MENDEL & THE GENE IDEA

14.1 Mendel used the scientific approach to identify two laws of inheritance

14.2 The laws of probability govern Mendelian inheritanceI. Intro

A. Scale 0-11. 0= will not occur2. 1= certain to occur

B. Important lesson of probability 1. Each event is independent of the next

a. Alleles of 1 gene segregate into gametes independently of another gene’s alleles

C. Probability can help us predict the outcome of the fusion of gametes

II. The multiplication and addition rules applied to monohybrid crossesA. Multiplication rule

1. Prediction of 2 independent events occurring simultaneouslya. Multiply all independent event probabilities

e.x Tossing pennies Event 1 Probability of tails = 1/2 Event 2 Probability of tails = 1/2 What is the probability of 2 coins flipped simultaneously of landing on tails

1/2 (event 1) x 1/2 (event 2)= 1/4

e.x F1 cross Rr X Rr R=round r= wrinkleEvent 1 Probability of egg receiving R = 1/2

Probability of egg receiving r = 1/2 Event 2 Probability of sperm receiving R = 1/2

Probability of sperm receiving r = 1/2 What is probability of egg + sperm = RR

1/2 (egg with R) X 1/2 (Sperm with R) = 1/4

B. Addition rule1. Probability of 1 of 2 or more mutually exclusive events will

occur is calculated by adding their individual probabilitiesa. Probability of 1 of 2 or more related but independent events

will occur is calculated by adding their individual probabilities

e.x Rr X RrWhat is probability of heterozygous round (Rr)

Rr = 1/4 rR = 1/4

1/2

e.x Cross = Rr X rrWhat is probability of RrWhat is probability of rr

1/2 Rr

1/2 rr

III. Solving complex genetics problems with the rules of probabilityA. Dihybrids

1. Cross = YyRr X YyRra. What fraction of offspring would be predicted to have

YyRR1. Step 1- due to independent assortment you can deal

with the 2 genes separately a. Set up a monohybrid cross for each

2. Step 2- Now use the laws of probability 1/2 Yy x 1/4 RR = 1/8 YyRR or 2/16

1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

1/4 rr

2. Practice= TTQq X TtQqa. What is the frequency of the genotype TTQq in the F2

generationB. Trihybrids

1. Cross = QqTtRr X Qqttrra. What fraction of offspring would be predicted to exhibit

the recessive phenotype for at least 2 of the three characteristics 1. Step 1- List all possible genotypes of offspring fulfilling

conditionqqttRRqqttRrqqTTrrqqTtrrQQttrrQqttrrqqttrr

2. Step 2- List all possible genotypes based on crossqqttRRqqttRrqqTTrrqqTtrrQQttrrQqttrrqqttrr

3. Monohybrid punnett square

1/4 QQ

1/2 Qq

1/4 qq

1/2 Tt

1/2 tt

1/2 Rr

1/2 rr

4. Implement multiplication and addition rulesqqttRr 1/4 (probability of qq) X 1/2 (tt) X 1/2 (Rr) = 1/16qqTtrr 1/4 X 1/2 X 1/2 = 1/16Qqttrr 1/2 X 1/2 X 1/2 = 1/8 or 2/16QQttrr 1/4 X 1/2 X 1/2 = 1/16qqttrr 1/4 X 1/2 X 1/2 = 1/16

Chance of at least 2 recessive traits = 6/16 or 3/8 14.3 Inheritance patterns are often more complex than predicted by

simple Mendelian geneticsI. Extending Mendelian genetics for a single gene

A. Degrees of dominance 1. Complete dominance

a. Mendel’s workb. One allele overshadows/masks the otherc. Homozygous dominant & heterozygous phenotypically the

same

2. Incomplete dominancea. Offspring are phenotypically intermediate between 2

parents1. Heterozygous flowers produce less red pigment than red

homozygote

3. Codominancea. Both alleles of a gene are expressed phenotypicallyb. ABO blood grouping

B. Relationship between dominance & phenotype1. How is dominance achieved

a. Alleles=nucleotide sequence proteins function1. Individual alleles do not interact2. Dominance or recessive is achieved through allele

expressionb. Ex Mendel’s peas

1. Round (dominant) & wrinkled (recessive)a. Round allele codes for enzymeb. Wrinkled allele codes for defective enzyme

c. Tay-sachs disease1. Disease manifests when enzymes cannot breakdown

certain lipids in the braina. Seizures, blindness, degeneration of motor & mental

performance, & death2. Homozygous dominant & heterozygous = no

manifestation3. Homozygous recessive = manifestation

2. Dominance/recessive a matter of viewpointa. Tay-sachs disease

1. Organismal level dominant/recessive2. Biochemical level incomplete dominance

a. Homozygous dominant = complete functional enzyme production

b. Heterozygous = functional enzyme & nonfunctional enzyme production but enough function to prevent manifestation

c. Homozygous = complete nonfunctional enzymeC. Frequency of dominance

1. Dominant allele not always the higher frequencya. Polydactyly

D. Multiple alleles1. Blood groups (ABO)

E. Pleiotropy1. 1 gene affecting multiple phenotypes2. Garden pea gene for flower color also influences seed color

II. Extending Mendelian genetics for two or more genesA. Epistasis

1. Gene at 1 locus alters the phenotypic expression of a gene at a second locus

2. Ex Mice – Black (B) dominant to brown (b) coat colora. However, a different gene for

color (C) controls the release of the pigments needed for hair color1. bb = brown but if ccbb will

be albino

B. Polygenic inheritance1. Multiple genes controlling a particular phenotype

a. Phenotype exists as a continuum1. Quantitative characters2. Height & skin color

III. Nature and nurture: The environmental impact on phenotype