1. 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 inheritance I. Intro A. Scale 0-1 1. 0= will not occur 2. 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

2. II. The multiplication and addition rules applied to monohybrid crosses A. Multiplication rule 1. Prediction of 2 independent events occurring simultaneously a. 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

3. e.x  F 1 cross Rr X Rr R=round r= wrinkle Event 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

4. B. Addition rule 1. Probability of 1 of 2 or more mutually exclusive events will occur is calculated by adding their individual probabilities a. Probability of 1 of 2 or more related but independent events will occur is calculated by adding their individual probabilities e.x  Rr X Rr What is probability of heterozygous round (Rr) Rr = 1/4 rR = 1/4 1/2

5. e.x  Cross = Rr X rr What is probability of Rr What is probability of rr 1/2 Rr 1/2 rr

6. III. Solving complex genetics problems with the rules of probability A. Dihybrids 1. Cross = YyRr X YyRr a. What fraction of offspring would be predicted to have YyRR 1. 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

7. 2. Practice= TTQq X TtQq a. What is the frequency of the genotype TTQq in the F2 generation B. Trihybrids 1. Cross = QqTtRr X Qqttrr a. 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 condition qqttRR qqttRr qqTTrr qqTtrr QQttrr Qqttrr qqttrr

8. 2. Step 2- List all possible genotypes based on cross qqttRR qqttRr qqTTrr qqTtrr QQttrr Qqttrr qqttrr 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

9. 4. Implement multiplication and addition rules qqttRr 1/4 (probability of qq) X 1/2 (tt) X 1/2 (Rr) = 1/16 qqTtrr 1/4 X 1/2 X 1/2= 1/16 Qqttrr 1/2 X 1/2 X 1/2= 1/8 or 2/16 QQttrr 1/4 X 1/2 X 1/2= 1/16 qqttrr 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 genetics I. Extending Mendelian genetics for a single gene A. Degrees of dominance 1. Complete dominance a. Mendel ’ s work b. One allele overshadows/masks the other c. Homozygous dominant & heterozygous phenotypically the same

10. 2. Incomplete dominance a. Offspring are phenotypically intermediate between 2 parents 1. Heterozygous flowers produce less red pigment than red homozygote

11. 3. Codominance a. Both alleles of a gene are expressed phenotypically b. ABO blood grouping

12. B. Relationship between dominance & phenotype 1. How is dominance achieved a. Alleles=nucleotide sequence  proteins  function 1. Individual alleles do not interact 2. Dominance or recessive is achieved through allele expression b. Ex  Mendel ’ s peas 1. Round (dominant) & wrinkled (recessive) a. Round allele codes for enzyme b. Wrinkled allele codes for defective enzyme c. Tay-sachs disease 1. Disease manifests when enzymes cannot breakdown certain lipids in the brain a. Seizures, blindness, degeneration of motor & mental performance, & death 2. Homozygous dominant & heterozygous = no manifestation 3. Homozygous recessive = manifestation

13. 2. Dominance/recessive a matter of viewpoint a. Tay-sachs disease 1. Organismal level dominant/recessive 2. 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 enzyme C. Frequency of dominance 1. Dominant allele not always the higher frequency a. Polydactyly D. Multiple alleles 1. Blood groups (ABO)

14. E. Pleiotropy 1. 1 gene affecting multiple phenotypes 2. Garden pea gene for flower color also influences seed color II. Extending Mendelian genetics for two or more genes A. 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 color a. However, a different gene for color (C) controls the release of the pigments needed for hair color 1. bb = brown but if ccbb will be albino

15. B. Polygenic inheritance 1. Multiple genes controlling a particular phenotype a. Phenotype exists as a continuum 1. Quantitative characters 2. Height & skin color

16. III. Nature and nurture: The environmental impact on phenotype