Patterns of Inheritance What patterns can be observed when traits are passed to the next generation?

Use of the Garden Pea for Genetics Experiments

Principles of Heredity Mendel’s Experiment with Peas Round seed x Wrinkled seed F1: All round seed coats F1 round plants x F1 round plants F2: 5474 round: 1850 wrinkled (3/4 round to 1/4 wrinkled)

Principles of Heredity Mendel needed to explain Why one trait seemed to disappear in the first generation. 2. Why the same trait reappeared in the second generation in one-fourth of the offspring.

Principles of Heredity Mendel proposed: Each trait is governed by two factors – now called genes. 2. Genes are found in alternative forms called alleles. 3. Some alleles are dominant and mask alleles that are recessive.

Principles of Heredity Mendel’s Experiment with Peas Round seed x Wrinkled seed RR rr F1: All round seed coats Rr Homozygous Dominant Homozygous Recessive Heterozygous

Homozygous parents can only pass one form of an allele to their offspring. R

Heterozygous parents can pass either of two forms of an allele to their offspring. R r R r

Principles of Heredity Additional Genetic Terms Genotype: alleles carried by an individual eg. RR, Rr, rr Phenotype: physical characteristic or appearance of an individual eg. Round, wrinkled

Mendel’s Principle of Genetic Segregation In the formation of gametes, the members of a pair of alleles separate (or segregate) cleanly from each other so that only one member is included in each gamete. Each gamete has an equal probability of containing either member of the allele pair.

Genetic Segregation Parentals: RR x rr F1 x F1: Rr x Rr R R r r

Genetic Segregation Genotypic Ratio: ¼ RR + ½ Rr + ¼ rr Phenotypic Ratio: ¾ Round + ¼ Wrinkled

Seven Traits used by Mendel in Genetic Studies

What Is a Gene? A gene is a segment of DNA that directs the synthesis of a specific protein. DNA is transcribed into RNA which is translated into protein.

Molecular Basis for Dominant and Recessive Alleles Dominant Allele Codes for a functional protein Recessive Allele Codes for a non-functional protein or prevents any protein product from forming

Principles of Heredity Mendel’s Experiment with Peas Round Yellow x Wrinkled Green F1: All round yellow seed coats F1 plants x F1 plants F2: 315 round, yellow 9/16 108 round, green 3/16 101 wrinkled, yellow 3/16 32 wrinkled, green 1/16

Principles of Heredity Mendel needed to explain Why non-parental combinations appeared in the F2 offspring. 2. Why the ratio of phenotypes in the F2 generation was 9:3:3:1.

Mendel’s Principle of Independent Assortment When gametes are formed, the alleles of one gene segregate independently of the alleles of another gene producing equal proportions of all possible gamete types.

Genetic Segregation + Independent Assortment Parentals: RRYY x rryy RY RY RY RY ry ry ry ry ry RY RrYy F1: 100% RrYy, round, yellow

F1 x F1: RrYy x RrYy RY Ry rY ry RY Ry rY ry ¼ RY ¼ Ry ¼ rY ¼ ry ¼ RY

F2 Genotypes and Phenotypes Round Yellow 1/16 RRYY + 2/16 RRYy + 2/16 RrYY + 4/16 RrYy Total = 9/16 R_Y_ Green 1/16 RRyy+ 2/16 Rryy Total = 3/16 R_yy Wrinkled Yellow 1/16 rrYY+ 2/16 rrYy Total = 3/16 rrY_ Wrinkled Green 1/16 rryy

Meiotic Segregation explains Independent Assortment

Solving Genetics Problems Convert parental phenotypes to genotypes Use Punnett Square to determine genotypes of offspring Convert offspring genotypes to phenotypes

Using Probability in Genetic Analysis 1. Probability (P) of an event (E) occurring: P(E) = Number of ways that event E can occur Total number of possible outcomes Eg. P(Rr) from cross Rr x Rr 2 ways to get Rr genotype 4 possible outcomes P(Rr) = 2/4 = 1/2

Using Probability in Genetic Analysis 2. Addition Rule of Probability – used in an “either/or” situation   P(E1 or E2) = P(E1) + P(E2) Eg. P(Rr or RR) from cross Rr x Rr 2 ways to get Rr genotype 1 way to get RR genotype 4 possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4

Using Probability in Genetic Analysis 3. Multiplication Rule of Probability – used in an “and” situation   P(E1 and E2) = P(E1) X P(E2) Eg. P(wrinkled, yellow) from cross RrYy x RrYy P(rr and Y_) = 1/4 x 3/4 = 3/16

Using Probability in Genetic Analysis 4. Conditional Probability: Calculating the probability that each individual has a particular genotype   Eg. Jack and Jill do not have PKU. Each has a sibling with the disease. What is the probability that Jack and Jill will have a child with PKU?

Using Probability in Genetic Analysis 4. Conditional Probability Jack is P_, Jill is P_ Parents of Jack or Jill: Pp x Pp P p P p PP Pp pp P(Pp) = 2 ways to get Pp 3 possible genotypes P(Jack is Pp) =2/3 P (Jill is Pp) = 2/3 X

Using Probability in Genetic Analysis 4. Conditional Probability P(child with PKU)= P(Jack is Pp) x P(Jill is Pp) x P(child is pp) = 2/3 x 2/3 x 1/4 = 1/9 P(child without PKU)= 1-1/9 = 8/9

Using Probability in Genetic Analysis To calculate probability of child without PKU, look at all possibilities for Jack and Jill. Jack Jill P_ child Probability 1/3 PP 1 1/9 2/3 Pp 2/9 3/4 3/9 Total=8/9

Using Probability in Genetic Analysis 5. Ordered Events: use Multiplication Rule For Jack and Jill, what is the probability that the first child will have PKU, the second child will not have PKU and the third child will have PKU? P(pp) x P(P_) x P(pp) = 1/9 x 8/9 x 1/9 = 8/729

Using Probability in Genetic Analysis 6. Binomial Rule of Probability – used for unordered events P = n! (asbt) s! t! a = probability of event X (occurrence of one event) b = probability of event Y = 1-a (occurrence of alternate event) n = total s = number of times event X occurs t = number of times event Y occurs (s + t = n)

Using Probability in Genetic Analysis 6. Binomial Rule of Probability ! = factorial= number multiplied by each lower number until reaching 1 5! = 5 x 4 x 3 x 2 x 1 1! =1 3! = 3 x 2 x 1 = 3 x 2! 0! = 1 2! = 2 x 1

Using Probability in Genetic Analysis 6. Binomial Rule of Probability Out of 3 children born to Jack and Jill, what is the probability that 2 will have PKU? n=3, a=1/9, s=2, b=8/9, t=1 3! (1/9)2(8/9)1= 3 x 2! (1/81) (8/9)= 24 2! 1! 2! 1! 729

Using Probability in Genetic Analysis The same result can be obtained using the multiplicative rule if all possible birth orders for families of three are considered: 1st child 2nd child 3rd child Probability PKU=1/9 No= 8/9 8/729 No=8/9 8/729 + 8/729 + 8/729 = 24/729

Chi-Square Goodness of Fit Test To evaluate how well data fits an expected genetic ratio

Chi-square Test for Goodness of Fit for 9:3:3:1 Ratio Phenotype Observed Number Expected Number (Fraction x Total) O-E (O-E)2 E Round, yellow 315 Round, green 108 Wrinkled, yellow 101 Wrinkled, green 32 Total 556 9/16 x 556 = 313 2 4 .0128 3/16 x 556 = 104 4 16 .154 3/16 x 556 = 104 -3 9 .087 1/16 x 556 = 35 -3 9 .257 X2= .511 df=degrees of freedom= number of phenotypes – 1 = 4-1=3 p value from table on page 1-17: p>.5 from table in Pierce: .975 > p >.9 Data supports hypothesis for any p>0.05

X2 p