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

2. Use of the Garden Pea for Genetics Experiments

3. 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)

4. 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.

5. 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.

6. 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

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

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

9. 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

10. 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.

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

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

13. Seven Traits used by Mendel in Genetic Studies

14. 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.

15. 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

16. 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 / round, green / wrinkled, yellow 3/ wrinkled, green 1/16

17. 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.

18. 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.

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

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

21. 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

22. Meiotic Segregation explains Independent Assortment

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

24. 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 ways to get Rr genotype possible outcomes P(Rr) = 2/4 = 1/2

25. 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 ways to get Rr genotype way to get RR genotype possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4

26. 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

27. 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?

28. 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

29. 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

30. 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

31. 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

32. 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)

33. 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! ! = 1 2! = 2 x 1

34. 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! ! 1!

35. 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 = 24/729

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

37. 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

38. X2 p