2. Heredity!!!

3. Known for Creating the science of genetics
Born Johann Mendel
20 July 1822
Austrian Empire (Czech Republic)
Died 6 January 1884 (aged 61)
Nationality Austrian
Fields Genetics
Institutions St Thomas's Abbey
Alma mater: University of Olomouc
University of Vienna
Known for Creating the science of genetics

9. Mendel!
He became a friar because it enabled him to obtain an education without having to pay for it himself! (think yourself at this time next year)
Mendel became a priest in 1847 and got his own parish in He did not enjoy working as a parish priest and instead got a job as a high school teacher in (Yes, he washed out as a priest)
 Mendel worked as a substitute high school teacher. (In 1850, he failed the oral part, the last of three parts, of his exams to become a certified high school teacher).

10. Two years later, after completing his studies, he returned to the monastery in 1854 and took a position as a physics teacher at a school at Brünn, where he taught for the next 16 years (still classified as a sub).
In 1856, he took the exam to become a certified teacher and again failed the oral part. (Come on people. Even Mr. Pickett passed that test).
He eventually became head of the abbey and gave up his teaching career (and his scientific studies). He was dead before his work became widely known!

11.  Mendel's pea plant experiments conducted between 1856 and 1863 established most of the rules of heredity, now referred to as the laws of Mendelian inheritance.
The profound significance of Mendel's work was not recognized until the turn of the 20th century (more than three decades later) with the independent rediscovery of these laws.

12. We still use Mendel’s terms today!

14. Unfortunately he didn’t create the Punnett Square!
Mendel Punnett

15. Who was Punnett?
 Although he went to Cambridge University as a medical student, Punnett graduated with a zoology degree in After graduation, Punnett continued at Cambridge as a researcher. He did work on the morphology of nemertine (ribbon) worms. Punnett has two species of marine worms named after him, Cerbratulus punnetti, Punnettia splendia!
Really, Oh Wow! Not one but 2 worms! I think I
want a career in science too!

16. Maybe your questions are more important than you think!
In 1908, Punnett was asked at a lecture to explain why recessive phenotypes still persist — if brown eyes were dominant, then why wasn't the whole country becoming brown-eyed? Punnett couldn't answer the question to his own satisfaction. He in turn asked his friend the mathematician, G. H. Hardy. Out of this conversation came the Hardy-Weinberg Law which calculates how population affects genetic inheritance.

17. Hardy Weinberg principles:
evolution will not occur in a population if seven conditions are met:
1. mutation is not occurring
2. natural selection is not occurring
3. the population is infinitely large
4. all members of the population breed
5. all mating is totally random
6. everyone produces the same number of offspring
7. there is no migration in or out of the population
IN NATURE ALL 7 OF THESE ARE NEVER TRUE AT ONE TIME!
THEREFORE EVOLUTION IS ALLWAYS HAPPENING!

18. Measuring Evolution of Populations

19. 5 Agents of evolutionary change
Mutation
Gene Flow
Non-random mating
Genetic Drift
Selection

20. Populations & gene pools
Concepts
a population is a localized group of interbreeding individuals
gene pool is collection of alleles in the population
remember difference between alleles & genes!
allele frequency is how common is that allele in the population
how many A vs. a in whole population

21. Evolution of populations
Evolution = change in allele frequencies in a population
hypothetical: what conditions would cause allele frequencies to not change?
non-evolving population
REMOVE all agents of evolutionary change
very large population size (no genetic drift)
no migration (no gene flow in or out)
no mutation (no genetic change)
random mating (no sexual selection)
no natural selection (everyone is equally fit)

22. FIRST POPULATION Q square = 0.3 (That is Q square = 3/10)

23. FIRST POPULATION EAT 3 SAD

24. SECOND POPULATION 2X7

25. SECOND POPULATION 2X7

26. SECOND POPULATION Eat 3

27. THIRD POPULATION Double the 11

28. Then eat 3 and double the 19 remaining to get the 4 population of 38

29. Hardy-Weinberg equilibrium
Hypothetical, non-evolving population
preserves allele frequencies
Serves as a model (null hypothesis)
natural populations rarely in H-W equilibrium
useful model to measure if forces are acting on a population
measuring evolutionary change
G.H. Hardy (the English mathematician) and W. Weinberg (the German physician) independently worked out the mathematical basis of population genetics in Their formula predicts the expected genotype frequencies using the allele frequencies in a diploid Mendelian population. They were concerned with questions like "what happens to the frequencies of alleles in a population over time?" and "would you expect to see alleles disappear or become more frequent over time?"
G.H. Hardy
mathematician
W. Weinberg
physician

30. if you cross heterozygotes you get:
AA + 2Aa + aa from a Punnett Square
In Math terms this is the same as:
p2 + 2pq + q2

31. Hardy-Weinberg theorem
Counting Alleles
assume 2 alleles = B, b
frequency of dominant allele (B) = p
frequency of recessive allele (b) = q
frequencies must add to 1 (100%), so:
p + q = 1 BB Bb bb

32. Hardy-Weinberg theorem
Counting Individuals
frequency of homozygous dominant: p x p = p2
frequency of homozygous recessive: q x q = q2
frequency of heterozygotes: (p x q) + (q x p) = 2pq
frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1 BB Bb bb

33. H-W formulas Alleles: p + q = 1 Individuals: p2 + 2pq + q2 = 1 B b BB

34. H-W formulas Individuals: p2 + 2pq + q2 = 1
WHEN WE LOOK AT AN ACTUAL TRAIT IN A POPULATION, WHICH OF THE CATS CAN WE ACTUALLY SEE?
Individuals: p2 + 2pq + q2 = 1 BB Bb bb BB Bb bb

35. Using Hardy-Weinberg equation
population: 100 cats
84 black, 16 white
How many of each genotype?
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): = 0.6
p2=.36
2pq=.48
q2=.16 BB Bb bb
Must assume population is in H-W equilibrium!
What are the genotype frequencies?

36. Using Hardy-Weinberg equation
p2=.36
2pq=.48
q2=.16
Assuming H-W equilibrium BB Bb bb
Null hypothesis
p2=.20
p2=.74
2pq=.10
2pq=.64
q2=.16
q2=.16
Sampled data 1:
Hybrids are in some way weaker.
Immigration in from an external population that is predomiantly homozygous B
Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats.
Sampled data 2:
Heterozygote advantage.
What’s preventing this population from being in equilibrium. bb Bb BB
Sampled data
How do you explain the data?
How do you explain the data?

37. Application of H-W principle
Sickle cell anemia
inherit a mutation in gene coding for hemoglobin
oxygen-carrying blood protein
recessive allele = HsHs
normal allele = Hb
low oxygen levels causes RBC to sickle
breakdown of RBC
clogging small blood vessels
damage to organs
often lethal

38. Sickle cell frequency High frequency of heterozygotes
1 in 5 in Central Africans = HbHs
unusual for allele with severe detrimental effects in homozygotes
1 in 100 = HsHs
usually die before reproductive age
Sickle Cell:
In tropical Africa, where malaria is common, the sickle-cell allele is both an advantage & disadvantage. Reduces infection by malaria parasite.
Cystic fibrosis:
Cystic fibrosis carriers are thought to be more resistant to cholera:
1:25, or 4% of Caucasians are carriers Cc
Why is the Hs allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…

39. Malaria
Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

40. Heterozygote Advantage
In tropical Africa, where malaria is common:
homozygous dominant (normal)
die or reduced reproduction from malaria: HbHb
homozygous recessive
die or reduced reproduction from sickle cell anemia: HsHs
heterozygote carriers are relatively free of both: HbHs
survive & reproduce more, more common in population
Hypothesis:
In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite.
Frequency of sickle cell allele & distribution of malaria

41. Any Questions??

42. Hardy-Weinberg Lab Data
Mutation
Gene Flow
Genetic Drift
Selection
Non-random mating

43. Hardy Weinberg Lab: Equilibrium
Original population
Case #1 F5
18 individuals
36 alleles
p (A): 0.5
q (a): 0.5
total alleles = 36
p (A): (4+4+7)/36 = .42
q (a): (7+7+7)/36 = .58 AA 4 Aa 7 aa 7 AA
.25 Aa
.50 aa
.25 AA
.22 Aa
.39 aa
.39
How do you explain these data?

44. Hardy Weinberg Lab: Selection
Original population
Case #2 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (9+9+6)/30 = .80
q (a): (0+0+6)/30 = .20 AA 9 Aa 6 aa AA
.25 Aa
.50 aa
.25 AA
.60 Aa
.40 aa
How do you explain these data?

45. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (4+4+11)/30 = .63
q (a): (0+0+11)/30 = .37 AA 4 Aa 11 aa AA
.25 Aa
.50 aa
.25 AA
.27 Aa
.73 aa
How do you explain these data?

46. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F10
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (6+6+9)/30 = .70
q (a): (0+0+9)/30 = .30 AA 6 Aa 9 aa AA
.25 Aa
.50 aa
.25 AA .4 Aa .6 aa
How do you explain these data?

47. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-1
6 individuals
12 alleles
p (A): 0.5
q (a): 0.5
total alleles = 12
p (A): (4+4+2)/12 = .83
q (a): (0+0+2)/12 = .17 AA 4 Aa 2 aa AA
.25 Aa
.50 aa
.25 AA
.67 Aa
.33 aa
How do you explain these data?

48. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-2
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (0+0+4)/10 = .4
q (a): (1+1+4)/10 = .6 AA Aa 4 aa 1 AA
.25 Aa
.50 aa
.25 AA Aa .8 aa .2
How do you explain these data?

49. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-3
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (2+2+2)/10 = .6
q (a): (1+1+2)/10 = .4 AA 2 Aa 2 aa 1 AA
.25 Aa
.50 aa
.25 AA .4 Aa .4 aa .2
How do you explain these data?

50. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA Aa aa p q AA
.25 Aa
.50 aa
.25
How do you explain these data?

51. Any Questions??