2nd year

1. GENETICS

2. Genetics
• The study of heredity.
heredity
• Gregor Mendel (1860’s) discovered the
fundamental principles of genetics by breeding
garden peas.
peas

3. Genetics
• Alleles
1. Alternative forms of genes.
2. Units that determine heritable traits.
3. Dominant alleles (TT - tall pea plants)
plants
a. homozygous dominant
4. Recessive alleles (tt - dwarf pea plants)
plants
a. homozygous recessive
5. Heterozygous (Tt - tall pea plants)
plants

4. Phenotype
• Outward appearance
• Physical characteristics
• Examples:
1. tall pea plant
2. dwarf pea plant

5. Genotype
• Arrangement of genes that produces the
phenotype
• Example:
1. tall pea plant
TT = tall (homozygous dominant)
2. dwarf pea plant
tt = dwarf (homozygous recessive)
3. tall pea plant
Tt = tall (heterozygous)

6. Punnett square
• A Punnett square is used to show the
possible combinations of gametes.
gametes

7. Breed the P generation
• tall (TT) vs. dwarf (tt) pea plants
T T
t
t

8. tall (TT) vs. dwarf (tt) pea plants
T T
Tt produces the
t Tt
F1 generation
t Tt Tt All Tt = tall
(heterozygous tall)

9. Breed the F1 generation
• tall (Tt) vs. tall (Tt) pea plants
T t
T
t

10. tall (Tt) vs. tall (Tt) pea plants
T t
produces the
TT Tt F2 generation
T
1/4 (25%) = TT
Tt tt 1/2 (50%) = Tt
t
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype

11. Monohybrid Cross
• A breeding experiment that tracks the inheritance
of a single trait.
• Mendel’s “principle of segregation”
a. pairs of genes separate during gamete
formation (meiosis).
b. the fusion of gametes at fertilization pairs
genes once again.

12. Homologous Chromosomes
eye color locus eye color locus
B = brown eyes b = blue eyes
This person would
have brown eyes (Bb)
Paternal Maternal

13. Meiosis - eye color
B
B sperm
B
Bb
haploid (n)
b
diploid (2n) b
b
meiosis I meiosis II

14. Monohybrid Cross
• Example:
Example Cross between two heterozygotes
for brown eyes (Bb)
BB = brown eyes
B b male
Bb = brown eyes gametes
bb = blue eyes
B
Bb x Bb
b
female gametes

15. Monohybrid Cross
B b
1/4 = BB - brown eyed
B BB Bb 1/2 = Bb - brown eyed
Bb x Bb 1/4 = bb - blue eyed
b Bb bb
1:2:1 genotype
3:1 phenotype

16. Dihybrid Cross
• A breeding experiment that tracks the inheritance
of two traits.
• Mendel’s “principle of independent assortment”
a. each pair of alleles segregates independently
during gamete formation (metaphase I)
b. formula: 2n (n = # of heterozygotes)

17. Independent Assortment
• Question: How many gametes will be produced
for the following allele arrangements?
• Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq

18. Answer:
1. RrYy: 2n = 22 = 4 gametes
RY Ry rY ry
2. AaBbCCDd: 2n = 23 = 8 gametes
ABCD ABCd AbCD AbCd
aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes

19. Dihybrid Cross
• Example: cross between round and yellow
heterozygous pea seeds.
R = round RrYy x RrYy
r = wrinkled
Y = yellow
RY Ry rY ry x RY Ry rY ry
y = green possible gametes produced

20. Dihybrid Cross
RY Ry rY ry
RY
Ry
rY
ry

21. Dihybrid Cross
RY Ry rY ry
Round/Yellow: 9
RY RRYY RRYy RrYY RrYy
Round/green: 3
Ry RRYy RRyy RrYy Rryy
wrinkled/Yellow: 3
rY RrYY RrYy rrYY rrYy wrinkled/green: 1
ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio

22. Test Cross
• A mating between an individual of unknown genotype
and a homozygous recessive individual.
• Example: bbC__ x bbcc
BB = brown eyes
Bb = brown eyes
bC b___
bb = blue eyes
bc
CC = curly hair
Cc = curly hair
cc = straight hair

23. Test Cross
• Possible results:
bC b___
C bC b___
c
bc bbCc bbCc or bc bbCc bbcc

24. Incomplete Dominance
• F1 hybrids have an appearance somewhat in
between the phenotypes of the two parental
varieties.
• Example: snapdragons (flower)
• red (RR) x white (rr)
R R
RR = red flower r
rr = white flower
r

25. Incomplete Dominance
R R
produces the
r Rr Rr
F1 generation
r Rr Rr All Rr = pink
(heterozygous pink)

26. Codominance
• Two alleles are expressed (multiple alleles)
alleles
in heterozygous individuals.
individuals
• Example: blood
1. type A = IAIA or IAi
2. type B = IBIB or IBi
3. type AB = IAIB
4. type O = ii

27. Codominance
• Example: homozygous male B (IBIB)
x
heterozygous female A (IAi)
IB IB
IA I AI B I AI B
1/2 = IAIB
1/2 = IBi
i I Bi IBi

28. Codominance
• Example: male O (ii) x female AB (IAIB)
IA IB
i I Ai IBi 1/2 = IAi
1/2 = IBi
i I Ai IBi

29. Codominance
• Question:
Question If a boy has a blood type O and
his sister has blood type AB,
what are the genotypes and
phenotypes of their parents.
• boy - type O (ii) X girl - type AB (IAIB)

30. Codominance
• Answer:
IA i
I B I AI B Parents:
genotypes = IAi and IBi
phenotypes = A and B
i ii

31. Sex-linked Traits
• Traits (genes) located on the sex
chromosomes
• Example: fruit flies
(red-eyed male) X (white-eyed female)
red white

32. Sex-linked Traits
Sex Chromosomes
fruit fly
eye color
XX chromosome - female Xy chromosome - male

33. Sex-linked Traits
• Example: fruit flies
(red-eyed male) X (white-eyed female)
• Remember: the Y chromosome in males
does not carry traits.
RR = red eyed XR y
Rr = red eyed
rr = white eyed Xr
Xy = male
Xr
XX = female

34. Sex-linked Traits
XR y
Xr XR Xr Xr y
1/2 red eyed and female
1/2 white eyed and male
Xr XR Xr Xr y

35. Population Genetics
• The study of genetic changes in populations.
populations
• The science of microevolutionary changes in
populations.
populations
• Hardy-Weinberg equilibrium:
the principle that shuffling of genes that occurs
during sexual reproduction, by itself, cannot
change the overall genetic makeup of a population.
• Hardy-Wienberg equation: 1 = p2 + 2pq + q2

36. Question:
• How do we get this equation?
Answer: “Square” 1 = p + q
↓
12 = (p + q)2
↓
1 = p2 + 2pq + q2

37. Hardy-Wienberg equation
• Five conditions are required for Hardy-Wienberg
equilibrium.
1. large population
2. isolated population
3. no net mutations
4. random mating
5. no natural selection

38. Important
• Need to remember the following:
p2 = homozygous dominant
2pq = heterozygous
q2 = homozygous recessive

39. Question:
• Iguanas with webbed feet (recessive trait) make
up 4% of the population. What in the population
is heterozygous and homozygous dominant.
dominant

40. Answer:
1. q2 = 4% or .04 q2 = .04 q = .2
2. then use 1 = p + q
1 = p + .2 1 - .2 = p .8 = p
3. for heterozygous use 2pq
2(.8)(.2) = .32 or 32%
4. For homozygous dominant use p2
.82 = .64 or 64%

41. Hardy-Wienberg equation
1 = p2 + 2pq + q2
• 64% = p2 = homozygous dominant
• 32% = 2pq = heterozygous
• 04% = q2 = homozygous recessive
• 100%