Download BIO152 Course in Review

BIO152 Course in Review
Tutorial 11
December 8
12/8/2006
1
Tutorial 11-Outline
„
What we have covered!
12/8/2006
2
1
Biology studies the unity & diversity of life
12/8/2006
3
‘Recipe’ for life is based on DNA; all life
shares the same basic machinery for inheritance
12/8/2006
4
2
Major themes in BIO152
Foundations of biology
All life is related & populations have
changed over time = evolution by
natural selection
1.
2.
The cell is the basic unit of life & a cell comes from
other cells = cell theory
3.
DNA is the basic hereditary material =
chromosome theory of inheritance
12/8/2006
5
Chromosomal theory of inheritance
DNA is the macromolecule carrying the
heritable material.
DNA is arranged in chromosomes
Chromosomes behave according to
Mendel’s laws:
1.
2.
3.
1.
2.
12/8/2006
segregation of homologous chromosomes &
independent assortment of homologous
chromosomes
During meiosis I
6
3
Pattern and Process
What we see is the pattern.
How this pattern arose is the process.
12/8/2006
7
Chromosome theory of inheritance
„
What is the pattern?
segregation of homologous chromosomes &
independent assortment of homologous
chromosomes
„
What is the process?
meiosis
12/8/2006
8
4
Chromosome theory of inheritance
„
What is the pattern?
„
What is the process?
12/8/2006
9
BIO152H
„
„
„
Demonstrated how science provides a way to
study of life
Explored the pattern and process of
evolution
Studied genetics to explore how variable
traits arise and are inherited
12/8/2006
10
5
Evolution
Pattern
Species are not independent and unchanging
entities but are related to one another and
can change through time.
Process
Change in allele frequencies in species over
time due to natural selection, genetic drift,
and mutation
12/8/2006
11
Evolution: change in populations over
time due to
„
Variation
‰ Selection
„ Inheritance
‰Time
12/8/2006
12
6
Natural selection
More individuals are born than survive and
reproduce.
2. Individuals vary in characteristics which are
heritable.
3. Certain heritable characteristics help
individuals survive and reproduce better
than other characteristics.
Natural selection is NOT a random process.
1.
12/8/2006
13
Natural selection
„
Occurs when heritable variation in certain
traits leads to improved reproductive
success
= Biological fitness
Adaptation-is a trait which confers
reproductive success
12/8/2006
14
7
Important implications
1. At what level does natural selection happen?
a. Individuals
b. Populations
2. At what level does evolution happen?
a. Individuals
b. Populations
12/8/2006
15
Lecture 2-Need doesn’t help if you don’t have
the trait
Natural selection needs genetic variation, but
actually acts on variation in __________
12/8/2006
16
8
Individuals do not change, populations change
over time.
Change is not progressive, changes in
populations with time [ evolution] reflective
the environment needed to survive and
reproduce at that time
12/8/2006
17
Extension
Evolution depends on _____ variation.
Natural selection acts on _____ variation
Phenotypic
Genotypic
[answers may be used more than once]
12/8/2006
18
9
Hardy Weinberg
Suppose 49% of a remote mountain village can not
taste phenylthiocarbamide (PTC). If this population
conforms to Hardy-Weinberg expectations for this
gene, what is the frequency of the population that
must be [SHOW your calculations]
„
homozygous dominant for this trait? ________
„
heterozygous for this trait?
________
„
tasters of PTC?
12/8/2006
19
genotype frequencies:
p2 + 2pq + q2 = 1.0
What about phenotype frequencies?
Recessive trait: ?
Dominant trait: ?
allele frequencies
p+q=1
12/8/2006
20
10
genotype frequencies:
p2 + 2pq + q2 = 1.0
What about phenotype frequencies?
Recessive trait: q2
Dominant trait: p2 + 2pq
allele frequencies
p+q=1
12/8/2006
21
PTC problem
49% of a remote mountain village can not taste
phenylthiocarbamide (PTC).
What do we know to calculate the frequency
„
homozygous dominant for this trait?
________
„
heterozygous for this trait?
________
„
tasters of PTC?
12/8/2006
22
11
genotype frequencies:
p2 + 2pq + q2 = 1.0
q2 = 0.49 therefore, q = 0.7
p= 1-q = 1- 0.7 = 0.3
Homozygous dominant?
Heterozygous?
Tasters?
12/8/2006
23
genotype frequencies:
p2 + 2pq + q2 = 1.0
q2 = 0.49 therefore, q = 0.7
p= 1-q = 1- 0.7 = 0.3
Homozygous dominant = p2 = 0.09
Heterozygous= 2pq = 2(0.7 x 0.3) = 0.42
Tasters= p2 + 2pq OR 1 - q2 = 0.51
12/8/2006
24
12
Pedigree of PTC tasting in 2 related families
open symbols are tasters; closed are non tasters
12/8/2006
25
Pattern of inheritance?
What is the most likely
genotype of
I – 1: tt
II – 3: TT
III – 2: Tt
III – 4: Tt or TT
III – 11: TT
12/8/2006
26
13
the principle of segregation?
the principle of independent assortment?
a. Half of the gametes produced by an individual that is
AaBb will be AB.
b. Each gamete has an equal chance of getting either
allele for a gene.
c. Each gamete formed in an organism has two copies
of each gene.
d. Each gamete formed in an organism will have one
copy of each gene.
12/8/2006
27
the principle of segregation?
the principle of independent assortment?
a. Half of the gametes produced by an individual that is
AaBb will be AB.
b. Each gamete has an equal chance of getting either
allele for a gene. Independent Assortment
c. Each gamete formed in an organism has two copies
of each gene.
d. Each gamete formed in an organism will have one
copy of each gene. Segregation
12/8/2006
28
14
What is the relationship
Between the principle of independent
assortment and LINKAGE?
12/8/2006
29
Speciation
What are the three requirements?
12/8/2006
30
15
Under what circumstance
Would the accumulated genetic differences
precede stopping gene flow?
12/8/2006
31
12/8/2006
32
16
„
Soapberry bugs (Freeman Fig 25.8)
12/8/2006
33
12/8/2006
34
17
12/8/2006
35
The END!
Good luck on the final exam
&
Enjoy the rest of your BIO courses at UTM!
12/8/2006
36
18
Errors when chromosomes fail to separate
properly –nondisjunction
12/8/2006
37
12/8/2006
38
19