Download AP Biology Review Unit 5 and 6

Watch the following Bozeman Biology Episodes (Others under Unit 3)
028 - Cell Cycle, Mitosis & Meiosis
029 - Mendelian Genetics
030 - Advanced Genetics
032 - Signal Transmission
033 - Genotypes & Phenotype
036 - Evolution of Cell Communication
037 - Cell Communication
038 - Signal Transduction in Pathways
Overview of cell signaling
EXTRACELLULAR
FLUID
1 Reception
Plasma membrane
CYTOPLASM
2 Transduction
3 Response
Receptor
Activation
of cellular
response
Relay molecules in a signal transduction pathway
Signal
molecule
Figure 11.5
Example of Pathway
 Steroid hormones bind to intracellular receptors
Hormone
EXTRACELLULAR
(testosterone)
FLUID
Plasma
membrane
Receptor
protein
Hormonereceptor
complex
1 The steroid
hormone testosterone
passes through the
plasma membrane.
2 Testosterone binds
to a receptor protein
in the cytoplasm,
activating it.
3 The hormone-
DNA
mRNA
NUCLEUS
Figure 11.6
CYTOPLASM
receptor complex
enters the nucleus
and binds to specific
genes.
4
New protein
The bound protein
stimulates the
transcription of
the gene into mRNA.
5 The mRNA is
translated into a
specific protein.
 Other pathways regulate genes by
activating transcription factors that
turn genes on or off
Growth factor
Receptor
Phosphorylation
cascade
Reception
Transduction
CYTOPLASM
Inactive
transcription Active
transcription
factor
factor
P
Response
Figure 11.14
DNA
Gene
NUCLEUS
mRNA
4 Minute action potential
START HERE - RESTING POTENTIAL (Includes multiple other steps)
Na+ is high outside the cell
K+ is high inside the cell
The cell is polarized (negative on the inside and positive on the outside)
A signal is received
Na+ channels open
Na+ diffuses into the cell
K+ channels open
K+ diffuses outside the cell
Depolarization (May include multiple steps)
ACTION POTENTIAL (Includes multiple other steps)
Na/K pump - pumps Na+ OUT and K+ into the cell
Repolarization (Reset) (May include multiple steps)
Action potential moves to the axon terminus
Ca++ channels open
Ca++ moves into the cell
Vesicles with neurotransmitters move to synaptic membrane
Neurotransmitters diffuse from presynaptic axon through synapse
Neurotransmitters bind at postsynaptic membrane (dendrite) of another nerve cell
END HERE The signal is restarted
 Meiosis
 Crossing over
 4 daughter cells all genetically different
 Sodaria lab
Genetics
- Monohybrid cross, dihybrid cross, allele,
homozygous heterozygous, sex-linked, linked genes,
F1, F2, incomplete dominance, codominance
I.P.M.A.T.P.M.A.T
2n=4
interphase 1
prophase 1
metaphase 1
anaphase 1
n=2
prophase 2 metaphase 2 anaphase 2 telophase 2
telophase 1
error in Meiosis 1
error in Meiosis 2
all with incorrect number
1/2 with incorrect number
 Start assuming genes are on different chromosomes – independent
assortment
 Other options
 Sex-linked
 Linked genes
 Codominace
 Multiple alleles
 Others?
x
XR XR
Xr
XR
XR
XR Xr
XR Xr
x
X rY
Y
XRY
XRY
100% red eyes

XR
BINGO!
Xr
XR X r
XRY
XR
Y
XR XR
XRY
XR Xr
X rY
100% red females
50% red males; 50% white males
 Linked genes are found close to each other on the
SAME chromosome
 Linked genes do not assort independently of each
other
 2 different cell lines in cat
Cross a recessive fly (mutant for both
traits) with a fruit fly that is
heterozygous for both
A = wild type wings (normal)
a = mutant wings (vestigial)
B = wild type color (yellow)
b = mutant color (black)
Expected
Observed
50
33
Mutant (vestigial wings/black) 50
33
Vestigial Wings, Yellow
0
17
Normal, Black Body
0
17
Wild Type (normal
wings/yellow
 Dihybrid Cross with two plants RrYy x RrYy
 R = round
r = wrinkled
 Y = yellow y = green
 Write a null hypothesis:
 The following are the observed results: complete the chi-square test
to determine if the results support your null hypothesis:
Round Yellow
Peas
Round Green
Peas
Wrinkled Yellow
Peas
Wrinkled Green
Peas
219
81
69
31
Find the full explanation at this link
Phenotype
Observed (O)
Expected (E)
O–E
(O – E)2
Round Yellow Peas
219
225
-6
36
Round Green Peas
81
75
6
36
Wrinkled Yellow
Peas
69
75
-6
36
Wrinkled Green
Peas
31
25
6
36
Total
400
400
Phenotype
(O – E)2
E
(O – E)2 / E
Round Yellow Peas
36
225
0.16
Round Green Peas
36
75
0.48
Wrinkled Yellow Peas
36
75
0.48
Wrinkled Green Peas
36
25
1.44
Total
2.56
 Chi square = 2.56
 Degrees of freedom = # of groups -1 = 4-1 = 3
 P = .05
 Critical value = 7.81
 The chi-square values is less than the critical value so we
ACCEPT the null hypothesis. There is no statistical difference
between what we expected in the dihybrid cross and the
observed result.