Transcript AP Chapter 14 Lecture - TJ

CHAPTER 14 MENDEL & THE GENE IDEA
14.1 Mendel used the scientific approach to identify two laws of inheritance
14.2 The laws of probability govern Mendelian inheritance
I. Intro
A. Scale 0-1
1. 0= will not occur
2. 1= certain to occur
B. Important lesson of probability
1. Each event is independent of the next
a. Alleles of 1 gene segregate into gametes independently of
another gene’s alleles
C. Probability can help us predict the outcome of the fusion of
gametes
II. The multiplication and addition rules applied to monohybrid crosses
A. Multiplication rule
1. Prediction of 2 independent events occurring simultaneously
a. Multiply all independent event probabilities
e.x Tossing pennies
Event 1 Probability of tails = 1/2
Event 2 Probability of tails = 1/2
What is the probability of 2 coins flipped
simultaneously of landing on tails
1/2 (event 1) x 1/2 (event 2)= 1/4
e.x F1 cross Rr X Rr R=round r= wrinkle
Event 1 Probability of egg receiving R = 1/2
Probability of egg receiving r = 1/2
Event 2 Probability of sperm receiving R = 1/2
Probability of sperm receiving r = 1/2
What is probability of egg + sperm = RR
1/2 (egg with R) X 1/2 (Sperm with R) = 1/4
B. Addition rule
1. Probability of 1 of 2 or more mutually exclusive events will
occur is calculated by adding their individual probabilities
a. Probability of 1 of 2 or more related but independent events
will occur is calculated by adding their individual
probabilities
e.x Rr X Rr
What is probability of heterozygous round (Rr)
Rr = 1/4
rR = 1/4
1/2
e.x Cross = Rr X rr
What is probability of Rr
What is probability of rr
1/2 Rr
1/2 rr
III. Solving complex genetics problems with the rules of probability
A. Dihybrids
1. Cross = YyRr X YyRr
a. What fraction of offspring would be predicted to have
YyRR
1. Step 1- due to independent assortment you can deal
with the 2 genes separately
a. Set up a monohybrid cross for each
1/4 YY
1/4 RR
1/2 Yy
1/2 Rr
1/4 yy
1/4 rr
2. Step 2- Now use the laws of probability
1/2 Yy x 1/4 RR = 1/8 YyRR or 2/16
2. Practice= TTQq X TtQq
a. What is the frequency of the genotype TTQq in the F2
generation
B. Trihybrids
1. Cross = QqTtRr X Qqttrr
a. What fraction of offspring would be predicted to exhibit
the recessive phenotype for at least 2 of the three
characteristics
1. Step 1- List all possible genotypes of offspring fulfilling
condition
qqttRR
qqttRr
qqTTrr
qqTtrr
QQttrr
Qqttrr
qqttrr
2. Step 2- List all possible genotypes based on cross
qqttRR
qqttRr
qqTTrr
qqTtrr
QQttrr
Qqttrr
qqttrr
3. Monohybrid punnett square
1/4 QQ
1/2 Tt
1/2 Qq
1/2 tt
1/4 qq
1/2 Rr
1/2 rr
4. Implement multiplication and addition rules
qqttRr 1/4 (probability of qq) X 1/2 (tt) X 1/2 (Rr)
qqTtrr 1/4 X 1/2 X 1/2
Qqttrr 1/2 X 1/2 X 1/2
QQttrr 1/4 X 1/2 X 1/2
qqttrr 1/4 X 1/2 X 1/2
= 1/16
= 1/16
= 1/8 or 2/16
= 1/16
= 1/16
Chance of at least 2 recessive traits
= 6/16 or 3/8
14.3 Inheritance patterns are often more complex than predicted by
simple Mendelian genetics
I. Extending Mendelian genetics for a single gene
A. Degrees of dominance
1. Complete dominance
a. Mendel’s work
b. One allele overshadows/masks the other
c. Homozygous dominant & heterozygous phenotypically the
same
2. Incomplete dominance
a. Offspring are phenotypically intermediate between 2
parents
1. Heterozygous flowers produce less red pigment than red
homozygote
3. Codominance
a. Both alleles of a gene are expressed phenotypically
b. ABO blood grouping
B. Relationship between dominance & phenotype
1. How is dominance achieved
a. Alleles=nucleotide sequence proteins function
1. Individual alleles do not interact
2. Dominance or recessive is achieved through allele
expression
b. Ex Mendel’s peas
1. Round (dominant) & wrinkled (recessive)
a. Round allele codes for enzyme
b. Wrinkled allele codes for defective enzyme
c. Tay-sachs disease
1. Disease manifests when enzymes cannot breakdown
certain lipids in the brain
a. Seizures, blindness, degeneration of motor & mental
performance, & death
2. Homozygous dominant & heterozygous = no
manifestation
3. Homozygous recessive = manifestation
2. Dominance/recessive a matter of viewpoint
a. Tay-sachs disease
1. Organismal level dominant/recessive
2. Biochemical level incomplete dominance
a. Homozygous dominant = complete functional
enzyme production
b. Heterozygous = functional enzyme & nonfunctional
enzyme production but enough function to prevent
manifestation
c. Homozygous = complete nonfunctional enzyme
C. Frequency of dominance
1. Dominant allele not always the higher frequency
a. Polydactyly
D. Multiple alleles
1. Blood groups (ABO)
E. Pleiotropy
1. 1 gene affecting multiple phenotypes
2. Garden pea gene for flower color also influences seed color
II. Extending Mendelian genetics for two or more genes
A. Epistasis
1. Gene at 1 locus alters the phenotypic expression of a gene at a
second locus
2. Ex Mice – Black (B) dominant
to brown (b) coat color
a. However, a different gene for
color (C) controls the release
of the pigments needed for
hair color
1. bb = brown but if ccbb will
be albino
B. Polygenic inheritance
1. Multiple genes controlling a particular phenotype
a. Phenotype exists as a continuum
1. Quantitative characters
2. Height & skin color
III. Nature and nurture: The environmental impact on phenotype