Transcript Principles of Heredity

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Patterns of Inheritance
What patterns can be observed
when traits are passed to the
next generation?
Use of the Garden Pea for Genetics Experiments
Principles of Heredity
Mendel’s Experiment with Peas
Round seed
x Wrinkled seed
F1: All round seed coats
F1 round plants
x F1 round plants
F2: 5474 round: 1850 wrinkled
(3/4 round to 1/4 wrinkled)
Principles of Heredity
Mendel needed to explain
1. Why one trait seemed to disappear
in the first generation.
2. Why the same trait reappeared in
the second generation in one-fourth
of the offspring.
Principles of Heredity
Mendel proposed:
1. Each trait is governed by two
factors – now called genes.
2. Genes are found in alternative
forms called alleles.
3. Some alleles are dominant and
mask alleles that are recessive.
Principles of Heredity
Mendel’s Experiment with Peas
Round seed
RR
Homozygous
Dominant
x Wrinkled seed
rr
Homozygous
Recessive
F1: All round seed coats
Rr
Heterozygous
Homozygous parents can only pass one
form of an allele to their offspring.
R
R
R
R
Heterozygous parents can pass either
of two forms of an allele to their offspring.
R
r
R
r
Principles of Heredity
Additional Genetic Terms
Genotype: alleles carried by an
individual eg. RR, Rr, rr
Phenotype: physical characteristic or
appearance of an individual
eg. Round, wrinkled
Mendel’s Principle of Genetic
Segregation
In the formation of gametes, the members of a
pair of alleles separate (or segregate) cleanly
from each other so that only one member is
included in each gamete.
Each gamete has an equal probability of
containing either member of the allele pair.
Genetic Segregation
Parentals:
RR x rr
R
R
R
R
r
r
r
r
Rr
Rr
Rr
F1 x F1:
Rr x Rr
Rr
R
½R
½r
r
R
½R
½ r
¼ RR ¼ Rr
¼ Rr
¼ rr
r
Genetic Segregation
Genotypic Ratio: ¼ RR + ½ Rr + ¼ rr
Phenotypic Ratio: ¾ Round + ¼ Wrinkled
Seven Traits used by Mendel in Genetic Studies
What Is a Gene?
• A gene is a segment of DNA that directs
the synthesis of a specific protein.
• DNA is transcribed into RNA which is
translated into protein.
Molecular Basis for Dominant and
Recessive Alleles
Dominant Allele
Codes for a functional
protein
Recessive Allele
Codes for a nonfunctional protein or
prevents any protein
product from forming
Principles of Heredity
Mendel’s Experiment with Peas
Round Yellow
x Wrinkled Green
F1: All round yellow seed coats
F1 plants
x F1 plants
F2: 315 round, yellow
108 round, green
101 wrinkled, yellow
32 wrinkled, green
9/16
3/16
3/16
1/16
Principles of Heredity
Mendel needed to explain
1. Why non-parental combinations
appeared in the F2 offspring.
2. Why the ratio of phenotypes in the
F2 generation was 9:3:3:1.
Mendel’s Principle of
Independent Assortment
When gametes are formed, the
alleles of one gene segregate
independently of the alleles of
another gene producing equal
proportions of all possible gamete
types.
Genetic Segregation +
Independent Assortment
Parentals:
RRYY
x
rryy
RY RY RY RY
ry ry
ry
RY
RrYy
F1: 100% RrYy, round, yellow
ry ry
F1 x F1:
RrYy
x
RY
Ry
¼ RY
rY
ry
¼ Ry
RrYy
RY Ry
¼ rY
rY
ry
¼ ry
1/16 RRYY 1/16 RRYy 1/16 RrYY 1/16 RrYy
¼ RY
¼ Ry
¼ rY
¼ ry
1/16 RRYy 1/16 RRyy 1/16 RrYy 1/16 Rryy
1/16 RrYY 1/16 RrYy
1/16 rrYY
1/16 rrYy
1/16 RrYy 1/16 Rryy
1/16 rrYy
1/16 rryy
F2 Genotypes and Phenotypes
Phenotypes
Round
Yellow
Round
Green
Wrinkled Yellow
Wrinkled Green
Genotypes
1/16 RRYY + 2/16 RRYy +
2/16 RrYY + 4/16 RrYy
Total = 9/16 R_Y_
1/16 RRyy+ 2/16 Rryy
Total = 3/16 R_yy
1/16 rrYY+ 2/16 rrYy
Total = 3/16 rrY_
1/16 rryy
Meiotic Segregation explains Independent Assortment
Solving Genetics Problems
1. Convert parental phenotypes to
genotypes
2. Use Punnett Square to determine
genotypes of offspring
3. Convert offspring genotypes to
phenotypes
Using Probability in Genetic
Analysis
1. Probability (P) of an event (E) occurring:
P(E) = Number of ways that event E can occur
Total number of possible outcomes
Eg. P(Rr) from cross Rr x Rr
2 ways to get Rr genotype
4 possible outcomes
P(Rr) = 2/4 = 1/2
Using Probability in Genetic
Analysis
2. Addition Rule of Probability – used in
an “either/or” situation
P(E1 or E2) = P(E1) + P(E2)
Eg. P(Rr or RR) from cross Rr x Rr
2 ways to get Rr genotype
1 way to get RR genotype
4 possible outcomes
P(Rr or RR) = 2/4 + 1/4 = 3/4
Using Probability in Genetic
Analysis
3. Multiplication Rule of Probability –
used in an “and” situation
P(E1 and E2) = P(E1) X P(E2)
Eg. P(wrinkled, yellow) from cross RrYy x RrYy
P(rr and Y_) = 1/4 x 3/4 = 3/16
Using Probability in Genetic
Analysis
4. Conditional Probability:
Calculating the probability that
each individual has a
particular genotype
Eg. Jack and Jill do not have PKU.
Each has a sibling with the disease.
What is the probability that Jack and
Jill will have a child with PKU?
Using Probability in Genetic
Analysis
4. Conditional Probability
Jack is P_, Jill is P_
Parents of Jack or Jill: Pp x Pp
P
p
P
PP
Pp
p
Pp
pp
X
P(Pp) = 2 ways to get Pp
3 possible genotypes
P(Jack is Pp) =2/3
P (Jill is Pp) = 2/3
Using Probability in Genetic
Analysis
4. Conditional Probability
P(child with PKU)=
P(Jack is Pp) x P(Jill is Pp) x P(child is pp) =
2/3 x 2/3 x 1/4 = 1/9
P(child without PKU)= 1-1/9 = 8/9
Using Probability in Genetic
Analysis
To calculate probability of child without PKU,
look at all possibilities for Jack and Jill.
Jack
Jill
P_ child
Probability
1/3 PP
1/3 PP
1
1/9
1/3 PP
2/3 Pp
2/3 Pp
1/3 PP
1
1
2/9
2/9
2/3 Pp
2/3 Pp
3/4
3/9
Total=8/9
Using Probability in Genetic
Analysis
5. Ordered Events: use Multiplication
Rule
For Jack and Jill, what is the probability that
the first child will have PKU, the second child
will not have PKU and the third child will have
PKU?
P(pp) x P(P_) x P(pp) =
1/9 x 8/9 x 1/9 = 8/729
Using Probability in Genetic
Analysis
6. Binomial Rule of Probability – used for
unordered events
P = n! (asbt)
s! t!
a = probability of event X (occurrence of one event)
b = probability of event Y = 1-a
(occurrence of alternate event)
n = total
s = number of times event X occurs
t = number of times event Y occurs (s + t = n)
Using Probability in Genetic
Analysis
6. Binomial Rule of Probability
! = factorial= number multiplied by each
lower number until reaching 1
5! = 5 x 4 x 3 x 2 x 1
3! = 3 x 2 x 1 = 3 x 2!
2! = 2 x 1
1! =1
0! = 1
Using Probability in Genetic
Analysis
6. Binomial Rule of Probability
Out of 3 children born to Jack and Jill,
what is the probability that 2 will have PKU?
n=3, a=1/9, s=2, b=8/9, t=1
3! (1/9)2(8/9)1= 3 x 2! (1/81) (8/9)= 24
2! 1!
2! 1!
729
Using Probability in Genetic
Analysis
The same result can be obtained using the
multiplicative rule if all possible birth orders
for families of three are considered:
1st child
2nd child
3rd child
Probability
PKU=1/9 No= 8/9 PKU=1/9
No=8/9
PKU=1/9 PKU=1/9
8/729
8/729
PKU=1/9 PKU=1/9 No=8/9
8/729
8/729 + 8/729 + 8/729 = 24/729
Chi-Square Goodness of Fit
Test
To evaluate how well data fits an expected genetic ratio
2
(observed
number
expected
number)
2
x 
expected number
Chi-square Test for Goodness of Fit for 9:3:3:1 Ratio
Phenotype
Observed Expected
Number
Number
(Fraction x Total)
O-E
(O-E)2 (O-E)2
E
Round, yellow
315
9/16 x 556 = 313
2
4
.0128
Round, green
108
3/16 x 556 = 104
4
16
.154
Wrinkled, yellow 101
3/16 x 556 = 104
-3
9
.087
Wrinkled, green
32
1/16 x 556 = 35
-3
9
.257
Total
556
df=degrees of freedom= number of phenotypes – 1 = 4-1=3
p value from table on page 1-17: p>.5
from table in Pierce: .975 > p >.9
Data supports hypothesis for any p>0.05
X2= .511
p
X2
Sex Determination
Sex Chromosomes: homologous
chromosomes that differ in size and
genetic composition between males and
females
Human
Chromosome
X
Y
Size and
gene number
Larger
2142 genes
Smaller
326 genes
Genetic
Composition
Multiple genes Single gene influences
unrelated
gender: TDF = testis
to gender
determining factor
Sex Determination
Autosome: any chromosome that is not
a sex chromosome
Humans have 22 pairs of autosomes
and 1 pair of sex chromosomes
Human Karyotype showing
homologous chromosome pairs
Sex Determination
Female
XX x
Male
XY
½X
½X
½X
¼ XX
¼ XX
½Y
¼ XY
¼ XY
Phenotypic Ratio of Offspring
½ Female + ½ Male
X-linked Genes
Hemophilic Male
XhY
Non-hemophilic
Female
(father is hemophilic)
x
XHXh
½ XH
½
Xh
½Y
½ Xh
¼ XHXh ¼ XhXh
¼ XHY
¼ XhY
Phenotypic Ratio of Offspring
¼ hemophilic males + ¼ non-hemophilic males
¼ hemophilic females + ¼ non-hemophilic females
Terms Specific to
Sex-linked Genes
Females
Homogametic: only X
found in gametes
Can be homozygous or
heterozygous for traits on Xchromosome
Males
Heterogametic: either X or
Y in gametes
Hemizygous for traits on the
X-chromosome
Patterns of Inheritance
•
•
•
•
•
Autosomal Dominant
Autosomal Recessive
X-linked Dominant
X-linked Recessive
Y-Linked Inheritance
Pedigree for Huntington’s Disease,
an Autosomal Dominant Trait
Pedigree for Albinism, an Autosomal Recessive Trait
Pedigree for Colorblindness,
an X-linked Recessive Trait