Patterns of Inheritance

1
Patterns of Inheritance
What patterns can be observed when traits are
passed to the next generation?
2
Use of the Garden Pea for Genetics Experiments
3
Principles of Heredity
Mendels Experiment with Peas
4
Principles of Heredity

• Mendel needed to explain
• 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.

5
Principles of Heredity

• Mendel proposed
• 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.

6
Principles of Heredity
Homozygous Dominant
Homozygous Recessive
Heterozygous
7
Homozygous parents can only pass one form of an
allele to their offspring.
8
Heterozygous parents can pass either of two
forms of an allele to their offspring.
R
r
R
r
9
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
10
Mendels 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.

11
Genetic Segregation
Parentals RR x rr
F1 x F1 Rr x Rr
R R r r
R r R r
½ R ½ r
r r
¼ RR
¼ Rr
R R
½ R ½ r
Rr
Rr
¼ Rr
¼ rr
Rr
Rr
12
Genetic Segregation
Genotypic Ratio ¼ RR ½ Rr ¼ rr Phenotypic
Ratio ¾ Round ¼ Wrinkled
13
Seven Traits used by Mendel in Genetic Studies
14
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.

15
Molecular Basis for Dominant and Recessive Alleles
16
Principles of Heredity
Mendels Experiment with Peas
17
Principles of Heredity

• Mendel needed to explain
• Why non-parental combinations appeared in the F2
  offspring.
• 2. Why the ratio of phenotypes in the F2
  generation was 9331.

18
Mendels 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.

19
Genetic Segregation Independent Assortment
Parentals RRYY x rryy
ry
RrYy
RY
20
F1 x F1 RrYy x RrYy
RY Ry rY ry RY Ry rY ry
¼ RY ¼ Ry ¼ 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
21
F2 Genotypes and Phenotypes
22
Meiotic Segregation explains Independent
Assortment
23
Solving Genetics Problems

• Convert parental phenotypes to genotypes
• Use Punnett Square to determine genotypes of
  offspring
• Convert offspring genotypes to phenotypes

24
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
25
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
26
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
27
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?
28
Using Probability in Genetic Analysis

• 4. Conditional Probability
• Jack is P_, Jill is P_
• Parents of Jack or Jill Pp x Pp

P(Pp) 2 ways to get Pp 3 possible
genotypes P(Jack is Pp) 2/3 P (Jill is Pp)
2/3
X
29
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
30
Using Probability in Genetic Analysis
To calculate probability of child without PKU,
look at all possibilities for Jack and Jill.
31
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
32
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 occurst number of
times event Y occurs (s t n)
33
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 1! 1
• 3! 3 x 2 x 1 3 x 2! 0! 12!
  2 x 1

34
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?

n3, a1/9, s2, b8/9, t1
3! (1/9)2(8/9)1 3 x 2! (1/81) (8/9) 24 2!
1! 2! 1!
729
35
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

8/729 8/729 8/729 24/729
36
Chi-Square Goodness of Fit Test
To evaluate how well data fits an expected
genetic ratio
37
Chi-square Test for Goodness of Fit for 9331
Ratio
9/16 x 556 313
2
4
.0128
4
16
3/16 x 556 104
.154
9
3/16 x 556 104
-3
.087
1/16 x 556 35
-3
9
.257
X2 .511
dfdegrees of freedom number of phenotypes 1
4-13 p value from table on page 1-17 pgt.5
from table in Pierce .975 gt p gt.9 Data
supports hypothesis for any pgt0.05
38
(No Transcript)
39
Sex Determination

• Sex Chromosomes homologous chromosomes that
  differ in size and genetic composition between
  males and females

40
Sex Determination

• Autosome any chromosome that is not a sex
  chromosome
• Humans have 22 pairs of autosomesand 1 pair of
  sex chromosomes

41
Human Karyotype showing homologous chromosome
pairs
42
Sex Determination
Female Male XX x XY
½ X ½ X
¼ XX
¼ XX
½ X ½ Y
¼ XY
¼ XY
Phenotypic Ratio of Offspring ½ Female ½ Male
43
X-linked Genes
Hemophilic Male Non-hemophilic
Female
(father is hemophilic)
XhY x XHXh
½ XH ½ Xh
¼ XHXh
¼ XhXh
½ Xh ½ Y
¼ XHY
¼ XhY
Phenotypic Ratio of Offspring ¼ hemophilic males
¼ non-hemophilic males ¼ hemophilic females ¼
non-hemophilic females
44
Terms Specific toSex-linked Genes
45
Patterns of Inheritance

• Autosomal Dominant
• Autosomal Recessive
• X-linked Dominant
• X-linked Recessive
• Y-Linked Inheritance

46
Pedigree for Huntingtons Disease, an Autosomal
Dominant Trait
47
Pedigree for Albinism, an Autosomal Recessive
Trait
48
(No Transcript)
49
Pedigree for Colorblindness, an X-linked
Recessive Trait