Patterns of Inheritance

Patterns of InheritancePatterns of Inheritance

What patterns can be observed What patterns can be observed when traits are passed to thewhen traits are passed to thenext generation?next generation?

Use of the Garden Pea for Genetics ExperimentsUse of the Garden Pea for Genetics Experiments

Principles of HeredityPrinciples of Heredity

F2: 5474 round: 1850 wrinkledF2: 5474 round: 1850 wrinkled (3/4 round to 1/4 wrinkled) (3/4 round to 1/4 wrinkled)

Mendel’s Experiment with PeasMendel’s Experiment with Peas

Round seed x Wrinkled seed Round seed x Wrinkled seed

F1: All round seed coatsF1: All round seed coats

F1 round plants x F1 round plantsF1 round plants x F1 round plants


Mendel needed to explainMendel needed to explain

1.1. Why one trait seemed to disappear Why one trait seemed to disappear in the first generation.in the first generation.

2. Why the same trait reappeared in 2. Why the same trait reappeared in the second generation in one-fourth the second generation in one-fourth of the offspring. of the offspring.


Mendel proposed:Mendel proposed:

1.1. Each trait is governed by two Each trait is governed by two factors – now called genes.factors – now called genes.

2. Genes are found in alternative 2. Genes are found in alternative forms called alleles. forms called alleles.

3. Some alleles are dominant and 3. Some alleles are dominant and mask alleles that are recessive. mask alleles that are recessive.


HomozygousHomozygousDominantDominant

HomozygousHomozygousRecessiveRecessive

HeterozygousHeterozygous


Round seed x Wrinkled seed Round seed x Wrinkled seed RR rrRR rr

F1: All round seed coatsF1: All round seed coats RrRr

R R R R

Homozygous parents can only pass one Homozygous parents can only pass one form of an allele to their offspring. form of an allele to their offspring.

R r R r

Heterozygous parents can pass either Heterozygous parents can pass either of two forms of an allele to their offspring. of two forms of an allele to their offspring.


Additional Genetic TermsAdditional Genetic Terms

Genotype: alleles carried by an Genotype: alleles carried by an individual eg. RR, Rr, rrindividual eg. RR, Rr, rr

Phenotype: physical characteristic or Phenotype: physical characteristic or appearance of an individual appearance of an individual eg. Round, wrinkledeg. Round, wrinkled

Mendel’s Principle of Genetic Mendel’s Principle of Genetic

SegregationSegregation

In the formation of gametes, the In the formation of gametes, the members of a members of a pair of alleles separatepair of alleles separate (or segregate) cleanly (or segregate) cleanly from each other so that only one member is from each other so that only one member is included in each gamete.included in each gamete.

Each gamete has an equal probability of Each gamete has an equal probability of containing either member of the allele pair.containing either member of the allele pair.

Genetic Segregation Genetic Segregation

Parentals: Parentals: RR x rrRR x rr

R R r r R R r r

RR

RR

r r r r

F1 x F1: F1 x F1: Rr x RrRr x Rr

R r R r R r R r

½ R½ R

½ r½ r

½ R ½ r ½ R ½ r ¼ RR¼ RR ¼ Rr¼ Rr

¼ Rr¼ Rr ¼ rr¼ rr

RrRr RrRr

RrRr RrRr

Genetic Segregation Genetic Segregation

Genotypic Ratio: ¼ RR + ½ Rr + ¼ rrGenotypic Ratio: ¼ RR + ½ Rr + ¼ rr

Phenotypic Ratio: ¾ Round + ¼ WrinkledPhenotypic Ratio: ¾ Round + ¼ Wrinkled

Seven Traits used by Mendel in Genetic StudiesSeven Traits used by Mendel in Genetic Studies

What Is a Gene?What Is a Gene?

• A gene is a segment of DNA that directs A gene is a segment of DNA that directs the synthesis of a specific protein.the synthesis of a specific protein.

• DNA is transcribed into RNA which is DNA is transcribed into RNA which is translated into protein.translated into protein.

Molecular Basis for Dominant and Molecular Basis for Dominant and Recessive AllelesRecessive Alleles

Dominant AlleleDominant Allele Codes for a functional Codes for a functional proteinprotein

Recessive AlleleRecessive Allele Codes for a non-Codes for a non-functional protein or functional protein or prevents any protein prevents any protein product from formingproduct from forming



Round Yellow x Wrinkled GreenRound Yellow x Wrinkled Green

F1: All round yellow seed coatsF1: All round yellow seed coats

F2: 315 round, yellow 9/16F2: 315 round, yellow 9/16 108 round, green 3/16 108 round, green 3/16 101 wrinkled, yellow 3/16 101 wrinkled, yellow 3/16 32 wrinkled, green 1/16 32 wrinkled, green 1/16

F1 plants x F1 plantsF1 plants x F1 plants


Mendel needed to explainMendel needed to explain

1.1. Why non-parental combinations Why non-parental combinations appeared in the F2 offspring.appeared in the F2 offspring.

2. Why the ratio of phenotypes in the 2. Why the ratio of phenotypes in the F2 generation was 9:3:3:1.F2 generation was 9:3:3:1.

Mendel’s Principle of Mendel’s Principle of Independent Assortment Independent Assortment

When gametes are formed, the When gametes are formed, the alleles of one gene segregate alleles of one gene segregate independently of the alleles of independently of the alleles of another geneanother gene producing equal producing equal proportions of all possible gamete proportions of all possible gamete types. types.

Genetic Segregation + Genetic Segregation + Independent AssortmentIndependent Assortment

Parentals: Parentals: RRYY x rryyRRYY x rryy

RY RY RY RY ry ry ry ryRY RY RY RY ry ry ry ry

ryry

RYRY RrYyRrYy

F1: 100% RrYy, round, yellowF1: 100% RrYy, round, yellow

F1 x F1: RrYy x RrYy

RY Ry rY ry RY Ry rY ry

¼ RY ¼ Ry ¼ rY ¼ ry

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

1/16 RrYy 1/16 Rryy 1/16 rrYy 1/16 rryy

F2 Genotypes and PhenotypesF2 Genotypes and Phenotypes

PhenotypesPhenotypes GenotypesGenotypes

RoundRound

YellowYellow

1/16 RRYY + 2/16 RRYy + 1/16 RRYY + 2/16 RRYy +

2/16 RrYY + 4/16 RrYy 2/16 RrYY + 4/16 RrYy

Total = 9/16 R_Y_Total = 9/16 R_Y_

Round Round

GreenGreen

1/16 RRyy+ 2/16 Rryy1/16 RRyy+ 2/16 Rryy

Total = 3/16 R_yyTotal = 3/16 R_yy

Wrinkled YellowWrinkled Yellow 1/16 rrYY+ 2/16 rrYy1/16 rrYY+ 2/16 rrYy

Total = 3/16 rrY_Total = 3/16 rrY_

Wrinkled GreenWrinkled Green 1/16 rryy1/16 rryy

Meiotic Segregation explains Independent AssortmentMeiotic Segregation explains Independent Assortment

Solving Genetics ProblemsSolving Genetics Problems

1.1. Convert parental phenotypes to Convert parental phenotypes to genotypesgenotypes

2.2. Use Punnett Square to determine Use Punnett Square to determine genotypes of offspringgenotypes of offspring

3.3. Convert offspring genotypes to Convert offspring genotypes to phenotypesphenotypes

Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis

1. Probability (P) of an event (E) occurring:1. Probability (P) of an event (E) occurring:

P(E) = P(E) = Number of ways that event E can occurNumber of ways that event E can occur

Total number of possible outcomesTotal number of possible outcomes

Eg. P(Rr) from cross Rr x RrEg. P(Rr) from cross Rr x Rr 2 ways to get Rr genotype 2 ways to get Rr genotype 4 possible outcomes 4 possible outcomes P(Rr) = 2/4 = 1/2 P(Rr) = 2/4 = 1/2


2. 2. Addition RuleAddition Rule of Probability – used in an of Probability – used in an “either/or”“either/or” situation situation

Eg. P(Rr or RR) from cross Rr x Rr Eg. P(Rr or RR) from cross Rr x Rr 2 ways to get Rr genotype 2 ways to get Rr genotype 1 way to get RR genotype 1 way to get RR genotype 4 possible outcomes 4 possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4 P(Rr or RR) = 2/4 + 1/4 = 3/4

P(EP(E11 or E or E22) = P(E) = P(E11) + P(E) + P(E22))


3. Multiplication3. Multiplication Rule Rule of Probability – used in an of Probability – used in an “and”“and” situation situation

Eg. P(wrinkled, yellow) from cross RrYy x RrYy Eg. P(wrinkled, yellow) from cross RrYy x RrYy

P(rr and Y_) = 1/4 x 3/4 = 3/16 P(rr and Y_) = 1/4 x 3/4 = 3/16

P(EP(E11 and E and E22) = P(E) = P(E11) X P(E) X P(E22))


4. Conditional Probability: C4. Conditional Probability: Calculating alculating the probability that each individual the probability that each individual has a particular genotypehas a particular genotype

Eg. Jack and Jill do not have PKU. Eg. Jack and Jill do not have PKU. Each has a sibling with the disease. Each has a sibling with the disease. What is the probability that Jack and What is the probability that Jack and Jill will have a child with PKU?Jill will have a child with PKU?

PP

pp

P pP p

PPPP PpPp

PpPp ppppXX


4. Conditional Probability 4. Conditional Probability

Jack is P_, Jill is P_ Jack is P_, Jill is P_

Parents of Jack or Jill: Pp x Pp Parents of Jack or Jill: Pp x Pp

P(Pp) = P(Pp) = 2 ways to get Pp2 ways to get Pp 3 possible genotypes3 possible genotypes

P(Jack is Pp) =2/3P(Jack is Pp) =2/3P (Jill is Pp) = 2/3P (Jill is Pp) = 2/3


4. Conditional Probability 4. Conditional Probability

P(child with PKU)= P(child with PKU)=

P(child without PKU)= 1-1/9 = 8/9P(child without PKU)= 1-1/9 = 8/9

P(Jack is Pp) x P(Jill is Pp) x P(child is pp) = P(Jack is Pp) x P(Jill is Pp) x P(child is pp) =

2/3 x 2/3 x 1/4 = 1/92/3 x 2/3 x 1/4 = 1/9


JackJack JillJill P_ childP_ child ProbabilityProbability

1/3 PP1/3 PP 1/3 PP1/3 PP 11 1/91/9

1/3 PP1/3 PP 2/3 Pp2/3 Pp 11 2/92/9

2/3 Pp2/3 Pp 1/3 PP1/3 PP 11 2/92/9

2/3 Pp2/3 Pp 2/3 Pp2/3 Pp 3/43/4 3/93/9

Total=8/9Total=8/9

To calculate probability of child without PKU, To calculate probability of child without PKU, look at all possibilities for Jack and Jill. look at all possibilities for Jack and Jill.


5. Ordered Events: use Multiplication 5. Ordered Events: use Multiplication Rule Rule For Jack and Jill, what is the probability that For Jack and Jill, what is the probability that the first child will have PKU, the second child the first child will have PKU, the second child will not have PKU and the third child will have will not have PKU and the third child will have PKU?PKU?

P(pp) x P(P_) x P(pp) = P(pp) x P(P_) x P(pp) =

1/9 x 8/9 x 1/9 = 8/7291/9 x 8/9 x 1/9 = 8/729


6. 6. Binomial Rule of Probability Binomial Rule of Probability – used for– used for unordered eventsunordered events

a = probability of event X (occurrence of one event) a = probability of event X (occurrence of one event) b = probability of event Y = 1-ab = probability of event Y = 1-a (occurrence of alternate event) (occurrence of alternate event) n = total n = total s = number of times event X occurss = number of times event X occurst = number of times event Y occurs (s + t = n)t = number of times event Y occurs (s + t = n)

P = P = n!n! (a (assbbtt)) s! t!s! t!


6. 6. Binomial Rule of ProbabilityBinomial Rule of Probability

! = factorial= number multiplied by each ! = factorial= number multiplied by each lower number until reaching 1 lower number until reaching 1

5! = 5 x 4 x 3 x 2 x 1 5! = 5 x 4 x 3 x 2 x 1 1! =11! =1

3! = 3 x 2 x 1 = 3 x 2! 0! = 13! = 3 x 2 x 1 = 3 x 2! 0! = 12! = 2 x 1 2! = 2 x 1


6. 6. Binomial Rule of Probability Binomial Rule of Probability

Out of 3 children born to Jack and Jill,Out of 3 children born to Jack and Jill, what is the probability that 2 will have PKU? what is the probability that 2 will have PKU?

3! 3! (1/9) (1/9)22(8/9)(8/9)11= = 3 x 2! 3 x 2! (1/81) (8/9)= (1/81) (8/9)= 24242! 1! 2! 1! 7292! 1! 2! 1! 729

n=3, a=1/9, s=2, b=8/9, t=1n=3, a=1/9, s=2, b=8/9, t=1


The same result can be obtained using the The same result can be obtained using the multiplicative rule if all possible birth orders multiplicative rule if all possible birth orders for families of three are considered:for families of three are considered:

11stst child child 22ndnd child child 33rdrd child child ProbabilityProbability

PKU=1/9PKU=1/9 No= 8/9No= 8/9 PKU=1/9PKU=1/9 8/7298/729

No=8/9No=8/9 PKU=1/9PKU=1/9 PKU=1/9PKU=1/9 8/7298/729

PKU=1/9PKU=1/9 PKU=1/9PKU=1/9 No=8/9No=8/9 8/7298/729

8/729 + 8/729 + 8/729 = 24/7298/729 + 8/729 + 8/729 = 24/729

Chi-Square Goodness of Fit Test

number expected

number) expected-number (observed 22x

To evaluate how well data fits an expected genetic ratio

Chi-square Test for Goodness of Fit for 9:3:3:1 RatioChi-square Test for Goodness of Fit for 9:3:3:1 Ratio PhenotypePhenotype Observed Observed

NumberNumberExpected Expected

Number Number

(Fraction x Total)(Fraction x Total)

O-EO-E (O-E)(O-E)22 (O-E)(O-E)22

EE

Round, yellowRound, yellow 315315

Round, greenRound, green 108108

Wrinkled, yellowWrinkled, yellow 101101

Wrinkled, greenWrinkled, green 3232

TotalTotal 556556

9/16 x 556 = 3139/16 x 556 = 313

3/16 x 556 = 1043/16 x 556 = 104

3/16 x 556 = 1043/16 x 556 = 104

1/16 x 556 = 351/16 x 556 = 35

22

44

-3-3

-3-3

44

1616

99

99

.0128.0128

.154.154

.087.087

.257.257

XX22= .511= .511

df=degrees of freedom= number of phenotypes – 1 = 4-1=3df=degrees of freedom= number of phenotypes – 1 = 4-1=3

p value from table on page 1-17: p>.5p value from table on page 1-17: p>.5 from table in Pierce: .975 > p >.9from table in Pierce: .975 > p >.9

Data supports hypothesis for any pData supports hypothesis for any p>>0.050.05

Patterns of Inheritance

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