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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. Round seed x Wrinkled seed. F1: All round seed coats. F1 round plants x F1 round plants. - PowerPoint PPT Presentation
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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?
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
Principles of HeredityPrinciples of Heredity
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.
Principles of HeredityPrinciples of Heredity
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.
Principles of HeredityPrinciples of Heredity
HomozygousHomozygousDominantDominant
HomozygousHomozygousRecessiveRecessive
HeterozygousHeterozygous
Mendel’s Experiment with PeasMendel’s Experiment with Peas
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.
Principles of HeredityPrinciples of Heredity
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
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
Principles of HeredityPrinciples of Heredity
Mendel’s Experiment with PeasMendel’s Experiment with Peas
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
Principles of HeredityPrinciples of Heredity
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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))
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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))
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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.
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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!
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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
Using Probability in Genetic Using Probability in Genetic AnalysisAnalysis
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