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Chpt. 14 Mendelian Genetics
Chpt. 14 Mendelian Genetics
How are traits passed from parent to offspring?
Chpt. 14 Mendelian Genetics
WE … know it isgenes on chromosomes…
Chpt. 14 Mendelian GeneticsMENDEL
had no idea what a chromosome even was!!!!!
Chpt. 14 Mendelian Genetics
18431843 - - entered entered an an AugustiniAugustinian an
monastemonastery.ry.
Chpt. 14 Mendelian Genetics18461846 - -
assigned to assigned to a High a High School as a School as a teacher… teacher… didn’t pass didn’t pass the teacher the teacher test. BACK test. BACK TO THE TO THE MONASTERYMONASTERY
Chpt. 14 Mendelian Genetics1851 - 531851 - 53
Tried to Tried to “find “find himself”, himself”, sent to sent to Univ. of Univ. of ViennaVienna
Chpt. 14 Mendelian Genetics
1851 - 1851 - 18531853 two two profs. profs. influencinfluenced him ed him and he and he began began researchresearch..
Statistics Professor and a Botany Professor
Chpt. 14 Mendelian Genetics1851 - 531851 - 53
His His research research was was “controver“controversial”he sial”he FLUNKED FLUNKED OUT!OUT!
Mendel became an emotional wreck!!
Chpt. 14 Mendelian Genetics
1854 1854 - - BACK to BACK to the the monastery, monastery, HOWEVER, HOWEVER, this time…this time…
Chpt. 14 Mendelian Genetics
18541854 - - the the Monsignor Monsignor allowed allowed him to him to research research inheritaninheritance…ce…
Mendel’Mendel’ss
MethodsMethods::
Worked Worked
with with Pea Plants
observed several characters
Either - Or traits
Mendel’sMendel’s
Methods:Methods:Cross-Cross-pollinatedpollinated two two contrastincontrasting g truetrue breedingbreeding pea pea varietiesvarieties
pg. 567 anatomypg. 567 anatomy
PP
FF11 4:04:0
FF22 3:1 3:1
True breeding
PP
FF11 4:04:0
FF22 3:1 3:1
hybridization
True breeding
Mating two true-breeding Mating two true-breeding varietiesvarieties
White White did did
notnot disappeadisappearr in F1! in F1! PurplePurple is just is just DOMINANTDOMINANT over over whitewhite
We now knowWe now know::Alternative versions of a gene
Results - Results - SummarySummary•In all crosses, the F1
generation shows only one of the traits regardless of which was male or female (4:0 ratio).
•The other trait reappeared in the F2 at ~25% (3:1 ratio).
Mendel’s Laws Mendel’s Laws ==•Alternate versions of genes ((allelesalleles)) account for variations.•Organisms inherit two alleles (one from each parent)- for each character.•When alleles differ, the dominant is fully expressed.
Mendel’s Laws Mendel’s Laws ==
•Each pair of alleles separate during gamete formation by chance- 50% probability that each version will be the one
involved in the fertilization.
Aa
A
a
Germ cell
after meiosis II
A a
P P
LawLaw of of SegregationSegregation
p p
-->
VOCABULARY
-->
TestcrosTestcrosss==
Cross of a phenotypically dominant with a homozygote recessive
TestcrosTestcrosss==
Cross of a suspected heterozygote with a homozygote recessive
ex: T_ x tt
If TT: all offspring will be dominant
If Tt: offspring will be1 Dominant :
1 recessive
6 Mendelian 6 Mendelian Crosses are Crosses are Possible Possible (these are your (these are your
expectedexpected):):Cross Genotype PhenotypeTT X tt all Tt all Dominant
Tt X Tt 1TT:2Tt:1tt 3 Dom: 1 ResTT X TT all TT all Dominanttt X tt all tt all Recessive
TT X Tt 1TT:1Tt all DominantTt X tt 1Tt:1tt 1 Dom: 1 Res
Sorry,but College Board who runs AP, says you must memorize this… so does Hobby, so does your Freshman collegiate Bio instructor!!
Lets try it with one trait:COLOR
What if you are tracking more than
ONE trait???
Do the traits “travel” together when gametes are
formed?
In other words, does seed color & seed shape always stay together?
Does the “R” allele always stay with the “Y” allele?
LawLaw ofof IndependentIndependent AssortmentAssortment
Mendel’sMendel’sLawLaw ofof IndependentIndependent AssortmentAssortment
•Each pair/types of alleles segregates into gametes independently.
•Color and seed shape are not a “package
deal”!`
Dihybrid CrossDihybrid Cross
Cross with two genetic traits.Need 4 letters to code for the cross.
ex: TtRrEach Gamete - Must get 1 letter for each trait.
ex. TR, Tr, tR, tr = possible gametes
How big of a Punnett Square? Number of Gametes possible
Critical to calculating the results of higher level crosses.
Look for the number of heterozygous traits.
Equation to Equation to determine number of gametes possible:determine number of gametes possible:
The formula 2The formula 2nn can be used, can be used, where “n” equals the number of where “n” equals the number of heterozygousheterozygous traitstraits possible in a possible in a cross.cross.
ex: TtRr, n=2ex: TtRr, n=2
2222 or 4 different kinds of or 4 different kinds of gametesgametes are possible. are possible.
TR, tR, Tr, trTR, tR, Tr, tr
Dihybrid CrossDihybrid Cross
TtRr X TtRrTtRr X TtRr each parent can produce 4 each parent can produce 4 types of gametes.types of gametes.
TR, Tr, tR, trTR, Tr, tR, tr
cross is a 4 X 4 with 16 cross is a 4 X 4 with 16 possible offspring.possible offspring.
FOIL
Results of Results of TtRr X TtRrTtRr X TtRr
9 Tall, Red flowered3 Tall, white flowered3 short, Red flowered1 short, white flowered
Or: 9:3:3:1Or: 9:3:3:1
CommentRatio of Tall to short is 3:1
(b/c Tt X Tt)Ratio of Red to white is 3:1
(b/c Rr X Rr)
The cross is really a The cross is really a productproduct of of the ratio of each trait multiplied the ratio of each trait multiplied together. (3:1) x (3:1)together. (3:1) x (3:1) FOIL
PROBABILITY
RULES OF INHERITANCE
&&
RuleRule ofof MultiplicationMultiplication--
used when used when determiningdetermining
chances thatchances that twotwo independentindependent eventsevents w/ w/
occur together occur together simultaneously.simultaneously.
RuleRule ofof MultiplicationMultiplication-- twotwo independentindependent
eventsevents w/ occur w/ occur together together
simultaneously.simultaneously.twotwo independentindependent
eventsevents ex. height & ex. height & flower colorflower color
RuleRule ofof MultiplicationMultiplication--
1.Determine the Determine the probability ofprobability of each independent event (ex. allele in each
sperm and allele in each egg)
2. multiply the multiply the probabilities of probabilities of those independent those independent events.events.
RuleRule ofof MultiplicationMultiplication--
ex. ex.
P generationP generation = Pp x = Pp x PpPp
““what is the what is the probability of a probability of a whitewhite offspring??” offspring??”
Can only get this one way…
(pp(pp))
Rule of Multiplication-
1. P generation = Pp x Pp
2. chance of Ovum p = chance of Ovum p = 1/2 1/2
1/4 pp
Pp
p
P
chance of Sperm p = chance of Sperm p = XX 1/21/2
RuleRule ofof MultiplicationMultiplication--
used when used when determiningdetermining
chances thatchances that twotwo independentindependent eventsevents w/ w/
occur together occur together simultaneously.simultaneously.
The probability of getting a tall offspring is ¾ (Tt x Tt).
The probability of getting a red offspring is ¾ (Rr x Rr).
The probability of getting a TALL red offspring is ¾ x ¾ = 9/16
Example: TtRr X TtRr
RuleRule ofof MultiplicationMultiplication--
1. YyRr is in a germ YyRr is in a germ cell: cell: “What is “What is the probability that the probability that a a gametegamete will be will be YR?”YR?”
2. 1/21/2 a chance a chance for Yfor Y 1/4 YRxx 1/21/2 a chance for R a chance for R
RuleRule ofof MultiplicationMultiplication--
used when used when determiningdetermining
chances thatchances that twotwo independentindependent eventsevents w/ w/
occur together occur together simultaneously.simultaneously.
ButBut,, what about, what about, for example, for example, heterozygote heterozygote parents?parents? Yy x YyYy x YyWhat is the probability that these parents would produce a heterozygote offspring?
RuleRule ofof AdditionAddition--
Used when determining the probability of an event… an event that can occur can occur inin 22 oror moremore differentdifferent waysways.
RuleRule ofof AdditionAddition--
1. Compute the probability for each way in which eacheach event can happen.2. AddAdd the probabilities for each of those ways.
RuleRule ofof AdditionAddition--
P1 = Yy x Yy
ex. “what is the probability of a heterozygous offspring?”
Ovum Y = 1/2 Sperm y = 1/2Ovum y = 1/2 Sperm Y = 1/2
RuleRule ofof AdditionAddition--
1/4 + 1/4 =
2/4 = 1/21/2There is a 1/2 chance There is a 1/2 chance
for a Yy offspringfor a Yy offspring
Ovum Y = 1/2 x Sperm y = 1/2
Ovum y = 1/2 x Sperm Y = 1/2
RuleRule ofof AdditionAddition--
Used when determining the probability of an event… an event that can occur can occur inin 22 oror moremore differentdifferent waysways.
RulesRules ofof
additionaddition andand multiplicamultiplicationtion::
Variations Variations on Mendelon Mendel
1. Incomplete Dominance2. Codominance3. Multiple Alleles4. Epistasis5. Polygenic Inheritance
Law ofLaw ofIncompleIncomple
te te DominancDominanc
ee
LawLaw ofof IncompleteIncomplete DominanceDominance
Multiple allelesMultiple alleles
Any of a set of three or more alleles, only
two of which can be present in a
diploid organism.
Multiple alleles
Multiple alleles
clumped
clumped
clumped
clumped
clumped
clumped
clumped
Comment
•Rh blood factor is a separate factor from the ABO blood group.
•Rh+ = dominant•Rh- = recessive•A+ blood = dihybrid trait
Epistasi
Epistasi
ss
Epistasis•When 1 gene locus alters the expression of a second locus.
ex: •1st gene: C = color, c = albino
•2nd gene: B = Brown, b = black
Epistasi
Epistasi
ss
“Labs”
In Gerbils too!In Gerbils too!CcBb X CcBbBlack X Black
F1 = 9 black (C_B_) 3 brown (C_bb) 4 albino (cc__)
Result
•Ratios often altered from the expected.
•One trait may act as a recessive because it is “hidden” by the second trait.
Problem•Wife is type A•Husband is type AB•Child is type OQuestion - Is this possible?Comment - Wife’s boss is type O
Bombay Effect
•Epistatic Gene on ABO group.
•Alters the expected ABO outcome.
•H = dominant, normal ABO•h = recessive, no A,B,
reads as type O blood.
Genotypes•Wife: type A (IA IA , Hh)•Husband: type AB (IAIB, Hh)•Child: type O (IA IA , hh)
Therefore, the child is the offspring of the wife and her husband (and not the boss).
Bombay - Detection
• When ABO blood type inheritance patterns are altered from expected.
PolygenPolygenicic
Several genes affect one character
ex. skin color, height
Result Result •Mendelian ratios fail.•Traits tend to "run" in families.
•Offspring often intermediate between the parental types.
•Trait shows a “bell-curve” or continuous variation.
PleiotroPleiotropypy
one gene, one gene, multiple multiple phenotypic phenotypic effectseffects
PleiotroPleiotropypyCystic Fibrosis
one gene, one gene, multiple multiple phenotypic phenotypic effectseffects
Some DOMINANT genes, Some DOMINANT genes, are not often are not often expressed in a expressed in a
populationpopulation
Some DOMINANT genes, Some DOMINANT genes, are not often are not often expressed in a expressed in a
populationpopulation
Does the “R” allele always stay with the “Y” allele?
= = normal normal malemale= normal= normal
femalefemale
= = affecteaffected d malemale
= affected= affected femalefemale
Reproductive Reproductive partnerspartners
siblingssiblings
Can determine:Can determine:1) Autosomal recessive disorder2) Autosomal dominant disorder
Can determine:Can determine:3) Sex-Linked disorder
Genetic ScreeningGenetic Screening
•Risk assessment for an Risk assessment for an individual inheriting a individual inheriting a trait.trait.
•Uses probability to Uses probability to calculate the risk.calculate the risk.
General FormalGeneral FormalR = F x M x D
R = riskF = probability that the female carries the gene.
M = probability that the male carries the gene.
D = Disease risk under best D = Disease risk under best conditionsconditions..
ExampleExample•Wife has an albino Wife has an albino parent.parent.
•Husband has no albinism Husband has no albinism in his pedigree.in his pedigree.
•Risk for an albino Risk for an albino child?child?
Risk Risk CalculationCalculation
•Wife = probability is 1.0 that Wife = probability is 1.0 that she has the allele.she has the allele.
•Husband = with no family Husband = with no family record, probability is near 0.record, probability is near 0.
•Disease = this is a recessive Disease = this is a recessive trait, so risk is Aa X Aa trait, so risk is Aa X Aa = .25= .25
•R = 1 X 0 X .25R = 1 X 0 X .25•R = 0R = 0
Carrier Carrier RecognitionRecognition•Fetal TestingFetal Testing–AmniocentesisAmniocentesis–Chorionic villi Chorionic villi samplingsampling
•Newborn ScreeningNewborn Screening
Summary
• Know the Mendelian crosses and their patterns.
• Be able to work simple genetic problems (practice).
• Watch genetic vocabulary.• Be able to read pedigree charts.
• Be able to recognize and work with some of the “common” human trait examples.