mendel and the gene idea chp 14. blending model vs. particulate model genetic material mixes like...
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Mendel and the Gene IdeaChp 14
Blending Model vs. Particulate Model
•genetic material mixes like paint
•Parent’s traits inseparable
•Discreet inheritable units
•Traits retain separate identities
Figure 14-01
Gregor Mendel
•Austrian Monk
•Did not know about DNA, genes, or chromosomes!
•Tried to prove particulate model of inheritance using pea plants
Web Lab: Mendel and his Peas
Advantages of working with pea plants:•Many different varieties
•Character = inheritable feature (GENE)•Trait = variation of that character (ALLELE)
•Can control plant matings (paintbrush)
LE 14-2
Removed stamensfrom purple flower
Transferred sperm-bearing pollen fromstamens of whiteflower to egg-bearing carpel ofpurple flower
Carpel Stamens
Parentalgeneration(P)
Pollinated carpelmatured into pod
Planted seedsfrom pod
Examinedoffspring:all purpleflowers
Firstgenerationoffspring(F1)
•Started with true-breeding varieties = HOMOZYGOUS•If allowed to self-pollinate, all offspring had the same traits as their parent plant
•Then, cross-pollinated two different true-breeding varieties = HYBRIDIZATION
Hybridization
LE 14-5_1
Appearance:Genetic makeup:
Gametes:
P Generation
F1 Generation
PurpleflowersPP
Whiteflowerspp
pP
Purple flowersPp
Appearance:Genetic makeup:
Monohybrid Cross – tracks a single character/gene
hybrid
Mendel’s Law of Segregation: Parental alleles separate (segregate) during gamete formation• occurs in Metaphase I of Meiosis I• only one allele per gamete
LE 14-5_2
Appearance:
P Generation
Genetic makeup:
Gametes
F1 Generation
Appearance:Genetic makeup:
Gametes:
F2 Generation
Purple flowersPp
P p1 21 2
P p
F1 sperm
F1 eggsPP Pp
Pp pp
P
p
3 : 1
PurpleflowersPP
Whiteflowerspp
P p
gametes
gametesPunnett Square shows possible offspring after fertilization
LE 14-6
Phenotype
Purple
Purple3
Purple
Genotype
PP(homozygous
Pp(heterozygous
Pp(heterozygous
pp(homozygous
1
2
1
Ratio 1:2:1
White
Ratio 3:1
1
R r
Rr
F1
F2 R rRr
rrRr
3 round: 1 wrinkled1 RR : 2 Rr : 1 rr
Law of Dominance: The dominant allele (trait) is fully expressed and the recessive allele has no noticeable effect
Recessive ≠ Bad
Test Cross
•Used to determine the genotype of an individual with a dominant phenotype
•Cross it with a homozygous recessive and observe offspring ratios
T T T
t
t
t
t
t
100% tall 100% Tt
Tt
Tt Tt
Tt
1 tall : 1 dwarf 1 Tt : 1 tt
Tt
Tt
tt
tt
Law of Independent Assortment: Each allele pair segregates independently of other allele pairs during Meiosis (Metaphase I).
Dihybrid Cross: tracks two genes (characters)
Tall purple
TtPp
TP Tp tP tp
TP
TP
Tp
Tp
tP
tP
tp
tp
TTPP
TTPp
TTPp
TtPP
TtPP
TtPp
TtPp
ttpp
TTpp
TtPp
TtPp
ttPP
Ttpp
Ttpp
ttPp
ttPp
9 tall purple3 tall white
3 dwarf purple1 dwarf white
12:4 or 3:112:4 or 3:1
Rules of Probability & Genetics
• Probability ranges from 0 – 1 • Rule of Multiplication: Used to determine
the chance of two or more independent events occurring together
1. Determine probability of each independent event2. Multiply probabilities together
Ex: What is the probability (chance) that when 2 coins are flipped they will both end up heads?
½ x ½ = ¼ (or .25)
½ ½
½
½
¼
¼
¼
¼
Multiplication RuleProbability Problems
• What is the probability (chance) that 2 dice will both show a 4 when rolled?
1/6 x 1/6 = 1/36
Ex (F1): T t P p x T t P p
F2: t p t p
½ x ½ x ½ x ½ = 1/16
t t p p
½ ½ ½ ½
T t
T TT Tt
t Tt tt
P p
P PP Pp
p Pp pp
Rule of Addition
• What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?
• Rule of Addition: used to determine the probability of an event that can occur in two or more different ways/combinations– Calculate probability for each possible
combination/way using Rule of Multiplication– Add the probabilities for each separate
combination to get the total probability
Rule of Addition
• What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?
• Possible combinations:
1 + 4 2 + 3 3 + 2 4 + 1 1/6 x 1/6 1/6 x 1/6 1/6 x 1/6 1/6 x 1/6
1/36 + 1/36 +
1/36 + 1/36 =
4/36 ( 1/9 )
A a
A AA Aa
a Aa aa
b b
B Bb Bb
b bb bb
C C
c Cc Cc
c Cc Cc
¾ ½ 1
= 3/8
x x
A a
A AA Aa
a Aa aa
b b
B Bb Bb
b bb bb
C C
c Cc Cc
c Cc Cc
A__ B__ C__ =
A__ B__ c c =
A__ b b C__ =
a a B__ C__ =
A a
A AA Aa
a Aa aa
b b
B Bb Bb
b bb bb
C C
c Cc Cc
c Cc Cc
A__ B__ C__ = 3/8
A__ B__ c c = 0
A__ b b C__ = 3/8
a a B__ C__ = 1/8
Sum = 7/8
Text book: pages 272 - 273
• All problems (#1-17) due on
• Tonight, work on problems:
# 2, 3, 4, 7, 8, & 10
Types of Dominance
• Complete Dominance – one allele complete masks the other; the heterozygous (Rr) and homozygous dominant (RR) have the same phenotype–The dominant allele usually codes for some protein/enzyme and the recessive allele codes for a defective protein/enzyme. Both are “expressed”, but only the dominant allele is functional and observable
Types of Dominance• Codominance = two alleles expressed
separately & both affect phenotype– Ex: Red & White rhododendron
“Roan” color in cattle
CR CR
CW CR CW CR CW
CW CR CW CR CW
Types of Dominance
• Incomplete Dominance = F1 hybrids (heterozygotes) have an intermediate phenotype– Results in a THIRD
phenotype– NOT the same as blending
(F2 show all phenotypes)– 1:2:1 phenotypic and
genotypic ratios in F2
Multiple Alleles = more than two alleles/varieties
IA & IB
i
are codominant
over
Fill in Interactive Question 14.6
Blood Type practice…
• Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:– Type O?– Type B? – Type A?– Type AB?
Blood Type practice…
• Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:– Type O? ¼ – Type B? ¼ – Type A? ¼ – Type AB? ¼
IB i
IA IAIB IAi
i IBi ii
Pleiotropy
• Pleiotropy = one gene has multiple phenotypic effects– Ex: cystic fibrosis & sickle-cell disease
Sperm
BC bC Bc bc
BbCcBBCcBbCCBBCC
BbCC bbCC BbCc bbCc
BbccBBccBbCcBBCc
BbCc bbCc Bbcc bbcc
BC
bC
Bc
bc
BbCc BbCc
14
14
14
14
14
14
14
14
916
316
416
Epistasis = (“force upon”) one gene affects the expression of another gene
Ex: coat color in mice
brownwhite
M__B__
mm __ __
¾ * ¾ = 9/16
¾ * ¼ = 3/16
¼ * 1 = 4/16 (¼)
aabbcc Aabbcc AaBbcc AaBbCc AABbCc AABBCc AABBCC
AaBbCcAaBbCc
20/64
15/64
6/64
1/64
Fra
ctio
n o
f p
rog
eny
Polygenic = additive effect of 2 or more genes on one phenotype
Ex: skin color height
# alleles + 1 = # phenotypes
AaBbCc = 25 cm
Seven (6 alleles + 1 = 7 phenotypes)
Nature vs. Nurture….
Acidic soil
Basic soil
Pedigree
• Family tree showing relationships between family members and the pattern of inheritance across the generations
Figure 14-16
Aa Aa
AaAA Aa
No – either he or his wife is AA, but can’t be certain
AA aa aa Aa A_ A_ A_
Aa Aa Aa Aa aa Aa Aa A_ A_ A_ A_ A_
Aa Aa
Recessive Disorders• Carriers = heterozygotes (Aa) have the
defective allele, but are not affected by disorder– “carry” allele, but do not express it– Can pass it on to offspring (50% chance)
¼
2/3
A a
A AA Aa
a Aa aa
Recessive Disorders
• Cystic Fibrosis
• Tay-Sachs Disease
• Sickle-cell Disease
?
Aa Aa Aa Aa
aa aaA_ A_A a
A AA Aa
a Aa aa
2/3 Aa x 2/3 Aa x ¼ aa = 4/36 = 1/9
Dominant Disorders
• RARE – nature will weed out (no carriers!)
• Achondroplasia Dwarfism
• Huntington’s Disease
LE 14-17a
Amniocentesis
Amnioticfluidwithdrawn
Fetus
A sample ofamniotic fluid canbe taken starting atthe 14th to 16thweek of pregnancy.
Centrifugation
Placenta Uterus Cervix
Fluid
Fetalcells
Biochemical tests can beperformed immediately onthe amniotic fluid or lateron the cultured cells. Biochemical
tests
Severalweeks
Karyotyping
Fetal cells must be culturedfor several weeks to obtainsufficient numbers forkaryotyping.
LE 14-17b
Chorionic villus sampling (CVS)
Placenta Chorionic villi
Fetus
Suction tubeinserted throughcervix
Fetalcells
Biochemicaltests Karyotyping and biochemical
tests can be performed onthe fetal cells immediately,providing results within a dayor so.
Severalhours
A sample of chorionic villustissue can be taken as earlyas the 8th to 10th week ofpregnancy.
Karyotyping