Chp 14 Mendel and the Gene Idea Blending

Chp 14 Mendel and the Gene Idea

Blending Model vs. Particulate Model • genetic material mixes like paint • Discreet inheritable units • Parent’s traits inseparable • 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 stamens from purple flower Transferred spermbearing pollen from stamens of white flower to eggbearing carpel of purple flower Parental generation (P) Carpel Stamens Pollinated carpel matured into pod Planted seeds from pod First generation offspring (F 1) Examined offspring: all purple flowers

• 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 P Generation Appearance: Genetic makeup: Gametes: Purple flowers PP White flowers pp P p F 1 Generation Appearance: Genetic makeup: Purple flowers Pp hybrid Monohybrid Cross – tracks a single character/gene

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 P Generation Appearance: Genetic makeup: Purple flowers PP White flowers pp P p Gametes F 1 Generation Appearance: Genetic makeup: Purple flowers Pp 1 Gametes: 2 1 P p 2 gametes F 1 sperm P p PP Pp Pp pp F 2 Generation P gametes F 1 eggs p 3 : 1 Punnett Square shows possible offspring after fertilization

LE 14 -6 3 Phenotype Genotype Purple PP (homozygous Purple Pp (heterozygous 1 2 1 Purple Pp (heterozygous White pp (homozygous Ratio 3: 1 Ratio 1: 2: 1 1

r R F 1 Rr F 2 R r Rr Rr rr 3 round: 1 wrinkled 1 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

100% tall 100% Tt 1 tall : 1 dwarf T T T 1 Tt : 1 tt t t 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)

TP TP Tp t. P tp Tall purple TP TTPp Tt. PP Tt. Pp Tp TTPp TTpp Tt. Pp Ttpp t. P Tt. Pp tt. PP tt. Pp tp Tt. Pp Ttpp tt. Pp ttpp 9 tall purple 3 tall white 3 dwarf purple 1 dwarf white 12: 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 event 2. 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)

Multiplication Rule Probability Problems • What is the probability (chance) that 2 dice will both show a 4 when rolled? 1/ 1/ = 1/ x 6 6 36

Ex (F 1): Tt. Pp x Tt. Pp ½ ½ F 2: t p t t T TT Tt tt P p P PP Pp pp p ½ x ½ x ½ = 1/16 ttpp T

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 1/ 1/ x 6 6 1/ 36 1/ + 1/ x 6 6 1/ 36 3+2 1/ 1/ x 6 6 + 1/36 + 4/ 1/ ) ( 36 9 4+1 1/ 1/ x 6 6 1/ 36 =

= 3/8 x ¾ A a A AA Aa aa 1 x ½ b b B Bb Bb b bb bb C C c Cc Cc

A__ B__ C__ = A__ B__ c c = A__ b b C__ = a a B__ C__ = A a A AA Aa aa b b B Bb Bb b bb bb C C c Cc Cc

A__ B__ C__ = 3/8 A__ B__ c c = 0 7 Sum = / A__ b b C__ = 3/8 a a B__ C__ = 1/8 A a A AA Aa aa b b B Bb Bb b bb bb 8 C C c Cc Cc

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 CW CR CR CR CW CW CR CW

Types of Dominance • Incomplete Dominance = F 1 hybrids (heterozygotes) have an intermediate phenotype – Results in a THIRD phenotype – NOT the same as blending (F 2 show all phenotypes) – 1: 2: 1 phenotypic and genotypic ratios in F 2

Multiple Alleles = more than two alleles/varieties IA & I B are codominant over i 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 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 AB? ¼ IB i IA I AI B I Ai i IB i ii

Pleiotropy • Pleiotropy = one gene has multiple phenotypic effects – Ex: cystic fibrosis & sickle-cell disease

Epistasis = (“force upon”) one gene affects the expression of another gene Bb. Cc Sperm 1 1 1 Ex: coat color in mice Bb. Cc 1 1 4 BC 1 4 b. C 1 4 Bc 1 4 bc 4 BC BBCC Bb. CC BBCc Bb. Cc 4 b. C Bb. CC bb. CC Bb. Cc bb. Cc 4 Bc BBCc Bb. Cc BBcc Bbcc 4 bc Bb. Cc bb. Cc Bbcc bbcc 9 16 3 16 4 16

Polygenic = additive effect of 2 or more genes on one phenotype aabbcc Aa. Bb. Cc Aa. Bbcc Aa. Bb. Cc AABBCc AABBCC 20/64 15/64 Fraction of progeny Ex: skin color height Aa. Bb. Cc 6/64 # alleles + 1 = # phenotypes 1/64

Aa. Bb. Cc = 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 aa aa Aa Aa A_ Aa Aa Aa AA A_ A_ Aa Aa Aa No – either he or his wife is AA, but can’t be certain

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 AA Aa Aa aa

Recessive Disorders • Cystic Fibrosis • Tay-Sachs Disease • Sickle-cell Disease

2/ 2/ Aa x ¼ aa = 4/ = 1/ Aa x 3 3 36 9 Aa Aa Aa aa A_ A_ ? Aa aa A a AA Aa Aa aa

Dominant Disorders • RARE – nature will weed out (no carriers!) • Achondroplasia Dwarfism • Huntington’s Disease

LE 14 -17 a Amniocentesis Amniotic fluid withdrawn Fetus A sample of amniotic fluid can be taken starting at the 14 th to 16 th week of pregnancy. Centrifugation Placenta Uterus Cervix Fluid Fetal cells Biochemical tests can be performed immediately on the amniotic fluid or later on the cultured cells. Fetal cells must be cultured for several weeks to obtain sufficient numbers for karyotyping. Biochemical tests Several weeks Karyotyping

LE 14 -17 b Chorionic villus sampling (CVS) A sample of chorionic villus tissue can be taken as early as the 8 th to 10 th week of pregnancy. Fetus Suction tube inserted through cervix Placenta Chorionic villi Fetal cells Biochemical tests Several hours Karyotyping and biochemical tests can be performed on the fetal cells immediately, providing results within a day or so.