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MENDELIAN GENETICSMENDELIAN GENETICS
GARDEN PEAS Advantages
• seeds easyto obtain
• characters easyto score
• crosses easilycontrolled
• short generationtime
• large numbers of progeny
• use of statisticsin analysis
• Unit factors (genes) exist in pairs (alleles)
• Dominance/Recessiveness
• Alleles segregate into gametes in equalfrequencies
• Alleles from different gene pairs assort independently into gametes
MENDEL'S POSTULATES
pure (true) breeding line- a group of genetically identical individuals that always produce offspring of the same phenotype when mated to each other
F1 generation
- first filial generation; the progeny resulting from the first cross in a series
F2 generation
- second filial generation; the progeny resulting from a cross of the F1 generation
monohybrid cross
- a genetic cross between two individuals involving only one character (e.g. AA x aa) in which the parents possess different alleles of the character
MONOHYBRID CROSS: PARENTAL AND F1
R_ = Round
R is dominant
rr = wrinkled
r is recessive
-------------------------------
F1 Gametes either R or r produced in equal frequencies
LAW OF EQUAL SEGREGATION
Parents
Gametes
F1 seeds
Gametes
(pure breeding)
CONCEPT OF DOMINANT AND RECESSIVE ALLELES
Round x wrinkled
F1 all Round (selfed or crossed to each other)
F2 3/4 Round; 1/4 wrinkled
Mendel concluded that one phenotype of the pea shape character (wrinkled) is hidden in the F1 and reappears in the F2.
Seed Parent
Rr
Pollen Parent
Rr
XF1
1/2 R 1/2 r 1/2 R 1/2 r
1/4 RR 1/4 Rr 1/4 Rr 1/4 rr
1/2 Rr
3/4 Round 1/4 wrinkled
Genotypic Ratio = 1: 2: 1 Phenotypic Ratio = 3: 1
Mendel explained his data by invoking the concept of genes:
- each pea plant has a pair of alleles for one character;one allele is dominant and the other is recessive
Mendel’s First Law - Law of Equal Segregation
- different alleles of one gene segregate into different gametes in equal frequencies (i.e. each gamete carries only one allele of a gene)
- the union of gametes to make a zygote is random (it doesn’t matter which allele is in each gamete)
Proof: testcross - crossing a homozygous recessive individual to anindividual of unknown genotype to determine the unknown genotype (e.g. R? x rr)
Parents RR x rr
F1 Rr (all round seeds)
F2 3:1 R_ : rr (phenotypic ratio)
Testcross (cross F1 individuals to homozygous recessive)
Rr x rr
Gametes R or r r
1:1 Rr : rr (phenotypic ratio)
If F1 were RR, all progeny from the test cross would be Rr(phenotypically round)
dihybrid cross
- a genetic cross involving two phenotypic characters in which the parents possess different alleles of each character (e.g. round, green x wrinkled, yellow peas)
DIHYBRID CROSS
R_ Round
rr wrinkled
Y_ Yellow
yy green
Parents
Gametes
F1 Progeny
F1 Gametes
Mendel’s Second Law - Independent Assortment
- different gene pairs assort independently into gametes
(pure breeding)
PHENOTYPES OF F2 PROGENY IN DIHYBRID CROSS
Phenotypic Ratio Phenotypes Alleles Present
9 round, yellow R_Y_3 round, green R_yy3 wrinkled, yellow rrY_1 wrinkled, green rryy
Punnett Square
The 9:3:3:1 phenotypic ratio observed in the F2 is created by the random superimpositionof two independent 3:1 phenotypic ratios.
3/4 R_ 3/4 Y_
1/4 rr 1/4 yy
9/16 R_Y_
3/16 R_yy
3/16 rrY_
1/16 rryy
F1 cross: RrYy x RrYy
MENDEL IGNORED FROM 1866-1900
• Mendel was an unknown researcher
• Darwin’s Origin of Species (1859)
- continuous not discontinuous variation
• Cytology of time could not explain hypothesis
- chromosomal theory of heredity developed by Theodore Boveri and Walter Sutton in early 1900's clearly showed link between chromosome segregation during meiosis and Mendel's unit factors (genes)
• Unable to replicate results in all organisms
MENDEL REDISCOVERED IN 1900
by Carl Correns, Hugo DeVries, Eric Von Tschermak
Using branch (forked) diagrams to determine expected phenotypic ratios from a cross
Parental cross RrYy x Rryy(R and Y are dominant alleles)
1/2 Y_ 3/8 R_Y_R_ 3/4
1/2 yy 3/8 R_yy
1/2 Y_ 1/8 rrY_rr 1/4
1/2 yy 1/8 rryy
• calculate frequencies for individual genes separately* because events are independent multiply probabilities
In the cross Aa bb CC Dd Ee x Aa Bb Cc dd Ee,what proportion of the progeny will have the genotypeAA bb CC dd EE?
• calculate frequencies for individual genes
Aa x Aa 1/4 of the progeny will be AA
bb x Bb 1/2 of the progeny will be bb
CC x Cc 1/2 of the progeny will be CC
Dd x dd 1/2 of the progeny will be dd
Ee x Ee 1/4 of the progeny will be EE
* because events are independent multiply probabilities
Probability of AA bb CC dd EE individuals = 1/4 x 1/2 x 1/2 x 1/2 x 1/4 = 1/128
In the cross AaBbCc x AaBbCC the expected number of
gametes = 2 x 2 x 2 = 8 and 2 x 2 x 1 = 4
genotypes = 3 x 3 x 2 = 18
phenotypes = 2 x 2 x 1 = 4
Assumes complete dominance and recessiveness for all gene pairs and independent assortment.
Don't blindly use 2n and 3n rules described in text; they can’t be used in situations such as the one above
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