mendel cultivated · 2021. 1. 20. · mendel cultivated pea plants and tested some plants he found...
TRANSCRIPT
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Mendel cultivated pea plants and tested some plants He found that true-breeding plants make offspring retained traits of the Parents every time. Experimental data, Mendel deduced that an organism has two genes ) for each inherited characteristic
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Site of
Gregor
Mendel’s
experimental
garden in the
Czech
Republic
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Mendel’s peas • Mendel looked at seven traits
or characteristics of pea plants:
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1. Flower color purple or white
2. Flower position axial or terminal
3. Stem length Tall or Dwarf
4 . Seed shape round or wrinkled
5. Seed color yellow or green
6. Pod shape inflated or constricted
7. Pod color green or yellow
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are used to show the mating of two parents and the possible offspring they can produce.
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STEPS:
TT and t t Cross
T T t t
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T t
T t
T t
T t
T T
t t
Genotypes: 100% T t
Phenotypes: 100% Tall plants
T T t t
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T T
T t
T t
t t
T t
T t
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• A heterozygous with a homozygous
T t t t
T
t
t
T t
t t
You can still use the shortcut!
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LAW PARENT CROSS OFFSPRING
DOMINANCE TT x tt tall x short
100% Tt Tall
SEGREGATION Tt x Tt tall x tall
75% tall 25% short
RrGg x RrGg round & green
x round & green
9/16 round seeds & green pods
INDEPENDENT ASSORTMENT
3/16 round seeds & yellow pods
3/16 wrinkled seeds & green pods
1/16 wrinkled seeds & yellow pods
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and
F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.
Example: snapdragons (flower)
Rr
Rr
Rr Rr
produces the F1 generation
All Rr = pink
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• Two alleles are expressed (multiple alleles) in heterozygous individuals.
: blood group
1. type A
2. type B
3. type AB =
4. type O =
= IAIA or IAi
= IBIB or IBi
IAIB
ii
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homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IAIB IBi
IAIB IBi
1/2 = IAIB
1/2 = IBi
IB
IA i
IB
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IAi IBi
IAi IBi
1/2 = IAi
1/2 = IBi
i
male Type O (ii) x
female type AB (IAIB) IA IB
i
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18
If a boy has a blood type O and his sister has blood type AB
boy - type O (ii) X girl - type
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IAIB
ii
Parents: genotypes =IAi and IBi phenotypes = A and B
IB
Answer:
IA i
i