1. Self-pollination: * is where the female part of the plant and the male part of the plant can cross and combine with each other. * An example is corn and pea plants. * the female ovum is fertilized by it’s own male anther pollen.
2. Cross Pollination (cross breeding):
when female and male genes come from different separate organisms.
Example; humans, most animals where sperm come from the male and the egg from the female.
3. Monohybrid Cross:
cross breeding program focusing only on ONE trait.
Examples:
A. maximizing fat content in jersey cows’ milk. B. mildew disease resistance in corn or wheat. D. absence of hip displasia in golden retriever dogs.
4. Linkage Cross: * Genes found on the same chromosome are said to be linked. * White hair, blue eyes and deaf are usually on the same chromosome for cats. * EG: when I got my cat, she had white hair and blue eyes. I was told that she would probably be deaf too since all those genes are found together on the same chromosome.
5. Hybridization: * A cross-breeding where you take extremely dissimilar varieties and cross them together. * the goal is to get the best possible combination between the two parents. * examples: a. children today are taller and smarter than their parents. b. cows that are more disease resistant and make more milk than their parents. c. corn plants that yield more per cob and have more uniform kernel size than their parents.
6. Test Cross: *Used to determine if a dominant trait is homozygous dominant or heterozygous dominant. * you always cross the unknown parent to a homozygous recessive one. * the resulting offspring will tell you if the unknown parent’s genotype was homozygous dominant or heterozygous.
EXAMPLE: In dogs, short hair is recessive (hh) and long hair is dominant (H). Sample cross 1: P1 = (HH) unknown parent X (hh) test cross P1 gametes: 100 % H X 100 h F1 = 100 % Hh Thus, all the dogs will have long hair in this case. IF all the dogs have long hair, this tells you that the unknown parent was homozygous dominant for long hair.
Sample cross 2: P1 = (Hh) unknown X (hh) test cross P1 gametes: (50% H + 50% h) X (100 % h) F1 = Hh 50%+hh 50%
If the kids are all 50% dominant trait + 50% recessive trait, then the unknown parent was heterozygous.
*obviously, this method of crossing works best with animals and plants that have a short breeding period.
Monohybrid Crosses - Sample Problem:
We want corn that is disease resistant to black rot.
P1 = parent generation.
F1 = First offspring of the parents.
F2 = Second generation from the parents.
Steps to find the % abundance of F1 Generation :
Step 1: List the genotypes of the P1 (parents)
Step 2: Determine the percent abundance of the P1 sex cells for each parent.
Step 3: Convert the percentages to decimals by dividing by 100.
Step 4: Cross each gamete of the female times all the gametes of the male to get the F1 genotypes.
Step 5: When you cross gametes to get the F1 genotypes, multiply the percent abundances of female times male of the parents.
Step 6: Add all the percent abundances of the F1 genotypes together. This should = 100% (which is 1).
Step 7. Collect like genotypes and add them together.
Step 8. Convert the genotypes to phenotypes if needed.
Sample: If mom was heterozygous for night blindness and dad was homozygous recessive for the same trait, what are the percent abundances of the F1 phenotypes?
1. Self-pollination:
* is where the female part of the plant and the male part of the plant can cross and combine with each other.
* An example is corn and pea plants.
* the female ovum is fertilized by it’s own male anther pollen.
2. Cross Pollination (cross breeding):
3. Monohybrid Cross:
- cross breeding program focusing only on ONE trait.
- Examples:
A. maximizing fat content in jersey cows’ milk.B. mildew disease resistance in corn or wheat.
D. absence of hip displasia in golden retriever dogs.
4. Linkage Cross:
* Genes found on the same chromosome are said to be linked.
* White hair, blue eyes and deaf are usually on the same chromosome for cats.
* EG: when I got my cat, she had white hair and blue eyes. I was told that she would probably be deaf too since all those genes are found together on the same chromosome.
5. Hybridization:
* A cross-breeding where you take extremely dissimilar varieties and cross them together.
* the goal is to get the best possible combination between the two parents.
* examples:
a. children today are taller and smarter than their parents.
b. cows that are more disease resistant and make more milk than their parents.
c. corn plants that yield more per cob and have more uniform kernel size than their parents.
6. Test Cross:
*Used to determine if a dominant trait is homozygous dominant or heterozygous dominant.
* you always cross the unknown parent to a homozygous recessive one.
* the resulting offspring will tell you if the unknown parent’s genotype was homozygous dominant or heterozygous.
EXAMPLE:
In dogs, short hair is recessive (hh) and long hair is dominant (H).
Sample cross 1:
P1 = (HH) unknown parent X (hh) test cross
P1 gametes: 100 % H X 100 h
F1 = 100 % Hh
Thus, all the dogs will have long hair in this case. IF all the dogs have long hair, this tells you that the unknown parent was homozygous dominant for long hair.
Sample cross 2:
P1 = (Hh) unknown X (hh) test cross
P1 gametes: (50% H + 50% h) X (100 % h)
F1 = Hh 50%+hh 50%
*obviously, this method of crossing works best with animals and plants that have a short breeding period.
Monohybrid Crosses - Sample Problem:
We want corn that is disease resistant to black rot.
P1 = parent generation.
F1 = First offspring of the parents.
F2 = Second generation from the parents.
Steps to find the % abundance of F1 Generation :
Step 1: List the genotypes of the P1 (parents)
Step 2: Determine the percent abundance of the P1 sex cells for each parent.
Step 3: Convert the percentages to decimals by dividing by 100.
Step 4: Cross each gamete of the female times all the gametes of the male to get the F1 genotypes.
Step 5: When you cross gametes to get the F1 genotypes, multiply the percent abundances of female times male of the parents.
Step 6: Add all the percent abundances of the F1 genotypes together. This should = 100% (which is 1).
Step 7. Collect like genotypes and add them together.
Step 8. Convert the genotypes to phenotypes if needed.
Sample: If mom was heterozygous for night blindness and dad was homozygous recessive for the same trait, what are the percent abundances of the F1 phenotypes?
B = No night blindness. b = night blindness.
Step 1:
P1 Mom = Bb
P1 Dad = bb
Step 2:
Gametes of Mom = 50% B, 50% b.
Gametes of Dad = 100% b.
Step 3:
Gametes for Mom = (0.5B+0.5b)
Gametes for Dad = (1b)
Step 4 and 5:
F1 = (Mom)x(Dad)
= (0.5B+0.5b)x(1b)
= (0.5Bb+0.5bb)
Step 6: they add to 100 % or 1.0
Step 7/8:
F1 phenotypes =
50% Bb = no night blindness.
50% bb = night blindness.
Home work:
Do 5.1 Tutorial #1 Practice Questions #1 and #2