Heredity and probability

what makes pea plants a model organism?

distinct traits

many offspring

easy to cross with each other or themselves

take up little room

short generation time

gene

an inherited factor (encoded in the DNA) that helps determine a characteristic

allele

one of 2+ alternate forms of a gene

locus

specific spot on a chromosome where an allele is

genotype

set of alleles possessed by an organism 

homozygote

an individual that has 2 of the same alleles at a locus

heterozygote

an individual that has 2 different alleles at a locus

characteristic (character)

attribute/feature

phenotype (trait)

appearance/manifestation of a characteristic

parents

founding generation

F1

first filial generation (offspring of parents)

F2

second filial generation (offspring of F1)

cross

breeding of 2 individuals

self

breeding with self

monohybrid crosses

single trait/gene

Mendel’s monohybrid pea plant experiment

1. developed homozygous plants for 2 phenotypes

2. crossed the 2 phenotypes

3. F1 all had same phenotype (dominant)

4. self-crossed F1

5. F2 phenotype ratio was 3:1 (3 dominant to 1 recessive)

conclusions from Mendel’s monohybrid pea plant experiment

1. both parents carry 2 alleles for each trait

2. each parent passes 1 allele and the 2 inherited alleles recombine in offspring

3. alleles can be dominant (round) or recessive (wrinkled)

principles of segregation (Mendel’s First Law)

1. each diploid organism has 2 traits per allele (one from mother and one from father)

2. allele pairs must segregate as gametes form

3. the likelihood of an allele being present in each gamete is 50% (1/2)

4. allele pairs reform at fertilization

Punnett square

grid where rows and columns are labeled based on the gametes produced by each parent

how is the probability of two independent events occurring together calculated?

multiplication (multiplication rule)

key indicator that the multiplication rule is required

“and”

how is the probability of two or more mutually exclusive events occurirng calculated?

adding the probability of each event (addition rule)

key indicator that the addition rule is required

“either” and “or”

the __________ method is easier than using a Punnett square when working with multiple loci

probability

If the probability of being blood type A is 1/8 and the probability of being blood type O is 1/2, what is the probability of being either blood type A or blood type O?

5/8 (addition rule)

phenotypic ratio = 3 : 1

Aa x Aa → ¾ A_ : ¼ aa

phenotypic ratio = 1 : 1

Aa x aa → ½ Aa : ½ aa

phenotypic ratio = uniform progeny

AA x AA → All AA

aa x aa → All aa

AA x aa → All Aa

AA x Aa→ All A_

genotypic ratio = 1 : 2 : 1

Aa x Aa → ¼ AA : ½ Aa : ¼ aa

genotypic ratio = 1 : 1

Aa x aa → ½ Aa : ½ aa

Aa x AA → ½ Aa : ½ AA

genotypic ratio = uniform progeny

AA x AA → All AA

aa x aa → All aa

AA x aa → All Aa

dihybrid crosses

2 different traits

Mendel’s dihybrid pea plant experiment

1. crossed 1 homozygous round-yellow with 1 homozygous wrinkled-green

2. F1 all had dominant phenotypes (round-yellow)

3. self-fertilized F1

4. F2 had new phenotype combinations and ratios (round-green, wrinkled-yellow) and ratio = 9 : 3 : 3 : 1

the Principle of Independent Assortment (Mendels second law/extension of the first law)

if alleles at different loci are on different chromosomes/far apart on same chromosome, they segregate independently, producing different combinations in the gametes

calculate the gametic ratios for heterozygote RrYy

“and” = multiplication rule

50% chance of each allele

50% R and 50% Y = 25% RY

50% R and 50% y = 25% Ry

50% r and 50% Y = 25% rY

50% r and 50% y = 25% ry

branch diagram

another way to show mathematical ratios in a multi-trait cross

steps of the branch diagram

1. split each trait into a monohybrid cross

2. use multiplication rule (each event is independent)

branch diagram example

¾ R x ¾ Y = 9/16 R_Y_

uses of the branch diagram

keep track of potential trait combinations

determine phenotypes/genotypes for traits

faster than Punnett for multiple trait crosses

if you have parents AaBB and aabb, what is the likelihood the offspring will be aaBb?

Parent 1 (AaBB): 50% a, 100% B. 0.5 × 1 = 0.5

Parent 2 (aabb): 100% a, 100% b. 1 × 1 = 1

0.5 aB x 1 ab = 0.5 aaBb