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HEREDITY = The passing of traits fromparents to offspring.
Transmitted by means of information
stored in molecules of DNA.
GENEITCS =Scientificstudy of heredity
◦ Based on knowledgethat traits aretransmitted bychromosomes.
◦ Our chromosomesmade up of ourgenes, which arepieces of DNA that
are
code for certain traits.
Humans have 46chromosomes inall your somatic(body) cells.
23chromosomesfrom yourmom and 23chromosomesfrom yourdad.
With MENDELIAN traits (the type of traitsthat Mendel studied), heterozygotes DONOT have a blend of the two alleles.
Instead, one type of allele dominates –
We show the characteristics of this
allele only – it is the DOMINANT trait.
The other version of the trait is still thereon half of our chromosomes (so we mightstill pass it on to our children, dependingon meiosis) BUT it DOES NOT affect usright now—it is the RECESSIVE trait.
Whether we are heterozygous(EX. Tt),Homozygous with the dominanttrait(EX. TT), orhomozygous with the recessive trait (EX. tt) it iscalled our GENOTYPE (type of genes thatwe have).
Which trait we physically show is our
PHENOTYPE (the type of allele that is
expressed).(EX. Blue eyes)
Mendel produced pure strains by allowingthe plants to self-pollinate for severalgenerations.
These strains were called the parental
generation or P1 strain.
Mendel cross-pollinated two strainstracked each trait through two
and
generations. (ex: TT x tt )
Trait = plant height
Alleles = T (tall), t (short)
P1 cross
T
= TT x tt
F1 Genotypic ratio =100% Tt
(4/4)
T
tF1 Phenotypic ratio
100% Tall(4/4)
=
t
Tt Tt
Tt Tt
The offspring ofT Tthis cross were all
t hybrids showingONLY thedominant trait &twere called the
First Filial or F1
generation.
Tt Tt
Tt Tt
Mendel then crossed two of his F1 plantsand tracked their traits; knowncross.
Trait = plant height
Alleles = T (tall), t (short)
as an F1
F1 cross = Tt x Tt
F2 Genotypic ratio =1/4 TT: 2/4 Tt: 1/4tt
T tT
F Phenotypic ratio =2
3 Tall: 1 short(3/4 tall; ¼ short)
t
TT Tt
Tt tt
When 2 hybrids were crossed, 75% (3/4)of the offspring showed the dominanttrait & 25% (1/4) showed the recessivetrait
Two hybrids ALWAYS create a
3(dominant trait): 1 (recessive trait) ratio.
The offspring of this cross were calledthe F2 generation.
Inheritable factors or genes areresponsible for all heritablecharacteristics.
Phenotype is based on genotype.
Each trait is based on two genes, onefrom the mother and the other from thefather.
True-breeding individuals are
homozygous (both alleles) are the same.
Formulated 3 laws of heredity in the early1860's.
1. Law of Dominance states that whendifferent alleles for a characteristic areinherited (heterozygous), the trait ofonly one (the dominant one) will beexpressed. The recessive trait'sphenotype only appears in true-breeding (homozygous) individuals.
Trait: Pod Color
Genotypes: Phenotype:
GGGreen Pod
GgGreen Pod
ggYellow Pod
2. The Law of Segregation = states thateachgenetic trait is produced by a pairof alleles
which separate (segregate)reproduction.
Rr
during
R r Explains the disappearance of a specifictrait in the F1 generation and itsreappearance in the F2 generation.
3. The Law of Independent Assortment=
states that each factor (gene) isdistributed (assorted) randomly andindependently of one another in theformation of gametes (egg or sperm).
RrYy
RY Ry rY ry
Explains that different traits are inheritedindependently, if on differentchromosomes
Ex: wrinkled seeds do not have to be
yellow. They can be green.
Ex: A gamete with RrYy
◦ R and r –gametes
Y and y –gametes
They can
separate into different
◦ Separate into different
◦ then recombine 4 ways toform gametes:
RY Ry rY ry
Use Punnett Squares to solve geneticproblems!
You cannot tell by looking at an organism that shows the dominant trait whether it is heterozygous (Rr) or homozygous(RR) for that trait
To determine the genotype of an organism
showing the dominant trait a test cross
would be done.
Test cross = the organism of unknowndominant genotype is crossed with ahomozygous recessive (rr) organism.
Black coat color in guinea pigs iscolor. Using aresults of
dominant over white coatPunnett square, show thecrossing a hybrid black with pure white.Then show the results of crossing ahybrid black and a hybrid black.
Black Trait – B (dominant)
White Trait – b (recessive)
Hybrid black – Bb
Pure white – bb
P1 cross:
B
Bb x bb
bGenotypes of F1 offspring
2/4 Bb2/4 bb
b Phenotypes of F1 offspring–
2 black : 2 white(or 50% black and 50%
white)b
Bb bb
Bb bb
__Bb_
_
__B_b
_
P1 cross: x
B Genotypes of F1 offspring
B1/4BB: 2/4Bb:1bb
Phen/4otypes of F1
offspring–
b 3Black: 1white
BB Bb
Bb bb
1.) In pigs, the white color (W) is dominant;the black color (w) is recessive. UsingPunnett squares, show the expectedresults of the following crosses.(smartboard)
__W_W_x
__w_w_
F G:
P1 cross:
100% Ww1W W
w 100% WhiteF P:1
w
Ww Ww
Ww Ww
__W_w_
__W_w_
P1 cross: x
W w 1WW: 2Ww: 1wwF2 G:
W
3White: 1blackF2 P:
w
WW Ww
Ww ww
Because each parent and offspring areusing two traits, each onealleles, 2 for each trait.
Each gamete produced by
generations will contain 2
each trait.
should have 4
the P1
alleles, one for
Example: A plant that is heterozygous forbeing tall and having green seeds is crossedwith a homozygous yellow and short
Traits = seed color and plant height
Alleles G =green T = tall
g = yellow t = short
Cross: TtGg x ttgg
Determine the gametes produced by eachparent by using the FOIL method.
◦ TtGg produces 4 different gametes:
TG Tg tG tg
◦ ttgg produces only 1 gamete: tg
TG Tg tG tg
tg Phenotypes:
4 tall/green seeds4tall/yellow seeds4short/green seeds4short/yellow seedstg Ttgg ttGg ttggGenotypes:
4TtGg:4Ttgg:4ttGg:4ttgg
tg Ttgg ttGg ttgg
Ttgg ttGg ttggtg
TtGg Ttgg ttGg ttgg
TtGg Ttg
TtGg
TtGg
The father has black hair (heterozygous)and brown eyes (heterozygous) and themother has blonde hair and blue eyes.
Genotype
Genotype
of father =
of mother =
Probability “AND” rule: If you want to know the probability of 2 traits happening at the same time-you MULTIPLY the probability of the two traits together.
Smart board practice
