MENDELIAN GENETICS

Laws of Heredity

A. Origins of Genetics• Passing characteristics from

parent to offspring is called heredity

• Accurate study of heredity began with Austrian monk Gregor Mendel at his monastery gardens

• Mendel used different varieties of garden pea plant– Could predict patterns of

heredity which form modern-day genetics principles

– Garden peas have eight observable characteristics with two distinct traits that Mendel counted and analyzed with each cross or breeding

Characteristic Trait PicturesFlower color Purple v. white

Seed color Yellow v. green

Seed shape Round v. wrinkled

Seed coat color Gray v. white

Pod shape Round v. constricted

Pod color Green v. yellow

Flower position Axial v. top

Plant height Tall v. dwarf

Pea Characteristics & Traits

• In nature, pea plants self pollinate since both reproductive organs (male stamen [pollen] & female pistil) are internal–Mendel physically

removed stamens & dusted pistils with pollen from chosen plants to observe results by cross-pollination

• Parent plants are P generation• All offspring are F generation (from Latin filialis for

son/daughter)– F1 generation = first offspring (children)

– F2 generation = second offspring (grandchildren)

F1

F2

F1

F2F2

F1

P P

F2

• Cross-pollinated two pure-bred plants with one very different trait (purple vs. white flowers) in P generation– Examined each F1 plant’s

trait & counted them• Allowed F1 generation to

self-pollinate to produce F2 generation– Examined each F2 plant’s

trait & counted them

P

F1 F1

F2F2

Monohybrid Cross

• Mendel collected tons of data – his results are reproducible– Monohybrid cross of white flowers and purple

flowers in P generation produced 100% purple flowers in F1 generation

– Self-pollination of F1 produced 705 purple flowers & 224 white flowers in F2 generation

F1P F2

• Mendel’s ratios still hold true today– Crossing pure bred traits in monohybrid cross will

ALWAYS express only one trait in F1

– Self-pollinating F1 will ALWAYS result in 3:1 ratio• 705:224 = 3:1

B. Heredity Theories & Laws• Mendel knew all ideas about “blending”

characteristics was bogus–Developed four hypotheses:• 1. individual has two copies of gene, one

from each parent

From Dad From Mom

• 2. there exists alternative versions of genes called alleles represented by letters

From Dad From Mom

AbC

DeF

ABC

deF

• 3. if two different alleles occur together, one may be expressed while other is not – dominant and recessive– UPPERCASE alleles are dominant alleles» Trait gets expressed ALWAYS

– lowercase alleles are recessive alleles» Trait only gets expressed if dominant is not

present

B trait will be expressed

e trait will be expressed

From Dad From Mom

AbC

DeF

ABC

deF

–When both alleles are identical, individual is considered homozygous for that trait• Homozygous dominant = both

dominant (UPPERCASE)• Homozygous recessive = both

recessive (lowercase)–When alleles are different, individual is

considered heterozygous for that trait

AbC

DeF

ABC

deF

TT = ? Homozygous dominant

Tt = ? Heterozygous

tt = ? Homozygous recessive

XX = ? Homozygous dominant

rr = ? Homozygous recessive

QUIZ YOURSELF

• 4. when gametes (sperm/eggs or spores) are formed, alleles for each trait separate independently during meiosis–Occurs during anaphase

A a

HH

pP

b b

C. General Rules for Genes• Each gene is given allele letter– Letter is always first letter of dominant trait• Ex: yellow peas are dominant over green peas

Y = yellow, y = green

• Ex: purple flowers are dominant over white flowersP = purple, p = white

YY yyYy

PP Pp pp

• Ex: In roses, pink petals are dominant over white petals, and tall stems are dominant over short stems.P = pink, p = whiteT = tall, t = short

Cross a male heterozygous pink tall with a female homozygous white short

Male = PpTtFemale = pptt

• Some traits are dominant – only one dominant allele needed in genome to show phenotype

• Some traits are recessive – both recessive alleles needed to express phenotype

Dominant Trait Recessive TraitPolydactyl (P) Non-polydactyl (p)

No Hitchhiker’s Thumb (T) Hitchhiker’s thumb (t)Tongue rolling (R) No tongue rolling (r)

Free-hanging ear lobe (F) Attached ear lobe (f)Widow’s Peak (W) No widow’s peak (w)

Brown eyes (B) Blue or green eyes (b)Left thumb on top (L) Right thumb on top (l)

Plus four others!

Polydactyl (PP or Pp)

No Hitchhiker’s Thumb (T)

TT or Tt tt

Tongue Rolling (R)

RR or Rr rr

Free-Hanging Earlobes (F)

FF or Ff ff

Widow’s Peak (W)

WW or Ww ww

Brown Eyes (B)

BB or Bb

bb

Left Thumb on Top (L)

LL or Ll ll

Mid-Digit Hair (H)

HH or Hh hh

Cleft Chin (C)

CC or Cc cc

Dimples (D)

DD or Dd dd

Freckles (F)

FF or Ff ff

D. Laws of Heredity• During meiosis (forming haploid gametes from

diploid cells), chromatids separate during anaphase II– Law of segregation: two alleles for character

separate when gametes are formed

Male Parent

(Tt)Female Parent

(Tt)Alleles segregate

(separate) into gametes

T t tT

Alleles segregate (separate) into gametes

Alleles pair up in all combos

Alleles segregate (separate) into gametes

Alleles pair up in all combos

• Mendel studied whether different characteristics were inherited together or separately– Conducted dihybrid crosses where two traits are

studied– Concluded that traits NOT inherited together &

developed law • Law of independent assortment: alleles of

different genes separate independently during gamete formation in meiosis

Law of Independent AssortmentMale Parent

(TtBb)Female Parent (TtBb)

Traits separate independently

TB Tb

tBtb

TB TbtB tb

TTBB TTBb TtBb16

total!

E. Punnett Square• Easiest way to represent Laws of Segregation

and Independent Assortment is through Punnett Square– Cross a homozygous dominant yellow pea with a

green (homozygous recessive)– Genotypes: YY & yy (1 trait, 4 alleles = 4 combos)

Y Yyy YyYy

YyYyStep 1: separate alleles from genotypes & place on top & down side Step 2: determine possible combinations by crossing alleles

• Have to analyze findings from crossings– Genotypic ratio: – Phenotypic ratio:

Y Yyy YyYy

YyYy

4 Yy4 yellow peas

Curi Family Eye Color• My dad has green eyes• My mom has brown eyes– Knowing that I have brown eyes, what is my

GENOTYPE?• Brown is dominant (B)• Green is recessive (b)–Dad must be bb

b

B

B

Bb Bb

Bb Bb

b

• According to Punnett Square, all my parents’ children should have BROWN eyes–In reality, my brother has green eyes. What

does this mean?•Mom’s genotype must be Bb

bB Bb Bb

bb

This means that there is a 50% (2/4 or ½) chance that each child my parents had could have green eyes. I lucked out.

b

b bb

Monohybrid Cross Examples• 1. Aliens with two eyes are dominant over

aliens with one eye. Cross a heterozygous two-eyed male with a homozygous one-eyed female.

– Genotypic ratio:– Phenotypic ratio:

Tt

t tTtTt

tt tt

2:2 (2 Tt: 2 tt)2:2 (2 two eyes: 2 one eye)

• 2. Orange carrots are dominant over purple carrots. Cross a male purple carrot with a heterozygous orange carrot.

– Genotypic ratio:– Phenotypic ratio:

oo

O o

Oo

Oo

oo

oo2:2 (2 Oo: 2 oo)2:2 (2 orange: 2 purple)

• Dihybrid (two traits) cross can be trickier– Cross heterozygous purple flowers, heterozygous

yellow pea with another of the same.– Genotypes: PpYy & PpYy (2 traits, 8 alleles = 16

combos!)

Step 1: find possible gametes (two traits each!) for each parent by doing FOIL method (first, outside, inside, last) & place on top & down side

PYPy

pYpy

PY Py pY py

Step 2: determine possible combinations by crossing alleles, making sure same alleles are together

PPYY

PPYy

PpYY

PpYy

PPYy

PPyy

PpYy

Ppyy

PpYY

PpYy

ppYY

ppYy

PpYy

Ppyy

ppYy

ppyy

• Analyze results– Genotypic ratio:

– Phenotypic ratio:

PYPy

pYpy

PY Py pY pyPPYY

PPYy

PpYY

PpYy

PPYy

PPyy

PpYy

Ppyy

PpYY

PpYy

ppYY

ppYy

PpYy

Ppyy

ppYy

ppyy

1:2:2:4:1:2:1:2:1 (1 PPYY: 2 PPYy: 2 PpYY: 4 PpYy: 1 PPyy: 2 Ppyy: 1 ppYY: 2 ppYy: 1 ppyy)

9:3:3:1 (9 purple/yellow: 3 purple /green: 3 white/yellow: 1 white: green)

Simple dihybrid rules …• Always will be maximum of 4 phenotypes– Trait A vs. Trait B, Trait C vs. Trait D = 4

phenotypes• AC, AD, BC, BD

• Heterozygous AaBb vs. Heterozygous AaBb will always have same phenotypic ratio–9 AB, 3aaB, 3Abb, 1aabb = 9:3:3:1

• Heterozygous AaBb vs. Homozygous aabb will always have same phenotypic ratio– 4 AB, 4 aaB, 4 Abb, 4 aabb = 4:4:4:4

Dihybrid Cross Examples• 1. Red ants are dominant over black ants,

and long antennae are dominant over short antennae. Cross a black short antennae male with a heterozygous red long female.–Genotypes: –Gametes:

rl, rl, rl, rl & RL, Rl, rL, rlrrll & RrLl

• Genotypic ratio:• Phenotypic ratio:

rl

rl

rl

rl

rlRL Rl rLRrLl

RrLl

RrLl

RrLl

Rrll

Rrll

Rrll

Rrll

rrLl

rrLl

rrLl

rrLl

rrll

rrll

rrll

rrll

4:4:4:4 (4 RrLl: 4 Rrll: 4 rrLl: 4 rrll)4:4:4:4 (4 red/long: 4 red/short:

4 black/long: 4 black/short

• 2. Green frogs are dominant over brown frogs, and spots are dominant over no spots. Cross a heterozygous green spotted female with the same type of male. –Genotypes: –Gametes: GS, Gs, gS, gs & GS, Gs, gS, gs

GgSs & GgSs

• Genotypic ratio:• Phenotypic ratio:

GS

Gs

gS

gs

GS Gs gS gsGGSS

GGSs

GgSS

GgSs

GGSs

GGss

GgSs

Ggss

GgSS

GgSs

ggSS

ggSs

GgSs

Ggss

ggSs

ggss

1:2:2:4:1:2:1:2:1 9:3:3:1

Curi Family Eye Color & Ear Shape• My father has green eyes and free-hanging ear

lobes (homo or hetero?) while my mother has brown eyes (heterozygous) and free-hanging ear lobes (homo or hetero?). What are the possible outcomes for the children?–Know that my mother is heterozygous for brown

eyes since my brother has green eyes–What about free hanging ear lobes?• I have free hanging, but my brother and sister are

attached! What does that mean about my parents?–Both parents MUST be heterozygous for free-

hanging ears!

• Genotypes:• Gametes:

bF

bF

bf

bf

BF bfBf bFBbFF

BbFf

BbFF

BbFf

BbFf

Bbff

BbFf

Bbff

bbff

bbFF

bbFfbbFF

bbFf

bbFf

bbFf

bbff

??

?

???

BRO

SIS

me

bF, bf, bF, bf & BF, Bf, bF, bfbbFf & BbFf

• Getting ratios of genotypes & phenotypes is actually calculating probability– Probability: likelihood that particular event

(genotype or phenotype) will occur– Calculated by dividing number of predicted

outcomes by number of total outcomes– Ex: 3 peas are yellow, 1 is green• Words: • Ratios: • Decimals: • Percentages: • Fractions:

F. Probability

¾ yellow, ¼ green (add to 4/4)

3 out of 4 are yellow3:1 yellow

0.75 yellow, 0.25 green (add up to 1.0)75% yellow, 25% green (add to 100%)