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Section 10.1 Mendel’s Laws of Heredity pg. 253

Section 10.1 Mendel’s Laws of Heredity pg. 253. WHY MENDEL SUCCEEDED Gregor Mendel (1822-1884) Austrian Monk known as the “father of modern genetics”

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Section 10.1

Mendel’s Laws of Heredity pg. 253

WHY MENDEL SUCCEEDEDGregor Mendel (1822-1884)Austrian Monk known as the “father of modern genetics”Found that inheritance follows certain laws later known as Mendel’s Laws of InheritanceHeredity is the passing on of characteristics from parent to offspring

Inherited characteristics are called traits (factors)

The branch of biology that studies heredity is called genetics – which was founded upon the rediscovery of his work.

WHY MENDEL SUCCEEDEDMendel’s pea plant collection contained around 28,000 plants

Pea plants reproduce sexually by producing male and female sex cells (gametes)

Male gamete forms in pollen

Female gamete formed in ovary

Fertilization occurs when the male gamete unites with the female gamete

Pollination is the transfer of pollen to ovary in a plant – normal reproduction. Remove male parts

Snipping the stamen could prevent self-pollination

Flower parts

Quickly sketch and label this diagram

A similar figure is in the book on pg. 642

WHY MENDEL SUCCEEDEDCross-pollination is transferring pollen of one plant to the ovary of another plant• Mendel did this to get certain plants to breed with others to be sure of the parents• Mendel was very careful with all of his work

Figure 10.1

Mendel was able to create tall plants and short plants (purebreds)• He referred to the offspring of a purebred tall and a purebred short as a hybrid• Crossing a 6’ tall plant with a 2’ tall (short) plant resulted in all 6’ tall plants• Crossing the hybrid offspring resulted in 75% tall and 25% short

• P1 refers to the “Parental generation”

• F1 (“Filial”) refers to the offspring

Tall pea plant

All tall pea plants

3 tall: 1 short

P1

MENDEL’S MONOHYBRID CROSSES

F1

F2

Figure 10.2

Short pea plant

MENDEL’S MONOHYBRID CROSSESSo what does MONOHYBRID refer to?Referring to figure 10.3:• When Mendel crossed a purebred tall with a purebred short he got all tall plants• When he crossed a purebred purple flower with a purebred white flower he got all purple flowers• He referred to the trait that was observed in these cases as dominant• The trait that seemed to “disappear” he called recessive•Law of Dominance• Mendel concluded that these plants have “factors” that control each of the traits (color, shape, height)• We call these factors genes (parts of DNA)• Alternative forms of genes (tall vs. short or yellow vs. green) are known as alleles.

Recessive trait

Dominant trait

Seed shape

Seed color

Flower color

Flower position

Pod color

Pod shape

Plant height

round yellow purpleaxial (side) green inflated tall

wrinkled green whiteterminal

(tips) yellow constricted short

Figure 10.3

Below is a list of human traits, write them down.

Widow’s peak Tongue Rolling

Dimples Mid-digit Hair

Free Ear Lobes Straight Thumb

Freckles Bent Pinky Finger

Are they dominant or recessive? (Take a guess)

HUMAN TRAITS

TT

MENDEL’S MONOHYBRID CROSSES

These two alleles for each trait can be expressed as a single letter

For plant height we can use the letters “T” & “t”

Dominant allele is “T”

Recessive is “t”

Mendel’s purebred tall plants were “TT”

His purebred recessive plants were “tt”

Fill in the blanks in the figure to the right which t goes where?

Tall plant Short plant

All tall plants

F1

TT t t tT

P1

TT

MENDEL’S MONOHYBRID CROSSES

Mendel concluded that the allele (gene form) of tall plants was dominant to the allele for short plants

Confirming that the plants had two alleles for each trait (TT = Tall, Tt = Tall, or tt = short)

Knowing that traits are inherited from parents, he also concluded that these alleles are inherited

However a plant can only get one allele from each parent

The gametes (sex cells) contained either one or the other form of the gene (T or t)

The Law of Segregation states that every individual has two alleles of each gene and when gametes are produced, each gamete receives one of these alleles.

PHENOTYPES AND GENOTYPES

Phenotype refers to the organism’s physical characteristic (what you can see) Ex: Tall

Genotype refers to the organism’s genetic makeup (what you can’t see) Ex: TT or Tt

Homozygous represents two alleles that are the same (TT or tt)

Heterozygous organisms have different alleles (Tt)

How are we going to distinguish Homo and Hetero?

PUNNETT SQUARES1905 - Reginald Punnett devised an easy way to find expected genotype proportions of offspring from known parent genotypes based off Mendel’s laws

Heterozygous tall parent

T

T t

T

t

t

TT Tt

Tt tt

Heterozygous tall parent

tT

t

t tT

T

T

G= 50% Tt : 25% TT : 25% tt

P= 3 Tall : 1 short

Monohybrid Cross (one trait)

PUNNETT SQUARESMonohybrid crosses are easy to separate alleles according to Mendel’s Law of Segregation

If we have heterozygous parents (Tt X Tt) we can just separate the T from the t

For Dihybrid crosses, the gamete separation is a little tricky

If we have two parents that are heterozygous for seed shape (Rr) and seed color (Yy) their genotype is RrYy

To separate alleles into gametes we use the FOIL method from algebra

RrYy makes four different gametes

Using the FOIL method we get…RY Ry rY ry

Round Yellow

Round green

wrinkledYellow

wrinkledgreen

MENDEL’S DIHYBRID CROSSESMendel also crossed plants with two different traitsRound=R, wrinkled=r & Yellow=Y, green=yWhat is the genotype of a purebred (homozygous) plant with Round Yellow seeds?What is the genotype of a purebred (homozygous) plant with wrinkled green seeds? Purebred (homozygous) Round Yellow seeds X Purebred (homozygous) wrinkled green seeds

Result of F1…All plants had Round Yellow seeds

However crossing the F1 gives a 9:3:3:1 ratioWhich leads us to Mendel’s second law…The Law of Independent Assortment states that genes for different traits are inherited independently of each other.

RRYY

rryy

Round Yellow (RRYY) X wrinkled green (rryy)

Round Yellow wrinkled green

All Round Yellow

Round Yellow Round green wrinkled Yellow wrinkled green

9 3 3 1

P1

F1

F2

R_Y_ R_yy rrY_rryy

Gametes from RrYy parent

RY Ry ryrY

RY

Ry

rY

ry

RRYY

RRYy

RrYY RrYyRRYy

Rryy

RrYY

Rryy

RrYy

RrYy rrYY rrYy

RrYy

RRyy

rrYy rryy

Gametes from RrYy parent

Starting here what are the gametes?

PROBABILITYKnowing the parents genotype we can predict the probable offspring genotype and phenotype

What is the probability of having Rr offspring?

What is the probability of having Round offspring?

R r

R

r

RR Rr

Rr rr

75%

50%

PROBABILITYGiven the parents genotype and number of offspring, you should be able to predict the number of each genotype and phenotype.

PROBABILITY PROBLEMR=Round seeds & r=wrinkled seeds

1. P1 genotype: RR X rr

2. All of the F1 offspring will be ______.

3. Assume 140 F2 offspring are created from F1.

4. ________ will have their parents (F1) genotype.5. ________ will have Round seeds.6. ________ will have wrinkled seeds.

7. ________ will have the same genotype as the P1.

PROBABILITY PROBLEM (#2)

Rr (Round)

2. All of the F1 offspring will be ______.

P1=RR X rr (always put first parent on top of square)R R

r

r

Rr Rr

Rr Rr

2. All of the F1 offspring will be ___________.

PROBABILITY PROBLEM (#4)

70

4. ________ will have their parents (F1) genotype.

F1= All RrR r

R

r

RR Rr

Rr rr

4. ________ will have their parents (F1) genotype.

# of F2 Offspring = _____

Expected % of genotype (Rr) that is same as parents= ______

50% (2/4) of 140 = _____

140 X .5 = 70

140

50%70

PROBABILITY PROBLEM (#5)

105

5. ________ will have Round seeds.

R r

R

r

RR Rr

Rr rr

5. ________ will have Round seeds.

# of F2 Offspring = _____

Expected % of Round phenotype = ______

75% (3/4) of 140 = _____

140 X .75 = 105

140

75%105

PROBABILITY PROBLEM (#6)

35

6. ________ will have wrinkled seeds.

R r

R

r

RR Rr

Rr rr

6. ________ will have wrinkled seeds.

# of F2 Offspring = _____

Expected % of wrinkled phenotype = ______

25% (1/4) of 140 = _____

140 X .25 = 35

140

25%35

PROBABILITY PROBLEM (#7)

70

7. ________ will have the same genotype as the P1 (RR or rr).

R r

R

r

RR Rr

Rr rr

7. ________ will have the same genotype as the P1.

# of F2 Offspring = _____

Expected % of RR or rr genotype = ______

50% (2/4) of 140 = _____

140

50%70

Actual Results of Mendel’s Work