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Mendelian Genetics

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Gregor MendelThe Father of Genetics

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Gene

• A discrete unit of hereditary information consisting of a specific nucleotide sequence in DNA (or RNA in some viruses)

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Alleles• •Alleles are alternate forms of a gene.• •Examples: tall and short for plant height or purple

or white for flower color.• •Every trait has at least two alleles- one from each

parent.• •The location of an allele on a chromosome is

known as its locus (loci = plural form).

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Genotype

• •The letters that represent (symbolize) the trait being investigated. The genetic make-up of an organism.

• •Examples: Bb, BB, bb

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Phenotype• The actual representation of the genes.

The Physical appearance or traits in an organism resulting from its genetic makeup (what you see).

• Examples: tall, purple flower or white flower, blond hair, freckles, etc.

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Dominant

• The allele that is fully expressed in an organism (observed).

• •Represented by capital letters.

• •Tall = T

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Recessive

• •The allele that is masked by the dominant allele.

• •Represented by lower case letters.

• •Short = t

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Homozygous

• When both alleles (letters) are the same.

• BB = Homozygous Dominant

• bb = Homozygous recessive

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Heterozygous

• When the alleles (letters) are different.

• One upper case letter and one that is lower case.

• Bb = Heterozygous

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Example Problem• Round = R• wrinkled = r• If a plant has round seeds, do we

know what its genotype is?• It could be RR or Rr• If a plant has wrinkled seeds, do

we know what its genotype is?• Yes, it is rr.

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Punnett Square

• A method for finding predicted outcomes and probabilities for offspring from any cross.

• A chart for predicting the traits of offspring.

Some more terms:• P-generation is the parental generation.

• The p-generation produce the F1 generation.

• The F1 generation crossed with itself produces the F2 generation.

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Example Problem

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• In foxes, red coat color is determined by the dominant gene R; silver-black coat is determined by the recessive gene r. A homozygous (pure) red male is crossed with a silver-black female. (The P generation).

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1.What is the genotype of the female?

What are the genotype percentages of their offspring?

•First…make a Punnett square for showing your work

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Example Problem

r

r

R R

Rr

RrRr

Rr

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Assignment

• Section 32-3

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Assignment

33-4 & 33-5

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#1

b

b

B b

Bb

bbBb

bb

B = Brownb = blue

1 point

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#1

• 1/2 or 50% chance of blue-eyed.

• 1/2 or 50% chance of Brown-eyed.

1 point

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#2

t

t

T t

Tt

ttTt

tt

T = Tall

t = short

1 point

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#2 What fraction of offspring would be tall?

• 1/2 or 50% would be Tall.

1 point

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#3

w

W

W w

Ww

WwWW

ww

W = White

w = black

1 point

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#3a What fraction of the offspring will be white?

• 3/4 or 75% will be white.

1 point

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#3b What fraction of the offspring will be black?

• 1/4 or 25% will be black.

1 point

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#3c What fraction of each genotype will you get?

• 1/4 or 25% will be WW

• 1/2 or 50% will be Ww

• 1/4 or 25% will be ww.

1 point

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#3d What fraction of each phenotype will you get?

• 3/4 or 75% will be White.

• 1/4 or 25% will be black.

1 point

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#4

W

W

R R

RW

RWRW

RW

RR = RedWW = WhiteRW = Roan

1 point

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#4 Give the fraction of each color of offspring?

• 1/1 or 100% will be Roan - RW.

1 point

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#5

W

W

R W

RW

WWRW

WW

RR = RedWW = WhiteRW = Roan

1 point

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#5 Give the fraction of each color of offspring?

• 1/2 or 50% will be Roan - RW.

• 1/2 or 50% will be White - WW.

1 point

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#1a

g

G

G g

Gg

GgGG

gg

G = Green

g = red

1 point

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#1a Give the fraction of each genotype of

offspring?• 1/4 or 25% will be GG.

• 1/2 or 50% will be Gg.

• 1/4 or 25% will be gg.

1 point

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#1b

g

g

G g

Gg

ggGg

gg

G = Green

g = red

1 point

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#1b Give the fraction of each genotype of

offspring?

• 1/2 or 50% will be Gg.

• 1/2 or 50% will be gg.

1 point

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#2a What is the fraction of each phenotype?

• 3/4 or 75% will be Green.

• 1/4 or 25% will be red.

1 point

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#2b What is the fraction of each phenotype.

• 1/2 or 50% will be Green

• 1/2 or 50% will be red.

1 point

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# 3

•The mother had to be heterozygous or Bb since the couple had a blue eyed child.

b

B

b b

bb

BbBb

bb

1 point

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# 4

•There would be a 50% chance that the 2nd child from the couple would have a brown eyes.

1 point

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# 5• Spotted = S

• white = s

• Mother = ss since she is white.

• Father’s Genotype would be Ss.

• Father’s Phenotype would be Spotted

The couple had two spotted and two white kittens.

2 points

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# 6• Man has blue eyes - bb.

• G-Ma has blue eyes - bb.

• Woman has brown eyes - Bb.

• 50% of children would be Bb. (Brown)

• 50% of children would be bb. (Blue)

5 points

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Please put a score on top of their paper.

• Put the number correct out of 28

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Punnett Squares• Why are punnett squares useful?

• We can use a punnett square to predict the probable genotypes and phenotypes for offspring from a genetic cross.

• Genotype = What is inside the genes, the make-up.

• Phenotype = The outward expression of the genes.