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Mendelian Genetics

Transcript of Mendelian Genetics - classpages.warnerpacific.educlasspages.warnerpacific.edu/bdupriest/BIO...

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Mendelian Genetics

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Johann Gregor Mendel

*

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Monohybrid Crosses

• Breed a “true-breeding” plant

with a particular trait with a

true-breeding plant with the

contrasting trait

• Example: Breed plant with

purple flowers with a plant

with white flowers

• Not an example: Breed plant

with purple flowers with plant

with round seeds

Fig. 3-1

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Fig. 3-1

True-breeding

Round seed True-breeding

Wrinkled seed

Round seed

Is it true-breeding?

How to find out?

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Mendel’s First 3 Postulates

• “Unit factors” occur in pairs

• Dominance vs. recessiveness

• Segregation of unit factors during

gamete formation

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Figure 3-2

(Why do we use “D” or “d” for trait forms in this example?)

(What do pink boxes represent? Blue circles?)

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Monohybrid cross: 2nd look

Punnett Square

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Monohybrid cross approach

• 1) Determine genotype of each parent

• 2) Determine possible gametes produced

by each parent

• 3) Determine all possible combinations of

gametes genotypes of offspring

• 4) Determine frequency of genotypes and

phenotypes of offspring

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Practice problems

Consider this P1 cross: A true-breeding plant

with yellow seeds is crossed with a true-

breeding plant with green seeds.

What are the genotypes and gametes of the

parents?

What are the genotypes and gametes of the

F1 offspring?

What are the genotype and phenotype ratios

of the F2 generation?

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Practice problems After breeding true-breeding black cats with

true-breeding white cats, all offspring were

black. After breeding all those black cats

(F1), some of the resulting offspring were

white, and some were black. Assuming

standard Mendelian inheritance patterns,

and assuming 32 F2 cats were born, how

many would you expect to be

white? How many would you

expect to be black?

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Practice problems

A scientist bred two fruitflies together. They

appeared to be normal. 245 of the offspring

had no wings; 722 of the offspring appeared

normal. What were the likely genotypes of

the two fruitflies that were bred?

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How do you know if an individual with a

dominant phenotype is heterozygous or

homozygous?

How do you know if an individual that

appears normal is true-breeding or not?

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Testcross Analysis

• Testcross analysis allows geneticists to

determine whether dominant phenotype is due

to homozygous “AA” or heterozygous “Aa”

genotype

• Genetic cross is performed:

Unknown dominant x recessive

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Practice problems

A dog breeder crossed a big dog with a

small dog. Assume smallness is recessive in

this case. If the big dog is heterozygous,

what fraction of offspring would be big?

If the big dog is homozygous, what fraction

of offspring would be big?

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Practice problems

A dark brown ascomycote fungus was bred

with a light tan ascomycote. Assume light

tan is recessive, and dark brown is

dominant. If half of the resulting yeast are

dark brown, what is the genotype of each

parent?

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Mendel’s First 3 Postulates

• “Unit factors” occur in pairs

– “Unit factors” are called ____________

• Dominance vs. recessiveness

– Dominant traits are expressed in

heterozygotes; recessive traits are

expressed only in homozygous recessive

individuals

• Segregation of unit factors during

gamete formation

– Each gamete contains only one copy of

each gene/chromosome

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What happens if you consider more

than one character at a time?

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Law of Independent Assortment

• Unit factors for different traits segregate

independently from each other during gamete

formation

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Figure 3-5

Phenotypes

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Figure 3-7

Phenotypes: A slow way to

predict outcomes

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Figure 3-6

Phenotypes: A fast way to

predict outcomes

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Practice problems

Consider this P1 cross: A true-breeding plant

with purple flowers and axial buds is crossed

with a true-breeding plant with white flowers

and terminal buds.

What are the genotypes and gametes of the

parents?

What are the genotypes and gametes of the

F1 offspring?

What are the genotype and phenotype ratios

of the F2 generation?

Page 23: Mendelian Genetics - classpages.warnerpacific.educlasspages.warnerpacific.edu/bdupriest/BIO 250... · Mendelian Probabilities • Cross outcomes follow the multiplication rule –

Practice problems

A fruitfly F2 generation includes 95 flies

with brown eyes and normal (wild-type)

wings, 91 flies with red eyes and no wings, 280

flies with red eyes and wild-type wings, and 35

flies with brown eyes and no wings.

Calculate the ratios (decimal form) of these

numbers, then estimate the fractional equivalent.

Do these fractions line up with what you expect for

a dihybrid cross?

Which traits are dominant, which recessive?

What are the possible parental phenotypes?

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Trihybrid Genetic Cross • Trihybrid cross= three pairs of traits that assort

independently, such as AaBbCc

– For any one pair, phenotypic ratio = 3:1

– For any two pairs, phenotypic ratio = 9:3:3:1

• Trihybrid cross pattern of segregation and

independent assortment is identical to dihybrid

Phenotype Ratios: 27:9:9:9:3:3:3:1

Could theoretically be carried out for as many traits

as you want to study

Number of phenotype categories = 2x ,

where x = number of traits considered, and

when each trait has only two possible forms

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Mendelian Probabilities

• Using mathematical relationships to make

predictions is faster and allows for more possible

combinations than diagramming

• Only two rules to keep in mind

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Mendelian Probabilities

• Gamete formation follows Addition Rule

– Use when inputs are mutually exclusive (either-or)

• GG produces G gametes only

probability of producing G gamete = 1

• Gg produces G or g gametes (1:1)

probability of producing G gamete = ½

probability of producing g gamete = ½

probability of producing G or g gamete: (½ + ½ = 1)

• gg produces g gametes only

probability = 1

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Mendelian Probabilities

• Cross outcomes follow the multiplication rule

– Use when inputs are independent of each other

• GG X gg = Gg

probability of 1st parent passing on G = 1

probability of 2nd parent passing on g = 1, so…

overall probability of offspring Gg phenotype is

1 X 1 = 1

Page 28: Mendelian Genetics - classpages.warnerpacific.educlasspages.warnerpacific.edu/bdupriest/BIO 250... · Mendelian Probabilities • Cross outcomes follow the multiplication rule –

Mendelian Probabilities

• Cross outcomes follow the multiplication rule

– Use when inputs are independent of each other

• GG X Gg = ½ GG + ½ Gg

probability of 1st parent passing on G = 1

probability of 2nd parent passing on G = ½

probability of 2nd parent passing on g = ½ , so…

overall probability of offspring GG phenotype is

1 X ½ = ½ and

overall probability of offspring Gg phenotype is

1 X ½ = ½

Page 29: Mendelian Genetics - classpages.warnerpacific.educlasspages.warnerpacific.edu/bdupriest/BIO 250... · Mendelian Probabilities • Cross outcomes follow the multiplication rule –

Mendelian Probabilities

• Cross outcomes follow the multiplication rule

– Use when inputs are independent of each other

• Gg X Gg = ¼ GG + ½ Gg + ¼ gg

probability of 1st parent passing on G= ½

probability of 1st parent passing on g = ½

probability of 2nd parent passing on G = ½

probability of 2nd parent passing on g = ½ , so…

overall probability of offspring GG phenotype is

½ X ½ = ¼ and

overall probability of offspring gg phenotype is

½ X ½ = ¼ and…

Page 30: Mendelian Genetics - classpages.warnerpacific.educlasspages.warnerpacific.edu/bdupriest/BIO 250... · Mendelian Probabilities • Cross outcomes follow the multiplication rule –

Mendelian Probabilities

• Cross outcomes follow the multiplication rule

– Use when inputs are independent of each other

• Gg X Gg = ¼ GG + ½ Gg + ¼ gg

probability of 1st parent passing on G= ½

probability of 1st parent passing on g = ½

probability of 2nd parent passing on G = ½

probability of 2nd parent passing on g = ½ , so…

overall probability of offspring Gg phenotype is

½ X ½ = ¼ and

overall probability of offspring gG phenotype is

½ X ½ = ¼

but Gg is same as gG, so ¼ + ¼ = ½

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Probability Rules

• Addition Rule: The probability of obtaining one

or the other of two mutually exclusive events is

the sum of their individual probabilities

• Multiplication Rule: The probability of two

independent events occurring simultaneously

equals the product of their individual

probabilities

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Practice problems

Albinism (lack of melanin) is recessive.

What is the probability of an albino parent passing

on an “A” allele? Probability for the “a” allele?

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Practice problems

In pigs, drooping ears are recessive, while erect

ears are dominant. If a litter of piglets includes one

or more – but not all – with droopy ears, what does

that mean for the parents’ genotype?

If you cross a pig with erect ears with a pig with

droopy ears, what is the probability of obtaining a

piglet with erect ears? Droopy ears?

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Pedigree Analysis

• Purposes:

– To determine individual genotypes in human

families (low n)

– To predict the mode of transmission of single-

gene traits

• E.g., Huntington’s Disease, a dominant

progressive neurodegenerative disorder

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Pedigree Analysis: Dominance

Familial Inheritance of Huntington Disease

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Inheritance of Recessive Genes

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Practice Problems

• What is the probability that a 4th child born in

generation IV to the married cousins would

have the recessive phenotype?

• If individual 3 in generation IV mated with a

heterozygote, what is the probability the

offspring would have the

recessive phenotype?

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Practice Problems

• If individual 1 in generation I mated with an

individual showing the recessive phenotype,

what is the probability the offspring would have

the recessive phenotype?