Basic Mendelian Crosses

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Basic Mendelian Crosses

Purebreeding Parents

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Testcross

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F1 x F1 Cross

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Forked Line (Branch) Diagrams

Gametes possible from AaBbccDd individual

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Phenotypes possible from cross: AaBbCcDd x AaBbCcDd

p(A- B- C- D-) = 81/256

P(A- bb C- dd) = 9/256

Forked-Line (Branch) Diagrams

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Forked-Line (Branch) Diagrams

Genotypes possible from cross Aa Bb CC x Aa bb cc

1/2 Bb 1/1 Cc1/4 AA

1/2 bb 1/1 Cc p (AA bb Cc)?

1/2 Bb 1/1 Cc1/2 Aa

1/2 bb 1/1 Cc p (aa bb Cc)?

1/2 Bb 1/1 Cc1/4 aa

1/2 bb 1/1 Cc

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Mathematical Method

Answers specific question, not all inclusive

Individual with genotype Aa Bb cc Dd

What is the probability of gamete with abcd ?

p (a) = 1/2 p (b) = 1/2 p (c) = 1 p (d) = 1/2

p (a) and (b) and (c) and (d) = 1/2 * 1/2 * 1 * 1/2 = 1/8

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Mathematical Method

Cross between Aa Bb Cc x Aa Bb Cc

What is the probability of offspring with genotype:

Aa BB cc?

p (Aa) = 1/2 p (BB) = 1/4 p (cc) = 1/4

p (Aa) and (BB) and (cc) = 1/2 * 1/4 * 1/4 = 1/32

AA BB Cc?

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Mathematical Method

Cross between Aa Bb Cc Dd EE x aa Bb CC dd Ee

p (aa bb CC dd Ee)?

p (Aa Bb Cc Dd Ee)?

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Mathematical Method

When ratios for all genes are consistent:

Ex. Aa Bb Cc x Aa Bb Cc

# Phenotypes = (2)n 2 = # phenotypes for each gene n = # monohybrid crosses (genes)

Ex. (2)3 = 8

# Genotypes = (3)n 3 = # genotypes for each genen = # monohybrid crosses (genes)

Ex. (3)3 = 27

Calculating Number of Genotypes Possible

# genotypes possible = 3 n

3 = Genotypes possible for each gene - AA, Aa, aa

n = # heterozygous gene pairs

Calculating Number of Gametes Possible

# gametes possible = 2 n

2 = diploid with 2 copies of each gene/chromosome

n = # heterozygous gene pairs

Ex. AaBb : 2 2 = 4 possible (AB, Ab, aB, ab)

Ex. AaBbCc : 2 3 = 8 possible (ABC, ABc, Abc, etc.)

Human chromosomes : 2 23 = > 8 x 10 6

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Binomial Expansion: Uses

Predict comprehensive phenotypic ratios ( 9:3:3:1, etc.)

Determine probability of particular categories

Only applicable if:

Ratio for every trait (gene) is the same (ex. AaBb x AaBb - 3:1)

Not applicable if:

Ratios vary (ex. AaBb x Aabb - 3:1 for A-:aa; 1:1 for B-:bb)

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Binomial Expansion: Generating Comprehensive Ratios

Probabilities: p = dominant phenotype, q = recessive phenotype

N = # genes X = # of dominant (N-X) = # of recessives

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Example based on phenotypes from AaBb x AaBb cross

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A-B-: A-bb: aaB-: aabb

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Example for AaBbCcDdEe x AaBbCcDdEe cross

(p+q)5 = p5 + 5 p4q + 10 p3q2 + 10 p2q3 + 5 pq4 + q5

5! = 5 * 4 * 3 * 2 * 1 = 4! 1! (4*3*2*1) * 1

5! = 5* 4 * 3 * 2 * 1 = 3! 2! (3 *2*1) * (2*1)

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Binomial Expansion: Ways to Determine Coefficients

Pascal’s Pyramid

p q

p6 + p5q + p4q2 + p3q3 + p2q4 + pq5 + q6

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Binomial Expansion: Ways to Determine Coefficients

Direct method (shortcut):

Example: (p+q)4 =

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Binomial Expansion: Determining Phenotypic Ratios

Probability: p(A-B-C-) = 27/(43) = 27/64 p (A-B-cc) =p(two dominant and one recessive) =

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Binomial Expansion: Determining Phenotypic Ratios

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Binomial Expansion: Determining Probability

p = probability of one event (ex. girl)

q = probability of alternative event (ex. boy)

Probability that in N trials, you will get X girls and (N-X) boys

= N! (pX) (q)(N-X) Note: Only use coefficient X! (N-X)! when order is not considered

Example: probability of 2 girls and 4 boys in a family of 6?

6! (1/2)2 (1/2) 4 = 2! 4!

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Examples of Applying Binomial

Two brown-eyed parents mate and have a blue-eyed child.

What are the parents genotypes?

Which allele for color is dominant?

If two individuals with the same genotype had four children,what is the probability of them having all blue eyes?

What is the probability of them having two brown and two blue?

What is the probability of the first two being brown and the rest blue?

What is the probability of their fifth child having blue eyes?

In how many different orders could this occur?

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Sample Problem

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Answer to Sample Problem (Part 1)

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Answer to Sample Problem (Part 2)

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Pedigree Analysis

Symbols Used

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Pedigree Analysis Sample Pedigree

For rare conditions: assume those outside family are homozygous or hemizygous normal;

Genetic counselor:p (carrier in population)ex. CF carrier = 0.05

Arrow indicates propositus

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Pedigree for Autosomal Recessive Trait

Usually loss-of-function Albinism: absence of pigment in skin, eyes, hairCystic fibrosis: thick mucus that blocks lungs, glandsSickle cell anemia: abnormal hemoglobin, blockageXeroderma pigmentosum: No nucleotide excision repair

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Things to Look for with Autosomal Recessive Traits

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Pedigree for Autosomal Dominant Trait

Insufficient product, interference with normal, or gain-of-function Achondroplasia: dwarfism, defect in long bone growthBrachydactyly: shortened fingersHypercholesterolemia: high cholesterol, heart disease Huntington disease: nervous system degeneration

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Things to Look for with Autosomal Dominant Traits

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Are these autosomal traits dominant or recessive?

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Pedigrees for X- linked Traits

Recessive Color blindness: insensitivity to red or green lightHemophilia: defective clotting, A or B typeMuscular dystrophy: Duschenne type, muscle wasting

DominantHypophosphatemia: rickets (bowlegged)

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What to look for with X- linked Traits

Dominant Recessive Hemizygous

XAXA or XAXa XaXa XAY XaY

Recessive Dominant

Affected daughterAffected father

has affected fatherpasses it on to

and carrier motherall daughters

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What to look for with X- linked Recessive Traits

Mother can carry and pass on the recessive alleleMore males express these traitsMother to son inheritanceAffected daughters must have affected fathersCriss-cross pattern of inheritance

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Sample X- linked Recessive Pedigree

X = normal, X = abnormal

Males: 1/2 normal, 1/2 affected

Females: 1/2 normal, 1/2 carriers

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Sample X- linked Recessive Pedigree

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What to look for with X- linked Dominant Traits

Affected mother will pass it on to 50% both daughters and sonsAffected father will only pass it on to all daughters

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Sample X- linked Dominant Pedigree

X = normal, X = abnormal

Overall: 1/2 affected, 1/2 normal

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Y- linked (Holandric) Pedigrees

Traits are only seen in males and passed on from father to son.

X = normal, Y = abnormal

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Sample Pedigree Problem

Adherent Earlobes (recessive autosomal)

Genotypes: II 3 = I 1 and 2 = p II 1 is Aa = p III 2 is Aa = p III 2 and III 4 would have aa child =

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Huntington Disease (same family, later date)

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Autosomal trait (rare recessive)

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Sample Pedigree Problem from Lab

Probability of carrier? 1-4 8 9 1516 17

If 16 married 17, what is the probability of an affected child?

I

II

III

Documents

Basic Mendelian Crosses