Sickle Cell Anemia. Incomplete Dominance Sex linked traits

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Sickle Cell Anemia

Incomplete Dominance

Sex linked traits

Multiple Alleles All chicken have combs on their heads, but it does not always look the same. The comb is a fleshy growth on the top of the chicken's head. Both male and female chickens have combs, but the ones on the male are larger. Combs of different breeds may look different in shape and even in color. Chicken comb features:rrpp = singlerrP_ = peaR_pp = roseR_P_ = walnut 1. What must be the genotypes of the two parents for the outcome to always be a walnut offspring? (neither can be walnut to begin with) ____________ x _____________ = walnut (R_P_) 2. Show a Punnett square for the following cross and describe the phenotypic ratios. RrPp x RrPp 3. Show a punnett square for the following cross and describe the phenotypic ratios. rrpp x RrPp 4. Show a punnett square for the following cross and describe the phenotypic ratios. rrpp x rrPp

5. A rose crossed with a pea produces six walnut and five rose offspring. What must be the genotypes of the parents? Show the cross.

• Natural selection acts on individuals, but only populations evolve• Explain this statement

Hardy-Weinberg Principle

Hardy-Weinberg Principle

• Describes a non-evolving population• The gene pool does not change

– Gene pool - all alleles for all loci in the population

– Allele frequencies remain constant in non-evolving populations

• Why do this?

Allelic variation within a population can be modeled by the Hardy-Weinberg equations.

A Population in Hardy-Weinberg Equilibrium http://zoology.okstate.edu/zoo_lrc/biol1114/tutorials/Flash/life4e_15-6-OSU.swf

Fig. 23-6

The seven assumptions underlying Hardy–Weinberg equilibrium are as follows:

1. organisms are diploid2. only sexual reproduction occurs3. generations are non overlapping4. mating is random5. population size is infinitely large6. allele frequencies are equal in the sexes7. there is no migration, mutation or selection

Interpretation:A population with 80% dominant and 20% recessive

alleles that meets the conditions for Hardy-Weinberg

equilibriumwill pass 80% dominant and 20% recessive alleles to

the next generation

Frequencies of alleles

Alleles in the population

Gametes produced

Each egg: Each sperm:

80%chance

80%chance

20%chance

20%chance

q = frequency of

p = frequency of

CR allele = 0.8

CW allele = 0.2

Hardy-Weinberg equations:- p + q = 1- p2 + 2pq + q2 = 1

- Describes alleles in a gene pool- This is the equation for a trait with 2 alleles

- Can be used to predict genotypes- p – represents dominant allele- q – represents the recessive allele

• Worksheet 3, 4 and 5

SpermCR

(80%)

CW

(20 %

)

80% CR ( p = 0.8)

CW (20%)

20% CW (q = 0.2)

16% ( pq) CRCW

4% (q2) CW CW

CR

(80%

)

64% ( p2) CRCR

16% (qp) CRCW

Eg

gs

Worksheet: #6

Gametes of this generation:

64% CRCR, 32% CRCW, and 4% CWCW

64% CR   +    16% CR    =   80% CR = 0.8 = p

4% CW     +    16% CW   =   20% CW = 0.2 = q

64% CRCR, 32% CRCW, and 4% CWCW plants

Genotypes in the next generation:

Reality

• Hardy-Weinberg - hypothetical population

• In real populations, allele and genotype frequencies change over time

The winged trait is dominant.

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

A. What is the frequency of the "aa" genotype in the population?

B. What is the allele frequency of the "a" allele?

C. What is the frequency of the "A" allele in the population?

D. What is the frequencies of the genotype "Aa” in the population?

E. What is the frequencies of the two possible phenotypes if "A" is completely dominant over "a."