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Genetic Variation, Genotypes, and Disease

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1 Define the term "genetic variation."

If a gene or locus has two alleles (A and a) in a population, what are all the possible genotypes?

If the frequency of (A) allele = p, and the frequency of (a) allele is q, what is the frequency of all possible genotypes in this population?

2 What is the Hardy-Weinberg equilibrium?

In a population, a locus A has two alleles (A) and (a). The frequency f of (A) is f (A) = 0.6; what is the f (a)?

Using these frequencies, calculate the frequencies of all possible genotypes in a population in Hardy-Weinberg equilibrium.

3 In a population (Z), the frequencies of genotypes of a two allele locus (B and b) are f(BB) = 0.3, f(bb) = 0.6, f (Bb) = 0.1. Calculate the frequencies of both alleles.

Using the allele frequencies calculated in a., calculate the frequencies of all possible genotypes at this locus in a population after one generation of random mating.

Is the population (Z) in part a above in Hardy-Weinberg equilibrium? The population size is 1000.

4 Describe how you would use the Hardy-Weinberg equilibrium to calculate genotype frequencies of a locus with three alleles.
If the f(D) of an X-linked gene in a population = 0.8, what is the f(d) of the other allele at this locus?

What is the frequency of females' homozygous for d allele (XdXd)?

What is the frequency of males' hemizygous for the d allele (XdY)?

5 Explain why more males present with X-linked recessive diseases. Show your reasoning.

Why must the five assumptions/criteria apply to a population before we can say it is in Hardy-Weinberg equilibrium?

Take any one of these assumptions and explain in detail how it could disrupt the Hardy-Weinberg equilibrium for a particular gene locus.

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1. a. Define the term "genetic variation."
Genetic variation refers to the range of phenotypes caused by different genotypes. This variation comes from mutations in the genetic code, gene flow, nonrandom mating, natural selection, and assortment that occurs during meiosis.
Web reference:
http://evolution.berkeley.edu/evosite/evo101/IIICGeneticvariation.shtml

b. If a gene or locus has two alleles (A and a) in a population, what are all the possible genotypes?
AA, Aa, aa

c. If the frequency of (A) allele = p, and the frequency of (a) allele is q, what is the frequency of all possible genotypes in this population?
AA = p squared (p^2)
Aa = 2pq
aa = q^2

2. a. What is the Hardy-Weinberg equilibrium?
This equation allows to predict the frequency of genotypes and phenotypes within a population that conforms to the assumptions of the equation, which include no mutation, gene flow, genetic drift, nonrandom mating, or natural selection. Hardy-Weinberg states that the sum of allele frequency is equal to 1 (p+q=1) and the sum of the genotype frequencies also equals 1 (p^2+ 2pq + q^2 = 1). Additionally, these frequencies do not change in subsequent generations as long as the assumptions of the equilibrium are met.

Web reference:
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/Hardy_Weinberg.html
http://www.tiem.utk.edu/~gross/bioed/bealsmodules/hardy-weinberg.html

b. In a population, a locus A has two alleles (A) and (a). The frequency f of (A) is f (A) = 0.6; what is the f (a)?
Remember p+q=1 Therefore, if A=0.6 then a =1-0.6=0.4

c. Using these frequencies, calculate the frequencies of all possible genotypes in a population in Hardy-Weinberg equilibrium.
AA = p^2 = (0.6)^2 = 0.36
Aa = 2pq = 2(0.6)(0.4) = 2(0.24) = 0.48
q^2 = (0.4)^2 = 0.16

Check that p^2+ 2pq + q^2 = 1, 0.36+0.48+0.16= 1

3. a. In a population (Z), the frequencies of genotypes of a two ...

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See Also This Related BrainMass Solution

Analyzing Genetic Disease Through Several Generations

In this SLP you are tasked with analyzing the transmission of a genetic disease through several generations. Males are indicated by square, females by circles. Individuals with the disease are colored solid black. Speculate on the genetic basis by which this disease is inherited. If any known carriers (those carrying the gene, but not manifesting symptoms of the disease) can be identified, list the individuals.

(see the attached file for diagrams)

Hypothesis

Based on the information observed and collected from the pedigree above, the above pedigree is an example of a X-linked recessive rare conditions in human. The recessive allele a is the rare allele that caused the recessive disease phenotype. It is carried by the X chromosome of one of the parents, and is called the X-linked recessive allele. The X-linked recessive allele is passed down to only the sons of the parents , which further proved that the condition is an X-linked recessive condition. The X-linked recessive conditions here could be anything from hemophilia or color blindness. Both conditions are classical X-linked recessive conditions.

Methods and Deductive Reasoning:
Let say a = the recessive allele that cause the rare diseases hemophilia
Let say A = the dominant normal allele
The recessive allele a is on the X chromosome. The parents Martha and Ian must be carrier of the recessive allele a that cause hemophilia since both of them do not have the recessive conditions. From the mating of the parents, we can deduce the genotypes of the offsprings or descendants. Leela does not have the recessive conditions, but she must be a carrier of the recessive allele because her son have the recessive condition. Her genotype must be heterozygous for the recessive allele XAXa . Peri is a male that is not affected by the disease. He has a normal phenotype so he must have the genotype XAY. Steven has the rare recessive disorder so his genotype must be XaY. Both Sarah Jane and Donner do not have the recessive conditions, although they can be carrier. Their genotypes must be either XAXa or XAXA . Since Donner mated with Mickey and their two sons has the recessive conditions, Donner must be heterozygous with genotype XAXa. Mickey is a male that do not show the recessive condition. Therefore, his genotype must be XAY. Sarah Jane did not mated with anyone. She did not have the recessive conditions; we can conclude that her genotype must be either XAXa or XAXA.
Since Peri and Leela mated and their son have the recessive conditions, but their daughter does not. Their genotype is appropriate. Harry, the son of both Leela (genotype XAXa ) and Peri ( genotype XAY ) must have the recessive allele on his X chromosome. Harry genotype is XaY, indicating positive for the recessive condition. The daughter of Leela and Peri must have the genotype XAXA indicating no recessive condition. When Donna and Mickey mated, their offsprings are three sons (two sons has the recessive conditions) and one daughter with no recessive conditions. Donner must be a carrier with genotype XAXa and Mickey must have the genotype XAY. Since both of their sons Ben and Mike have the recessive conditions, their genotype must be XaY . Jack is a male that does not show the recessive condition;his genotype must be XAY. Since both Ben and Amy (offsprings of Donna and Mickey) mated , their one offspring daughter Rose has the recessive condition. Amy, the mother must be the heterozygous carrier with genotype XAXa . Ben, which has the recessive conditions, must have the genotype XaY. Their daughter Rose must be homozygous for the recessive allele and must have the genotype XaXa .

Conclusion
This pedigree is a classic example showing the inheritance of the X-linked recessive condition from the first generation (parents ) to the second generations and subsequently to the third generations. The recessive allele here can be any disorder that happen to be linked to the X chromosome. Here we decided to pick the recessive condition hemophilia, which is caused by a failure for blood to clot. The recessive allele arose from a spontaneous mutation in the protein Factor XIII that failed to function in blood clotting. The mutation in the allele a is either from the mother or the great grandparents of the mother that passed down to subsequent generations. The X-linked recessive allele condition can only be passed down through the X chromosome. The most important clues about the X-linked recessive inheritance is that both parents does not have to be affected, but are carrier of the X-linked recessive allele. The recessive allele on the X-linked chromosome is passed down and only affected male sons because males only have one X chromosome. The daughter can be heterozygous carrier and not be affected with the recessive condition because they have another X dominant chromosome that dominate over the expression of the recessive trait. Males cannot be carrier of the X-linked recessive trait; they can either do not have the recessive allele or be affected with the recessive allele. Females can be heterozygous carrier of the recessive alleles and not get the phenotype. Females can also be homozygous carrying only the dominant unaffected allele. The only scenario where the daughter can get the recessive conditions is when the mother is a carrier with genotype XAXa and the father has the recessive allele with genotype XaY. The daughter may get the Xa allele from the mother and the Xa allele from the affected father. The affected daughter genotype is XaXa as you can see in the mating of the two parents Ben and Amy with their affected daughter Rose.

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