Question 1. In Mendelian genetics, the focus is on allele transmission in one pa
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Question 1. In Mendelian genetics, the focus is on allele transmission in one pair of mating individuals. Population genetics, by contrast, looks at the collective interaction of a large number of mating pairs. Having a clear grasp of the distinction between these is very important for thinking about evolutionary problems. For a single mating pair, almost all students know the strategy for predicting offspring genotypes. For example, the rules of Mendelism predict that a mating of two heterozygous parents (Aa x Aa) will yield these offspring proportions:
1/4 AA, 1/2 Aa, and 1/4 aa
But what proportions of offspring genotypes are expected if we are analyzing mating in a population of 10,000,000 parents, instead of only two parents? Suppose that this population had 4,000,000 individuals with genotype "aa" and 6,000,000 with genotype "Aa", and that each individual chooses a mate completely at random. What is the expected proportion of offspring genotypes that will result?
One standard way to approach this problem would be to:
1. calculate the frequency of allele "A" (= p) and allele "a" (= q) among the parents
2. then apply the Hardy-Weinberg law to predict the relative number of offspring genotypes: (p2, 2pq, q2)
This approach works perfectly well, but the justification for this procedure may not be very clear. So, rather than using this method, please solve this problem by the following alternative method. This second approach takes longer, but it may help to build intuition about what happens when a population of many individuals mates at random. In brief, we enumerate what kinds of mating pairs can form, then we census the types of offspring that are produced by each kind of mating pair.
Step 1: List all the possible mating combinations that can take place. For each female parent in a mating pair, there are two possible genotypes ("aa" or "Aa"). For each of these, two possible male partner genotypes could be chosen. Thus, in this entire population, there are four possible mating combinations ("mating types"). List these. (note that if the parent population had had all three genotypes (AA, Aa, aa) present instead of only two genotypes, there would be more mating types to consider)
Step 2: Determine how common each of the four mating types will be. Let "i" be the relative abundance in the population for one parent genotype in a mating pair and "j" be the frequency associated with the other parent's genotype. If mating occurs at random in the whole population, a proportion "i x j" of all mating pairs is expected to take place between these two genotypes. Calculate the relative proportion for each of the four mating types identified in Step 1. Check your work by verifying that the sum of proportions for all four of the possible mating types must equal 1.0. In other words, every mating that takes place must fall into one of four categories, and the sum of frequencies for all four must be 1.0
Step 3: Each of the four different mating types will result in a particular set of offspring genotypes. Use the standard rules of Mendelism to analyze the offspring that will be produced by a single mating pair of each type. Make a table where rows=the four mating types, and columns= the three offspring genotypes. All 12 entries in the table will be from the set {0, 1/4, 1/2, 1}
Step 4: Think of the three possible offspring genotypes (AA, Aa, or aa) one at a time. One of these is produced by only one of the four mating types, and the others are produced by multiple different mating types. Find which matings produce each offspring genotype. For the whole population, the relative number of offspring belonging to a certain genotype can be inferred from the table in Step 3 if entries in each row are weighted by the frequency of that mating type (Step 2). Calculate the overall proportion for each offspring genotype by combining across all the mating types. Check your calculation by verifying that the sum of proportions for all three offspring genotypes is equal to 1.0.
Final Step: compare your prediction about the composition of offspring genotypes in this population to what the Hardy-Weinberg formula would yield. Calculate allele frequencies (p, q) among the parents (4,000,000 aa and 6,000,000 Aa), and then find p2, 2pq, and q2. Steps 1-4 and the Hardy-Weinberg formula are two alternative ways of performing exactly the same calculation, so each analysis can serve as a check on whether the other approach was done correctly.
Explanation / Answer
The number of individuals with genotype aa = 4,000,000 out of 10,000,000. This represents 40% of the individuals are homozygous aa or 0.4 is the frequency of homozygous aa.
The number of individuals with genotype Aa = 6,000,000. It represents 60% of the population are heterozygous, the frequency is 0.6.
The following table shows the genotype of the parents and genotype frequencies of the offspring:
Genotype of Mother
Genotype of Father
Frequency
Genotype of offspring
aa
Aa
AA
aa
aa
0.4 × 0.4
0.16
aa
Aa
0.4 × 0.6
0.12*
0.12*
Aa
aa
0.6 × 0.4
0.12
0.12
Aa
Aa
0.6 × 0.6
0.09**
0.18**
0.09**
Total
0.49
0.42
0.09
The cross between aa and Aa individual produces offspring of aa and Aa at the ratio of 0.5 : 0.5
The cross between Aa and Aa individuals produces offspring of aa, Aa, and AA at ratio of 0.24: 0.5: 0.25 respectively
Therefore, from the above table, the offspring will have the genotypes aa, Aa, and AA at a ratio of 0.49, 0.42, and 0.09 respectively.
aa = 0.49, Aa = 0.42, AA = 0.09
Calculate the same using Hardy-Weinberg equilibrium equation:
The frequency of a allele (q)2 is 0.4. Therefore, the value of q = 0.4 = 0.633
According to the equation p + q = 1, p + 0.633 = 1, p = 0.367
The genotype frequencies are calculated as p2 + 2pq + q2 = 1
(0.367)2 + 2 (0.367) (0.633) + (o.633)2 = 1
0.14 + 0.40 + 0.46 = 1
Therefore, according to the Hardy-Weinberg formula, the incidence of genotypes in the offspring is as follows:
aa = 0.46, Aa = 0.40, AA = 0.14
The values, when compared with the alternative calculation, differ slightly, but the deviation is not too high.
Genotype of Mother
Genotype of Father
Frequency
Genotype of offspring
aa
Aa
AA
aa
aa
0.4 × 0.4
0.16
aa
Aa
0.4 × 0.6
0.12*
0.12*
Aa
aa
0.6 × 0.4
0.12
0.12
Aa
Aa
0.6 × 0.6
0.09**
0.18**
0.09**
Total
0.49
0.42
0.09
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