If the frequency of the M allele in the human MN blood group system is 0. 65 in

Question

If the frequency of the M allele in the human MN blood group system is 0. 65 in a population at equilibrium, then the frequency of the N allele must be 0. 04. If a recessive disease is found in 50 out of 100. 000 individuals, what is the frequency of the heterozygote carriers for this disease? In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds?

Explanation / Answer

Part A

FALSE

According to the Hardy-Weinberg equillibrium the sum of the frequencies of alleles must be one.

Hence, the frequency of the allele N must be

1 - 0.65 + 0.35

Therefore, the given statement is false.

Part B

The correct answer is C = 0.043

Frequency of affected individual in the population =1/2000 means recessive homozygote

Recessive homozygote---------q^2 = 50/100000

                                                          = 0.0005

Hence, the frequency of Recessive allele q = 0.022

Normal allele p frquency = 1 - 0.022 = 0.978

carrier = 2pq = 2 x 0.97 x 0.022 = 0.04268

                                                    = 0.043

Part C

The correct option is Disruptive.

This selection involves the selection of homozygous types. The disruptive selection eventually divides the big population into two separate ones. One population will have small beaks and the other has large beaks.

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