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EleoNora [17]
3 years ago
14

The fish Astyanax mexicanus has populations in surface lakes and in caves. Cave populations lose their eyes through degeneration

and are unpigmented, while surface populations have normal eyes and pigment. The two groups can form hybrids when mated in the laboratory, but rarely interact in the wild. This is an example of a species currently undergoing hybridization.
A: hybridization.
B: immigration.
C: allopatric speciation.
D: sympatric speciation.
Biology
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

Option C. allopatric speciation.

Explanation:

Allopatric speciation is a takes place when a population is isolated geographically for example as they are in wild, the fish astyanax mexicanus's two distinguished population of cave and lakes mate rarely. In laboratory they are capable of mating and form hybrid as mentioned in question.

It is occurs in same species due to geographic isolation that prevent the interference in the gene flow of one population to other.

Thus, the correct answer is: option C.

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Convert 16.45 m into Kilometers
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The human MN blood type is determined by two codominant alleles, L(M) and L(N). The frequency of L(M) in Eskimos on a small Arct
tatiyna

Answer:

(a) Frequency of M = 0.64

    Frequency of N = 0.04

    Frequency of MN= 0.32

(b)  Expected frequencies of M = 0.648

     Expected frequencies of MN = 0.304

     Expected frequencies of N =  0.048

Explanation:

(a) If random mating takes place in the population, then  the expected frequencies are

f(L(M)) = p = 0.8

F(L(N)) = q

q= 1 - p

= 1 - 0.8

= 0.2

Frequency of M = p^2 = ( 0.8)^2 = 0.64

Frequency of N = q^2 = (1-p)^2 = (1 - 0.8)^2 = (0.2)^2 = 0.04

Frequency of MN = 2pq = 2 * 0.8 * 0.2 = 0.32

(b)

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f(L(M)L(M)) = p^2 + Fpq = (0.8)^2 + 0.05 * 0.8 * 0.2 = 0.648

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f(L(N)L(N)) = q^2 + Fpq = (0.2)^2 + ( 0.05 * 0.8 * 0.2) = 0.048

5 0
2 years ago
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