Suppose a species of bird called the red-crested warer has a plumage length that

Question

Suppose a species of bird called the red-crested warer has a plumage length that is controlled by a single gene. The Plm allele produces long plumage and is dominant over the Plm allele. One population exists in North America and in a seperate population in South America. The trait is in hardy-Weinberg equilibrium in each population.
length that is controlled by a Suppose a species of bird called the red-crested warbler has a single gene. The Plm allele produces long plumage and is dominant over the plm allele. One population exists in North Ameríca (NA) and a separate population exists in South America (SA). The trait is in Hardy- Weinberg equilibrium in each population. An island nature preserve brings in 112 NA warblers and 452 SA warblers. Out of the 112 NA birds, 55 have long plumage. Out of the 452 SA warblers, 75 have long plumage. After the birds from the combined the island population produces 1000 off to have long plumage. Round your answer to two significant figures Calculate the number of these populations mate r offspring that are e Number birds Tools x 10

Explanation / Answer

In north America there are:
112 warblers present

therefore total no. of alleles = 224

55 warblers have long plume

therefore,no. of short plumule birds will be:

112 - 55 birds = 57 short plume birds.

q2 =No. of alleles of short plumule/ total no. of alleles of NA warblers

=(2*57) / 224 = 0.5089
Freq(short plume allele) = q

Square root of q =square root of 0.5089

=0.7134
So, Freq(long plume allele) = p

= 1 - q = 0.2866
From this North American population we get

No. of Birds with long plumule will be Frequency of long plumule alleleXtotal no. of alleles
=0.2866 * 224 = 64 long plume alleles
And, 0.7134 * 224 = 160 short plume alleles

Similarly,

In South America
452 birds total no. of birds = 452

therefore, Total no. of alleles = 904

No. of birds with long plumule = 75
So, No. of short plume birds will be = 452 - 75

= 377 .
short allele = q2 which will be equal to No. of alleles wiyh short plumule/ Total no. of alleles

= (2*377) / 904 = 0.8341
and, Frequency of hort plume allele = q = square root of q2=0.9133
and, Frequency of long plume allele = p = 1 - q = 0.0867
From this South American population we get
long plume alleles =0.0867 * 904 = 78
and short plume alleles 0.9133 * 904 = 826

From the above calculations, we get that Blended population has
64 + 78 = 142 long plume alleles
160 + 826 = 986 short plume alleles
142 + 986 = 1128 total alleles
Therefore, p or long plumule allele frequency = 142/1128 = 0.1259
and short plumule frequency or q = 986/1128 = 0.8741
No. of offspring in new population = 1000.

Since the long plumage is dominant

So, the homozygote long plumage as well as heterozygote with one long plumage allele will be long plumage birds which will be equal to p2 + 2pq

and the population with long plumage in mixed population will be p2 + 2pqX total population

therefore, population size with long plumage = (p2 + 2pq) = 1000 * (0.1259^2 + 2*0.1259*0.8741) = 236 birds

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