Germination A. is the same process for both monocots and dicots. B. ends with dormancy. C. is the process that occurs when the s
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Answer:
Check the explanation
Explanation:
North American:
112 birds total => 224 alleles
112 - 55 birds = 57 short plume birds.
= (2*57) / 224 = 0.5089
Freq(short plume allele) = q = 0.7134
Freq(long plume allele) = p = 1 - q = 0.2866
From this North American population we get
0.2866 * 224 = 64 long plume alleles
0.7134 * 224 = 160 short plume alleles
South American:
452 birds total => 904 alleles
452 - 75 birds = 377 short plume birds.
= (2*377) / 904 = 0.8341
Freq(short plume allele) = q = 0.9133
Freq(long plume allele) = p = 1 - q = 0.0867
From this South American population we get
0.0867 * 904 = 78 long plume alleles
0.9133 * 904 = 826 short plume alleles
Blended population has
64 + 78 = 142 long plume alleles
160 + 826 = 986 short plume alleles
142 + 986 = 1128 total alleles
p = 142/1128 = 0.1259
q = 986/1128 = 0.8741
All right. The new population has the p and q from the blended population. The new population has 1000 individuals. The portion of long plumed will be homozygote long plumage + heterozygote, which is + 2pq, and you multiply that by the population size 1000 to get the final answer.
population size * ( + 2pq) = 1000 * (0.1259^2 + 2*0.1259*0.8741) = 236 birds