Answer:
B
Explanation:
All living beings are in fact descendants of a unique ancestor commonly referred to as the last universal common ancestor (LUCA) of all life on Earth, according to modern evolutionary biology. Common descent is an effect of speciation, in which multiple species derive from a single ancestral population.
The advancements helps us grow the information and availability of resources to learn more and more about wildlife and the ecosystems as they grow. :)
If the temp.rises then it will lead to excessive evaporation in hydrosphere, on lithosphere it willl cause extreme hot conditions and with the scarcity of water the animals, birds will die and also plants will be affected.Hydrosphere and lithosphere are included in biosphere so it willl altogether effect biosphere...
Answer:
P = f(TLTL) = 0,16
H = f(TLTS) = 0,48
Q = f(TSTS) = 0,36
Explanation:
Hello!
The allele proportion of any locus defines the genetic constitution of a population. Its sum is 1 and its values can vary between 0 (absent allele) and 1 (fixed allele).
The calculation of allelic frequencies of a population is made taking into account that homozygotes have two identical alleles and heterozygotes have two different alleles.
In this case, let's say:
f(TL) = p
f(TS) = q
p + q = 1
Considering the genotypes TLTL, TLTS, TSTS, and the allele frequencies:
TL= 0,4
TS= 0,6
Genotypic frequency is the relative proportion of genotypes in a population for the locus in question, that is, the number of times the genotype appears in a population.
P = f(TLTL)
H = f(TLTS)
Q = f(TSTS)
Also P + H + Q = 1
And using the equation for Hardy-Weinberg equilibrium, the genotypic frequencies of equilibrium are given by the development of the binomial:



So, if the population is in balance:



Replacing the given values of allele frecuencies in each equiation you can calculate the expected frequency of each genotype for the next generation as:



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