In this case, Natural selection operates on only one genetic locus. At which there are two alleles, one dominant to other,

##### Relative Chances of survival from Birth to adult

Genotype | chance of survival |

AA ,Aaaa | 1 1-s |

**Selection Coefficient** “s” is the selection coefficient here. selection coefficient is expressed as a reduction in fitness relative to the best genotype. If s=0.1 the chance of survival for aa will be 90%. AA and Aa has a 100% chance of survival. But in case if fitness chance is 50% then aa has survival chance as 45% ifs= 0.1 but for our convenience we used 100% as relative fitness.

##### Relative fitness

fitness is considered as 1 in the above case and fitness is always measured as relative fitness. if s=0 in any case it would follow Hardy Weinberg theorem as no natural selection is acting.

##### Changes in the gene frequency is controlled by fitness:

The dominance or recessivity of any genotype do not control the change in gene frequency it depends upon natural selection. If aa is best fitted the gene frequency of aa will be more.

##### Population change through time:

To find the population change over time consider allele A expressed have gene frequency p so the gene frequency of A in next-generation is p’ and the change in gene frequency is considered as

∆p=p-p’ to begin with algebric expression and numerical value consider change in gene frequency first at birth and then at adult stage At birth genotypes formed by random mating ( thorugh Hardy weinberg theorem) and then selection operates, aa individuals have lower chance of survival and their frequency among the adults is reduced see figtotal number of aa adult is less then birth and we have to divide adult number of each genotype by the total population size to express the adult numbers as frequencies comparable to frequensies at birth

Mean fitness= p2 + 2pq +q2(1-s)= 1-sq2

Dividing by mean ﬁtness in the algebraic case is the same as dividing by the population size after selection in the numerical example. Notice that now the adult genotype frequencies are not in Hardy–Weinberg ratios. If we tried to predict the proportion of aa from q2, remember the frequency of gene A = to frequency of AA and half Aa

The gene frequency of genotype A in next generation given as this

The genotype frequency of A increases in next generation as compare to previous one so p’ is greater than p

The Change in gene frequency is given as:

For example, if p=q=0.5 andaaindividuals have ﬁtness 0.9 compared with AAand Aaindividuals (s =0.1) then the change in gene frequency to the next generation will be(0.1×0.5×(0.5)2)/(1−0.1×(0.5)2)=0.0128; the frequency of Awill therefore increaseto 0.5128.

We can use this result to calculate the change in gene frequency between successivegenerations for any selection coefﬁcient (s) and any gene frequency.

##### Important interpretations from the case:

- A gene will increase in frequency until it fixed
- The higher selectioncoefﬁcient against the genotype, the A gene increases in frequency more rapidly.
- the frequency of Aslowsdown when it becomes common, and it would take a long time ﬁnally to eliminate the a gene. This is because the a gene is recessive. When is rare it is almost always found in Aa individuals, who are selectively equivalent to AA individuals: selection can no longer “see” the a gene, and it becomes more and more difﬁcult to eliminate them.
- we can also find “s” by the equation that we get from previous calculations as

Haldane (1924) ﬁrst produced this particular model of selection. One important feature of the model is that it shows how rapidly, in evolutionary time, natural selection can produce change.

Reference; This article is taken from Evolution book by Ridley