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Sagot :
25 babies would expect to have cystic fibrosis, a homozygous recessive condition.
As per the data given in the above question,
The data given in the question are as follow,
Total number of individuals is 2500.
Ans the frequency of the dominant allele is 0.9.
Also the population is at hardy-weinberg equilibrium.
Then how many babies would you expect to have cystic fibrosis?
So the frequency of dominate allele (p) + frequency of recessive (CF) allele (q)=1
[tex]0.9+q=1\\q=1-0.9[/tex]
Further solving we get,
q=0.1
Frequency of CF genotype [tex]=q^2=(0.1)^2=0.01 \:or1\%[/tex]
Therefore the expected no of babies with CF,
[tex]=0.01\times 2500\\=25[/tex]
A homozygous recessive disease called cystic fibrosis would be anticipated in 25 newborns.
For more such question on cystic fibrosis
https://brainly.com/question/2290129
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