<h2>Frequency of allele </h2>
Explanation:
Hardy Weinberg Equilibrium is used to calculate the allelic as well as genotypic frequency
Allelic frequency of dominant and recessive allele is represented by p and q respectively whereas genotypic frequency of dominant genotype is represented by 
and 
respectively
Given:
H allele (p) = hairy heffalump (dominant)
h allele (q) = hairless heffalump (recessive)
36% of heffalump population is hairless represents the % of recessive genotype, hh () =36%
Calculation of frequency of the h allele (q) :
Frequency of genotype hh () will be: 36/100=0.36 or 0.6*0.6
Frequency of h allele (q) will be 0.6
Answer:
I think its D. Artificial Selection
Explanation:
Sorry if this didn't helped you.
Answer:
The person has been dead for approximately 15,300 years
Explanation:
<u>Available data</u>:
- The half-life of carbon 14 is 5,600 years
- The human skeleton level of carbon 14 is 15% that of a living human
To answer this question we can make use of the following equation
Ln (C14T₁/C14 T₀) = - λ T₁
Where,
- C14 T₀ ⇒ Amount of carbon in a living body at time 0 = 100%
- C14T₁ ⇒ Amount of carbon in the dead body at time 1 = 15%
- λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
- T₀,₅ ⇒ The half-life of carbon 14 = 5600 years
- T₀ = 0
- T₁ = ???
Let us first calculate the radioactive decay constant.
λ = (Ln2)/T₀,₅
λ = 0.693/5600
λ = 0.000123
Now, let us calculate the first term in the equation
Ln (C14T₁/C14 T₀) = Ln (15%/100%) = Ln 0.15 = - 1.89
Finally, let us replace the terms, clear the equation, and calculate the value of T₁.
Ln (C14T₁/C14 T₀) = - λ T₁
- 1.89 = - 0.000123 x T₁
T₁ = - 1.89 / - 0.000123
T₁ = 15,365 years
The person has been dead for approximately 15,300 years
