Anthropologist use the term Enculturation to refer to the process of learning your culture, ordinarily as a child.
Ofcourse, <span>Cone-bearing plants</span>
Answer:
Option (b).
Explanation:
The frequency of alleles K and k in population are 0.6 and 0.4. The inbreeding coefficient is 0.3.
The heterozygote frequency can be calculated by the formula:
F = 
Here, K = 0.6, k= 0.4, F = 0.3 and H = approximate genotypic frequency of heterozygote Kk.
Put the values in the above formula
0.3 = 
H = 0.36 ≈ 0.34.
Thus, the approximate genotypic frequency of the heterozygote Kk is 0.34.
Hence, the correct answer is option (b).
The answer for this question is C, I just did my final for Biology and got a good score in it.
Hope it helps!
Answer:
A. Glucose:
As lactose is absent will follow the glucose metabolism, so the lac repressor will hold tightly with the operator which leads to the prevention of transcription as RNA polymerase can not bind to promoter so no lactose metabolism.
B. lactose:
In the case or wild mutant lactose metabolism will occur as lactose is present, allolactose binds to the lac repressor thus the operator and promoter are free so transcription starts. The same result will be in all test except in the case of a mutant repressor.
C. glucose and lactose:
In this case, diauxic growth will occur. This growth takes place when two sugars are present in the media. The preferred sugar will be consumed first That is glucose. Then a lag phase will come which will be followed by the second sugar metabolism that is lactose. the given two cases :
Lactose will be metabolized first, as operator site mutation prevents repressor protein binding and thus will lead to the constitutive synthesis of lac operon in test B.
The same thing will happen, as in this case also, constitutive synthesis of lac operon will happen in test C.
Cultures grown in the presence of both glucose and lactose are metabolizing lactose. In the test (b) and (c) will be responsible due to constitutive synthesis of lac operon will happen.
