Answer:
Explanation:
Given -
Total Population - 
Individuals with homozygous HbA/hbA genotype - 
Individuals with homozygous hbs/hbs genotype -
Individuals with heterozygous hba/hbs genotype - 
Let us assume the given population is in Hardy Weinberg' equilibrium
The frequency of individuals in the given population with homozygous HbA/hba genotype is equal to number of individuals with homozygous HbA/hba genotype divided by total population.
The frequency of hba allele is equal to
Answer:
Can someone pls answer my questions plssss?
Explanation:
Answer:
The correct answer is 2. 5'-AUGUCAGGUACGCCACAU-3'
Explanation:
The template strand for transcription is 3'-TACAGTCCATGCGGTGTA-5 and RNA polymerase will add complementary mRNA sequence against this strand in the opposite direction i.e., 5' to 3' direction.
In RNA, A(adenine) makes complementary base pairing with U(uracil), not with T(thymine) and the rest of the base-pairing is same as in DNA. Therefore the correct mRNA produced from the transcription of this DNA molecule by the help of RNA polymerase is 5'-AUGUCAGGUACGCCACAU-3'.
Then this mRNA sequence will be used by ribosomes to synthesize proteins during the translation process.
The answer is A, as bears don’t eat deer, but they both eat plants.
Heterotrophs are animals.
Autotrophs are plants.
Hope this helps!
