Answer:
Explanation:
Blue-white screening is a method for distinguishing proof of (recombinant bacteria). It depends on the capacity of ( B-galactosidase) to separate lactose. Blue-white tests exploit the molecule called (x-gel)_ which is like lactose in that it is severed by B-galactosidase. When separated, the (5-bromo-4-chloro-indoxyl) (same as past) turns _(_blue). In the event that uncleaved, which implies a non-function B-gal gene, the X-gal remains (white)_. Subsequently, a __(white) bacterial province implies the B-galactosidase gene isn't practical, and in this way there ___lacz__ a recombinant gene embedded into the vector.
Show the statement. (A picture or an example of the problem)
Answer:
The frequency of recessive allele should be 0.30.
Explanation:
According to Hardy-Weinberg equilibrium, sum of both dominant allele frequency (p) and recessive allele frequency (q) should be equal to 1 (p+q = 1).
Brown hair = dominant (p)
white hair = recessive (q)
49% mice are brown hair
so dominant genotype frequency = 0.49
According to Hardy-Weinberg principle, square root of the genotype (homozygous) is equal to allele frequency.
√p =√0.49 = 0.70
The dominant allele frequency is 0.7
Now, by this value we can find the recessive allele frequency by
p + q = 1
1 - 0.7 = 0.3
So, the recessive allele (white hair mice) frequency is 0.30.
<span>A scientific theory is a well-substantiated explanation of some aspect of the natural world that is acquired through the scientific method and repeatedly tested and confirmed through observation and experimentation.</span>
U can hear them talking about you you can see them laphing and snickering
