Answer:
246
Explanation:
Increased amount / current population = 972
Amount released = 108
time(t) = 2years
Using the exponential growth function :
f(t) = A*g^t
Where g = growth rate
To obtain the growth rate = g
972 = 108 * g^2
g^2 = 972 / 108
g^2 = 9
g = 3
To determine population after 9 months :
9 months to years = (9/12) = 0.75
t = 0.75
f(0.75) = 108*(3^0.75)
f(0.75) = 108 * 2.2795070
f(0.75) = 246.18676
= 246
I think your question all turn into lowercases, uppercases, and lowercases are important in genetics because it can differentiate between the dominant and recessive forms.
let me explain at least how to calculate the offspring percentage:
for example, you have Rr crossed with another Rr (R for wrinkled peas and r for smooth peas), you just have to match between the four letters, and you will have four possibilities:
R and R
R and r
r and R
r and r
you will have RR, Rr, Rr, and rr
if we convert into percentages, it will give:
25% RR
50% Rr (there's two Rr so 25 + 25)
25% rr
Hey can u break it down and give me some more info on the exact thing u have t right about i will do it for you...:)
Answer:
Each mutant would be mated to wild type and to every other mutant to create diploid strains. The diploids would be assayed for growth at permissive and restrictive temperature. Diploids formed by mating a mutant to a wild type that can grow at restrictive temperatures identify the mutation as recessive. Only recessive mutations can be studied using complementation analysis. Diploids formed by mating two recessive mutants identify mutations in the same gene if the diploid cannot grow at restrictive temperature (non-complementation), and they identify mutations in different genes if the diploids can grow at restrictive temperature (complementation).
Explanation:
Recessive mutations are those whose phenotypic effects are only visible in homo-zygous individuals. Moreover, a complementation test is a genetic technique used to determine if two different mutations associated with a phenotype colocalize in the same <em>locus</em> (i.e., they are alleles of the same gene) or affect two different <em>loci</em>. In diploid (2n) organisms, this test is performed by crossing two homo-zygous recessive mutants and then observing whether offspring have the wild-type phenotype. When two different recessive mutations localize in different <em>loci</em>, they can be considered as 'complementary' since the heterozygote condition may rescue the function lost in homo-zygous recessive mutants. In consequence, when two recessive mutations are combined in the same genetic background (i.e., in the same individual) and they produce the same phenotype, it is possible to determine that both mutations are alleles of the same gene/<em>locus</em>.
