The surface area will be:
S.A. = 6l²
S.A = 6(100 x 10⁻⁶)²
Volume = l³
Volume = (100 x 10⁻⁶)³
Surface area to volume ratio:
[6(100 x 10⁻⁶)²] / (100 x 10⁻⁶)³
S.A : Vol = 6 x 10⁴
Answer:
1. Evaporation
2. Lengthening of a river upslope.
3. Sand and gravel
4. Storm surge
5. Adds clean water to the drinking water supply
6. cuts down on the number of open pores
7. A confined aquifer is pumped for drinking water
Answer:
The recombination frequency between two genes exhibits a positive correlation with the distance between them, that is, farther they are, and more will be the chance of recombination. Thus, recombination frequency is used to signify distance among the two genes, for example, 1 percent recombination frequency demonstrates distance of 1 map unit.
Let us consider that the heterozygous female of genotype AaBb can generate four kinds of gametes, that is, AB, Ab, aB and ab. Of these, the two gametes are the outcomes of recombination, or it can be said that 50 percent are recombinants. Thus, it can be concluded that in case of two linked genes, the maximum probable recombination frequency is 50 percent.
This shows that any genes, which are distant than 50 map units will function as unlinked and will function as if they were on distinct chromosomes, and the frequency of recombinant frequency will be 50 percent.
In the given question, it is given that the map distance between the two genes is 80 map units, that is, more than 50 map units. The maximum probable recombinant offspring will be 50 percent of the entire offspring.
