The formal formula for calculating the number given a percentage is:
n = (t/100)*p, where
t is the total number of pies and p is the percentage of the total.
Therefore,
n = 1550/100 * 30 = 465
A short cut is to express the percentage as a fraction: 30% = 0.3, and simply multiply the total by this fraction:
1550 * 0.3 = 465
Answer: Their supporting structure would be most likely the MUSCULOSKELETAL SYSTEM of the multicellular organisms.
Explanation:
The musculoskeletal system found in multicellular organisms are made up of structures which provides support, protection and allows for various movements. These structures include:
--> Muscles
--> Cartilage
--> Ligaments
--> Tendons and
--> Joints
Furthermore, Microtubules and microfilaments are the major components of cytoskeleton. They are the structures that gives a cell it's shape and helps organise it's parts. As described in the question, they are elongated structures that cross the cell, giving SUPPORT to the cell, and carries organelles and objects throughout the cell.
Therefore, since the Microfilaments and Microtubules provides support and shape at the cellular level it can be most likened to the musculoskeletal system of multicellular organisms which performs the same functions.
Answer:
4.38%
Explanation:
Subtract the accepted value from the experimental value.
Take the absolute value of step 1.
Divide that answer by the accepted value.
Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
Answer:
The correct genotype of the two pure lines and the F1 is:
A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰ and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺
The number of additive alleles on each genotype are two and six respectively.
Explanation:
Locus( plural form . loci) are fixed point on a chromosome in which genes are located. These genes are specific genetic material or genotype.
Now;
If we decide to designate the allele of the four loci into either additive (⁺) or non-additive(⁰); we have the following :
Let's the allele of the four loci to be
A⁺/A⁰, B⁺/B⁰, C⁺/C⁰ and D⁺/D⁰
However, from the diagram below; we deduce that the correct genotype for the two pure lines and the F1 is as follows:
A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰ and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺ and the number of additive alleles on each genotype are two and six respectively.
The cross between both F1 traits will yield an heterozygous individual for the offspring. i.e A⁺A⁰B⁺B⁰C⁺C⁰D⁺D⁰ with only four additive allele
