Si el agua es abundante (no limitante), entonces las plantas pueden tener más estomas, lo que permite un mayor acceso al agua (y los iones de hidrógeno necesarios), y un mayor apoyo para los tejidos herbáceos.
Espero que esta respuesta sea correcta :)
This question is not complete
Complete Question:
Students investigated samples of amylase from 100 goats. 100 small filter paper discs were each soaked in a different sample of goat amylase. The students tested the activity of these amlyase samples using plain paper. Plain paper contains starch.
A circle of plain paper was placed into a Petri dish ad shown in the diagram below. Iodine solution was used to stain the starch in the plain paper.
When iodine solution reacts with the starch in the plain paper, what colour would you see?
Answer:
The colour that would be seen is blue black .
Explanation:
Amylase is an enzyme that is involved in the breaking down or hydrolyses of starch.
When Amylase reacts with iodine, it speeds up the rate of reaction of the hydrolysis of the starch present in the plain paper.
The colour that would be observed is a blue black colour and the starch in the paper would be broken down further more into simpler sugars.
Answer:
Explanation:
To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------40 % Parental (Not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k = 30%
<span>The three essential principles of test construction are listed below:
I. Standardization
II. Reliability
III. Validity
</span>Standardization- To standardize a test means that test is given to a large, representative sample of people, in order to establish the norms that future test takers are compared against (using a normal distribution)
Reliability - This principle requires that a test must produce reliable, consistent results when it is repeated. The reliability of a test can be verified using the test-retest method or the split-half method.
Validity - This refers to the ability of a test to measure what it was designed to measure.
Answer:
Consider the heterozygous oval, thick cell walled bacteria to have the alleles OoTT and the thin cell walled bacteria to have alleles oott. Results will be 50% oval, thick walled bacteria and 50% round, thick walled bacteria. This will be the F1 progeny.
When the oval, thick walled bacteria from the F1 progeny is cross bred with round, thick walled bacteria then 25 percent of the bacteria will be heterozygous oval, thick walled. 25 percent will be heterozygous oval and heterozygous thick walled. 25 percent will be round and thick walled. 25 percent will be round and thin walled.
