Answer:
Oxygen is necessary for the growth of all bacteria.
Explanation:
This statement is false, because not all bacteria require oxygen to produce energy and grow. The bacteria that need oxygen to grow are called aerobic bacteria. But there are also bacteria that can produce energy without oxygen, these are called anaerobic. Some bacteria can adapt to both environments, these are called facultative anaerobes bacteria.
Answer:
Cyanobacteria are microscopic organisms found in all kinds of water. They are single-celled organisms and produce their own food from sunlight via photosynthesis. Cyanobacteria are important to evolution because they developed the oxygen atmosphere we live in by producing waste from cyanobacteria. Plants also evolved from Cyanobacteria.
The answer is bacteremia. septic shock is what can result from bacteremia if gone untreated. Sepsis is a widespread infection that's life threatening. Usually if some bacteria is introduced into the blood stream the immune system takes care of it so for sepsis to occur it would have to be pretty large amounts of bacteria.
Answer:
Hydrogen bonding is caused by the tendency of some atoms in molecules to attract electrons more than their accompanying atom. This gives the molecule a permanent dipole moment – it makes it polar – so it acts like a magnet and attracts the opposite end of other polar molecules.
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%
