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Salsk061 [2.6K]
3 years ago
15

At room temperature, Mercury exist in the liquid state true or false​

Biology
1 answer:
Rina8888 [55]3 years ago
4 0

Answer:

True

Explanation:

Both Mercury and Bromine exist in the liquid state at room temperature.

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What is the process which scientists grade the work of other scientist before published?
Finger [1]
The process is called Peer-Reviewing. In Peer-Reviewing, scientists/academic peers "peer review" the work of a scientist to check for accuracy or any false information that might reduce the legitimacy of their study or change "how realistic/real" their study is.

p.s. This is also common sense because scientists make up an entire scientific community who base their scientific studies on consistency and accuracy. 

3 0
3 years ago
An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result
Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

                   

                    EW                   EW                     EW                   EW

ew               EWew               EWew               EWew               EWew      

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

All offspring are  (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine  the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If  EWew × eeww

we have;                 EW   &   ew    ×    ew  &    ew

           EW               ew

ew       EeWw          eeww

ew       EeWw          eeww

offspring that

are linked in   ⇒     EeWw    EeWw     &      eeww      eeww

F₂   will be

\frac{1}{2} = 50% of EeWw of the total 100 offspring in the F₂ cross

\frac{1}{2} = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes =  50% EeWw and 50% eeww.

For unlinked genes; If  EWew × eeww

if rearrangement occurs in EWew  and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

                       EW                    Ew                    eW                    ew

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

We have the following results for the unlinked genes

\frac{1}{4} = EeWw  25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = Eeww   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeWw   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeww    25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

3 0
3 years ago
PLEASE HELP I NEED TO GET THIS QUESTION RIGHT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
soldier1979 [14.2K]

Answer:

There was no oxygen in the flask.

Explanation:

Anaerobic organism do not need oxygen!!

5 0
3 years ago
Why does synthetically produced RNase refold incorrectly if the reducing agent is removed before urea removal?
olya-2409 [2.1K]
<h2>Disulfide bonds interaction </h2>

Explanation:

  • The high concentration of urea causes unfolding of protein and thus results in loss of function of a protein. The urea interacts with the protein and prevents the folding of a protein.
  • During oxidation, the disulfide bonds that are required for proper functioning and stabilization of protein are formed, while in the presence of urea, the disulfide bonds are not positioned correctly. The protein oxidation results in covalent modification of a protein that results in the change of physical and chemical properties of a protein.
  • The change of physical and chemical properties of protein after oxidation and in the presence of urea cannot be modified even after the removal of urea. Thus, protein does not fold properly.
  • Hence the right answer is "disulfide bonds are not positioned correctly rather than week bonding interaction"

6 0
3 years ago
Natural selection will only lead to evolution if it allows for an organism to:
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D)all of the above i think
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