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Eduardwww [97]
3 years ago
5

Need help etm class hard

Biology
1 answer:
Irina-Kira [14]3 years ago
5 0

Answer:

Liver

Thoracic

Xiphoid

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1. Do these results support the claim that application of the jewelweed is a good treatment for symptoms caused by poison ivy? E
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A claim is an unverified statement or theory which is made about something which may or may not be true.

<h3>Verification of a claim</h3>

For a claim to be validated, i.e, proven to be true (in science), then there needs to be tests and experiments which needs to be carried out to show that the claim is valid.

With this in mind, if for example if there's a claim that jewelweed helps to treat symptoms caused by poison ivy, then there needs to be a controlled experiment to see if this is true or not.

If after a series of experiments and it can be established without any iota of doubt, then the claim is valid and the scientific process will move to the next stage.

Please note that your question is incomplete so I gave you a general overview to give you a better understanding of the concept.

Read more about claim and evidence here:

brainly.com/question/10877313

8 0
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The Calvin cycle requires the energy from ______ and ______ which are produced in the light reactions.
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Why do experiments usually require a control?
zalisa [80]
Hey There!

Here is your answer:

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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
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