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V125BC [204]
2 years ago
5

Five friends looked at a piece of aluminum foil. They wondered what they would see if they could see inside an atom of aluminum.

This is what they said: Jeb: I think it would look like a tiny, dense ball with tiny particles tightly packed inside it. Jane: I think there would be a tiny, dense center surrounded by a lot of empty space where some particles are whizzing around. Hal: I think there would be a large center made up mostly of empty space wit particles whizzing around in it. It would be surrounded by a dense shell. Pick the friend that you agree with and explain why you agree. Be sure to include the friends' name and factual information to back up your explanation. Attachments
Biology
1 answer:
babunello [35]2 years ago
6 0

The answer is; Jane would be right.

The atom was discovered be more of empty space. The mass of an atom is mainly attributed to the mass of the nucleus which occupies a very small fraction of the size of the atom. The nucleus is Located in the center surrounded by orbiting electrons.  

The Geiger–Marsden experiment, where alpha particles are accelerated onto a thin aluminum foil proved this theory by Ernest Rutherford. Some of the particles were deflected by the foil and this was attributed to the dense nucleus of the aluminum atoms in the foil. The other particles passed through because most of the atoms are made of empty space.


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  1. we have to write down the genotype of the F1 female three times, changing the order of the genes each time.
  2. Then, we hypothesize what the double recombinant gametes would look like.
  3. When the theoretical double recombinants we obtain are the same as the ones observed in the F2, we know that <em>that </em>is the correct order of the genes.

In this problem, only if the middle gene is h+/h the double crossing over gives us the observed double recombinant gametes, therefore <u>hairless</u> is the middle gene.

\frac{s\ h^+\ c^+}{s^+\  h\ c}

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Distance \ [s-h]= \frac{number\ of\  recombinants \ [s-h]}{Total number of individuals}  * 100\\\\\\Distance \ [s-h]= \frac{32+38+12+18}{1000}  * 100\\\\Distance \ [s-h]= 10\  map\ units

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Distance \ [h-c]= \frac{130+140+12+18}{1000}  * 100\\\\Distance \ [h-c]= 30\  map\ units

<h3><u>The gene map for these genes is:</u></h3>

spineless -----------------hairless ---------------------------claret

                   10 m.u.                            30 m.u.

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