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mojhsa [17]
3 years ago
13

A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. How many bacteria are grown during t

he 12th hour?
Biology
1 answer:
pochemuha3 years ago
7 0
Here is a list showing how the bacteria grows:

Begin      -   15
1st hour   -   30
2nd hour  -   60
3rd hour   -  120
4th hour   -   240
5th hour   -   480
6th hour   -   960
7th hour   -   1,920
8th hour   -   3,840
9th hour   -   7,680
10th hour -   15,360
11th hour -   30,720
12th hour -   61,440

As you can see, the number of bacteria present is doubled at the end of each hour, as the question states. However, the question is not asking for the total number of bacteria at the end of the 12th hour; it is asking for the bacteria GROWN during the 12th hour.

Because there were 30,720 bacteria at the end of the 11th hour, and 61,440 at the end of the 12th, the number of bacteria that grew during that last hour must be 61,440 - 30,720 = 30,720 bacteria.

(P.S. the answer would be 61,440 if it was asking for the TOTAL number of bacteria at the end of the 12th hour)
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Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what freque
Dmitry [639]

Answer:

  • J K / j k = 20%
  • j k / j k = 20%
  • J k / j k = 30%
  • j K / j k = 30%                

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%                                

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Match the term to its description. Match Term Definition Astronomical body
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Switch A and D, but other than that your answers are correct.

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How would shoot and root growth be affected by a mutation that caused plants to lose the ability to polymerize glucose into star
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Answer:

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