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Sergeeva-Olga [200]
3 years ago
8

An example of natural selection is the red color of a male cardinal. The females of the species choose mates based on the vibran

t colors of the males' feathers. If females begin using different criteria than feather color when they choose mates, what would most likely happen to the color of the male cardinals over time? increased variation in the shades of red because the selection pressure has been relieved increased variation in the shades of red because the cardinals try different ways to impress the females decreased variation in the shades of red because no form of the trait is advantageous decreased variation in the shades of red because the only reason for variation was selection pressure
Biology
2 answers:
ziro4ka [17]3 years ago
7 0

Answer:

The correcta answer is opcion B). "<em>decreased variation in the shades of red because no form of the trait is advantageous</em>".

Explanation:

If females choose males for other traits than the red color of feathers, it is probably that the feathers would decrease the variation in the shades of red because no form of the trait is advantageous.

On the one hand, females do not choose males for this characteristic anymore, which means that this color is not advantaging for mating. On the competition with other males, they will have to develop the other trait that makes them look better genetically.

On the other hand, these bright colors make males very visible to predators, a pressure that will make the turn to less bright visible colors for better hiding and fewer chances to get caught.

Sidana [21]3 years ago
6 0

Answer:

decreased variation in the shades of red because no form of the trait is advantageous

Explanation:

The last option also seems correct, but selections is somehow aleatory. You cannot consider "reason" as a factor.

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The external iliac vein ________. the external iliac vein ________. empties directly into the inferior vena cava drains the pelv
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A diploid cell has two pairs of homologous chromosomes. How many different combinations of chromosomes could there be in the gam
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<h3><u>Answer;</u></h3>

There will be <u>4 different combination of chromosomes</u>

<h3><u>Explanation;</u></h3>
  • <em><u>A diploid cell is a cell that contains two sets of each chromosomes; thus a diploid cell with two pairs of homologous chromosomes has four chromosomes.</u></em>
  • <u><em>Gametes are produced through meiosis which is the type of cell division in which a parent cells divides into four daughter cells each with half the number of chromosomes as the parent, that is the cell are haploid.</em></u>
  • Therefore,<em><u> </u></em><em><u>a diploid cell with two pair of homologous chromosomes will produce four haploid cells after meiosis each with two chromosomes.</u></em>
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4 years ago
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3 years ago
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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%                                

3 0
3 years ago
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