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Rom4ik [11]
2 years ago
6

Which phase involves the pinching off of the cell using the cleavage furrow or cell plate?

Biology
1 answer:
IrinaK [193]2 years ago
3 0

The phase during which the cleavage furrow or cell plate pinches off the cell would be cytokinesis.

<h3>What is cytokinesis?</h3>

It is the division of the cytoplasm during cell division that results in two independent cells.

Cytokinesis occurs at the latter stage of cell division when separated chromatids have completed their migration to the poles.

Thereafter, the chromosomes at each pole decondense, nuclear envelope appears around each of them before the cleavage furrow pinches and the cytoplasm divides to result in 2 daughter cells.

Some agree that cytokinesis is part of the telophase stage of cell division.

More on cytokinesis can be found here: brainly.com/question/10606931

#SPJ1

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4 years ago
The order of the genes on a plant chromosome is A, B, C, where A and B are located 10 cM apart and B and C are located 3 cM apar
Nonamiya [84]

Answer:

  • Double Crossing-over prob = 0.3%
  • Simple Crossing-Over prob A-B = 9.7%
  • Simple Crossing-Over prob B-C = 2.7%
  • Total Crossing-Over prob = 12.7%

Explanation:

<u>Available data</u>:  

  • order of the genes on a plant chromosome is A, B, C
  • A and B are located 10 cM apart
  • B and C are located 3 cM apart

The genetic distance = recombination frequency x 100 expressed in map units (MU). One centiMorgan (cM) equals one map unit (MU).

We can use the provided map with the distances between genes to predict the occurrence of any kind of recombinant gamete.  

We know that there is a probability of a simple crossing over to occur between genes A and B, or between genes B and C. And we also know that there is a probability of double crossing-over occurrence.

We can get the probability of the double-crossing over by multiplying the recombination frequencies between A-B and B-C. So,

DCO prob = recombination frequency A-B x recombination frequency B-C

                 = 0.1  x  0.03 = 0.003 = 0.3%

DCO prob = 0.3%

According to this, we would expect to find

  • 0.15% AbC
  • 0.15% aBc

Now we need to know what are the probabilities of getting a simple crossing-over, SCO. We already have the genetic distances between genes. 10 cM equals 0.1 of recombination frequency, and 3cM equals 0.03 of recombination frequency. However, we can not estimate the probabilities of simple crossing over using only this information, because this data is including the probabilities of the double crossing-over occurrence. So we need to substrate this percentage.    

SCO prob A-B = 0.1 - 0.003 = 0.097 = 9.7%

SCO prob A-B = 9.7%

we would expect to find

  • 4.85% Abc
  • 4.85% aBC

SCO prob B-C = 0.03 - 0.003 = 0.027 = 2.7%

SCO prob B-C = 2.7%

we would expect to find

  • 1.35% ABc
  • 1.35% abC

The total Crossing-over probability, TCO, is then the sum of all the crossing over probabilities,

TCO= SCO A-B + SCO B-C + DCO

TCO prob = 9.7% + 2.7% + 0.3%

TCO = 12.7%

8 0
3 years ago
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