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mars1129 [50]
3 years ago
12

Homologous recombination is a unique feature of meiosis whose purpose is to increase genetic diversity during reproduction. What

does the process of homologous recombination entail, and how does this process achieve the purpose of increasing diversity? When does homologous recombination occur?
Biology
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

The correct answer will be-

1. Homologous recombination- crossing over

2. Occur-during pachytene of meiosis I of gamete formation.

Explanation:

The basis of diversity of organism present on earth lies in the sexual mode of reproduction possessed by the organism to produce offspring. The sex cells involved in the reproduction are formed by a process called meiosis.

During the formation of gametes, at the pachytene stage of meiosis I take place a mechanistically conserved process which leads to the exchange of DNA segment between non-sister chromatids called crossing over through a process known as homologous recombination. This exchange of genetic material in the homologous chromosomes increase the genetic diversity of the organism on the earth.

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In a scientific investigation of lakes, the depth of a lake is most likely to be measured in
sergey [27]

Answer:

D

Explanation:

Units of weight and volume won't be large enough to be meaningful. so liters and grams are too small. Find out the weight of a cubic meter of water and then estimate how many cubic meters there might be in a lake. Millions would probably be too small.

Miles are too big. Most lakes do not go to a depth of 1 mile, but you on on the right track. Distance is what you want.

Meters are the answer.

5 0
3 years ago
Read 2 more answers
how is translation initiated? view available hint(s)for part a hint 1for part a. how is translation initiated? the start codon s
Irina18 [472]

The Translation initiated is <u>Option D.All of the listed answers are correct. </u>

At the initiation of translation ribosomes and tRNA bind to the mRNA. tRNA is located at the first docking site of the ribosome. The anticodon of this tRNA is complementary to the start codon of the mRNA where translation begins. After binding to the mRNA, the ribosome initiates translation at the start codon AUG and moves the mRNA transcript one codon at a time until it reaches the stop codon.

When tRNA recognizes and binds to the corresponding codon in the ribosome, it transfers the corresponding amino acid to the end of the growing amino acid chain. tRNA and ribosomes then continue to decode the mRNA molecule until the entire sequence is translated into protein. tRNA acts as an adapter molecule during the translation process. Formerly known as soluble RNA or sRNA. As an adapter, it connects amino acids to nucleic acids.

Learn more about Translation initiated here:-brainly.com/question/7169341

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5 0
1 year ago
Which observation could lead to the conclusion
marusya05 [52]
<span>(3) It cannot perform metabolic processes.</span>
3 0
3 years ago
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What does vain mean in the poem?
Inessa05 [86]

Answer:

D. Useless

Explanation:

According to Dictionary.com, “in vain” is defined as “without real significance, value, or importance; baseless or worthless.”

7 0
3 years ago
Which of the following samples will have the greatest volume at STP?a. 22 g Neb. 22 g Hec. 22 g O2d. 22 g Cl2 e. All have the sa
sammy [17]

Answer:

22g of He

Explanation:

Volume of gas at STP is given as :

Volume occupied(dm³) = number of moles x 22.4dm³mol⁻¹

In order to solve this given problem, we would first find the number of moles from the different masses of atoms and molecules given.

Number of moles = \frac{mass }{Molar mass}

Now, let us obtain the molar mass of the given atoms and elements:

For Ne; molar mass is the atomic mass = 20gmol⁻¹

      He = 4gmol⁻¹

      O₂ = (16 x2)gmol⁻¹  = 32gmol⁻¹

      Cl₂ = (35.5x2)gmol⁻¹ = 71gmol⁻¹

Number of moles of Ne = \frac{22}{20} = 1.1mol

Volume occupied by Ne at STP = 1.1 x 22.4 = 24.64dm³

       

Number of moles of He = \frac{22}{4} = 5.5mol

Volume occupied by He at STP = 5.5 x 22.4 = 123.2dm³

Number of moles of O₂ = \frac{22}{32}  = 0.688mol

Volume occupied by O₂ at STP = 0.688 x 22.4 = 15.4mol

Number of moles of Cl₂ = \frac{22}{71} = 0.3099mol

Volume occupied by Cl₂ at STP = 0.3099 x 22.4 = 6.94dm³

5 0
3 years ago
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