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AURORKA [14]
4 years ago
13

HEEEEEELLLLLLLLLPPPPPPPPPP!!!

Biology
2 answers:
sp2606 [1]4 years ago
8 0

Answer:

A , how convection currents move in Earth’s interior

Explanation:

the arrows are the currents and the arrows are also showing how they move

FrozenT [24]4 years ago
3 0

Answer:

A. How convection currents move in Earth’s interior

Explanation:

it was this on E2020

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I apologize if the photo is too blurry.<br><br>I need help with biology ASAP plz!!
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The correct answer is c
7 0
3 years ago
Each body cell of a chimpanzee contains 48 chromosomes. After mitosis, how many chromosomes are present in each cell?
nekit [7.7K]

Answer:

During both mitosis and meiosis, DNA replicates first during S phase of interphase. Each copy (DNA molecule) is called chromatid. Before anaphase they remain together, joined by the centromere in the chromosome.

Part 1: How many chromatids and chromosomes are present at:

(a) anaphase of mitosis: During this phase sister chromatids split. We would have 48 chromosomes in each pole and 48 chromatides.

(b) anaphase I of meiosis: During this phase homologous chromosomes split, being a reductional division. In each pole we will have half the chromosomes we had after DNA replication. This is 24 chromosomes but 48 chromatides (remember they will split during anaphase II).

(c) anaphase II of meiosis: This is an equational division, we will have 24 chromosomes in each pole and 24 chromatides. Each chromatid is considered a chromosome.

(d) G1 prior to mitosis: During this phase DNA has not replicated yet and it is not condensed either. This formed is called chromatin. We will assign one chromatid for each chromosome. This is a somatic cell, so: 48 chromosomes and 48 chromatids.

(e) G2 prior to mitosis: After S phase, we have duplicated all chromosomes. We will assign two chromatids per chromosome: this is 96 chromatides and 48 chromosomes.

(f) G1 prior to meiosis: Before DNA duplication, 48 chromosomes, 48 chromatids.

(g) Prophase of meiosis I: After DNA replication, condensation of the chromatin takes place: 48 chromosomes, 96 chromatids.

Part 2: How many chromatids or chromosomes are present in:

(h) An oogonial cell prior to S phase: This is G1 phase, 48 chromosomes.

(i) A spermatide: This is the male haploid gametid, after meiosis: 24 chromosomes and 24 chromatids.

(j) A primary oocyte arrested prior to ovulation: They are arrested at prophase I of meiosis. This means their DNA is still duplicated and chromatides have not divided yet. 48 chromosomes and 96 chromatids.

(k) A secondary oocyte arrested prior to fertilization: They are halted at metaphase II of meiosis, meaning they have half the chromosomes (24) but chromatids are still together (48).

(l) A second polar body: They suffered the same process as the mature ovum but remained with little cytoplasm. They have 24 chromosomes and 24 chromatids.

(m) A chimpanzee sperm: They have completed both meiosis as well, they have 24 chromosomes and 24 chromatids.  

Explanation:

7 0
3 years ago
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Slurring words together at a low level of volume and pitch is called
telo118 [61]
Mumbling- slurring words together at a very low level of volume and pitch so that they are barely audible.
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2 years ago
Which group do humans belong to?
goldenfox [79]
Humans belong to the D) chordates group
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4 years ago
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Why did we need to measure the absorbance at two wavelengths during the beta-galactosidase lab?
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While the absorbance at 420 nm is related to the amount of o-nitrophenol produced, the absorbance at 600 nm is proportional to cell density, which aids in standardizing our estimations of enzyme activity.

<h3>Describe absorbance.</h3>

The amount of light absorbed by a solution is measured by its absorbance (A), often referred to as optical density (OD). The amount of light that may flow through a solution is called its transmittance.

<h3>How is the activity of beta-galactosidase determined?</h3>

The colorless ONPG substrate is changed by beta-Galactosidase into galactose and the chromophore o-nitrophenol, which results in a vivid yellow solution. The amount of substrate transformed at 420 nm can be calculated by measuring the solution's beta-galactosidase activity using a spectrophotometer or a microplate reader.

<h3>What is measured by the beta-galactosidase assay?</h3>

The -Gal Assay Kit gives users the tools they need to swiftly assess the amounts of active beta-galactosidase expressed in cells that have been transfected with plasmids encoding the lacZ gene.

learn more about absorbance here

<u>brainly.com/question/26830274</u>

#SPJ4

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