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Scrat [10]
3 years ago
10

How did the discovery of three categories of petite mutations in yeast lead researchers to postulate extranuclear inheritance of

colony size??
Biology
1 answer:
Nat2105 [25]3 years ago
4 0
<h3><u>Answer;</u></h3>

Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive petites followed non-Mendelian patterns that were consistent with the involvement of an extranuclear agent

<h3><u>Explanation</u>;</h3>
  • Mutations that yield defective mitochondria are expected to make cells grow much more slowly. These mutants were called petites to describe their small colonies compared to large wild type colonies.
  • Petite mutants could not grow when cells had an energy source requiring only metabolic activity of mitochondria - needed sugar as well which is part of glycolytic pathway.
  • Segregational petites, segregated in mendelian manner during meiosis. mutations cause defects in genes in cell nucleus encode proteins necessary for mitochondrial function.
  • Vegetative petite mutants do not segregate in mendelian manner; two types: neutral and suppressive; carry mutations in mitochondrial genome itself; when two yeast cells are mated, daughter cells inherit mitochondria from both parents.
  • Neutral petites lack most of their mitochondrial DNA; when mated with wildtype, the wildtype give their mitochondria so all cells display a normal phenotype.
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The answer is:
<span>Step 1 Genetic material duplicates.
</span>Step 2 Cell grows in size and other <span>organelles duplicate.
</span><span>Step 3 Cell wall or membrane forms.
</span>Step 4 Cells separate.

Binary fission is a type of cell division characteristic for prokaryotic cells. In this process, after the cell reaches its maximum size, the genetic material starts to duplicate. These two copies of genetic material are now connected to plasma membrane. The cell begins to grow in size and thus separate those two copies of genetic material. Other organelles are duplicated, too. Next, a new cell wall or cell membrane starts to grow in the middle of the cell leading to the separation of the cell. 
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3 years ago
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A 70-year-old man has a 90% blockage at the origin of the
bazaltina [42]

Option B

Middle colic arteries is the most likely additional  source of blood to the descending colon

<h3><u>Explanation:</u></h3>

The middle colic artery is a section of the superior mesenteric artery. Middle Colic provides the transverse colon, has a branch that runs to the right colon and branch that persists to the left colon that catches up with the left colic artery and proceeds as the marginal artery to provide the descending colon.

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How do ocean temperatures affect local weather?
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8 0
3 years ago
3. Look at the pictures beow and answer the following questions,
tankabanditka [31]

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5 0
3 years ago
A population is made up of individuals where 77 have the A1A1 genotype, 65 have the A1A2 genotype, and 123 have the A2A2 genotyp
lana [24]

Answer:

Frequency of allele A1- 0.41

Explanation:

In Hardy weinberg equilibrium,

P refers to the dominant allele

q refers to the recessive allele

The allele frequency will be p+q=1

The genotypic frequency is- P²+q²+2pq=1

P²= genotype of dominant trait ( A1A1)- 77

2pq= genotype of heterozygotes (2pq)- 65

q²= genotype of recessive trait (A2A2)-  123

Total number of offsprings= 77+  65+ 123

                                            = 265

Now to calculate allele frequency of A1=

\dfrac{\text{number of A1A1 offspring}}{ \text{total number of offspring}}+\dfrac{1}{2}(\dfrac{\text{number of A1A2 offspring}}{\text{total number of offspring}})

= 77/265 + 1/2( 65//265)

=  0.290+ 0.122

= 0.413

Thus, 0.41 is correct.

3 0
3 years ago
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