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lina2011 [118]
3 years ago
15

In taxonomy, each level of classification is referred to as a(an)

Biology
2 answers:
Neko [114]3 years ago
8 0
They are referred to lowering to one specific animal, includes their characteristics.

Domain
Kingdom
Phylum
Class
Order
Family
Genus
Species- the most specific
jok3333 [9.3K]3 years ago
3 0

rank or taxonomic catagory i think

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Do you see any correlation between the areas with high RNA-Seq read coverage (high peaks) and the different boxes in the tra-RA
Eddi Din [679]

Answer:

Yes, there is a correlation between the two areas. The areas with high RNA-Seq read coverage and the different boxes in the tra-RA isoform are related. The higher peaks are the translated regions (also known as the exons) while the smaller peaks are the untranslated regions (also known as the introns). That is the relationship between the areas.

Explanation:

Yes, there is a correlation between the two areas. The areas with high RNA-Seq read coverage and the different boxes in the tra-RA isoform are related. The higher peaks are the translated regions (also known as the exons) while the smaller peaks are the untranslated regions (also known as the introns). That is the relationship between the areas.

5 0
3 years ago
Does any one know what number 4 would be. Number 3 goes with number 4. <br> Please help.
natita [175]
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Chargaff discovered that: the tetranucleotide hypothesis was correct for any organism, the amount of A was equal to the amount o
emmainna [20.7K]

Answer:

the amount of A was equal to the amount of T

amount of C and the amount of G

5 0
4 years ago
An autosomal recessive disease has an incidence of 1/10,000. What is the approximate frequency of heterozygote carriers for this
goblinko [34]

Answer:

Option B

Explanation:

Given

Number of incidences or frequency of an autosomal recessive disease = \frac{1}{10000}

As per Hardy Weinberg's equation, frequency of a recessive genotype is q^{2}

q = \sqrt{\frac{1}{10000} }\\ q = \frac{1}{100}

As per first equation of Hardy Weinberg's -

p+q=1

so ,

p = 1-q\\p = 1-\frac{1}{100}\\p = \frac{99}{100}

p^2 = (\frac{99}{100})^2\\p^2 =\frac{9801}{1000}

As per second equation of Hardy Weinberg's -

p^{2} + q^{2} + 2pq=1

Substituting the given values in above equation, we get -

\frac{9801}{10000} + \frac{1}{100} + 2pq=1\\2pq = 1-(\frac{9801}{10000} + \frac{1}{100})\\2pq = 0.99%\\OR[tex]2pq=1%

2pq = \frac{1}{100}

Hence, option B is correct.

8 0
3 years ago
What determines the genotype of an organism
Brut [27]

The alleles that are passed from parents to offspring determines the genotype of an organism.

8 0
3 years ago
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