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MatroZZZ [7]
2 years ago
7

Patient presents with no menses and positive pregnancy test but an ultrasound reveals no uterine contents. an embryo has implant

ed on left ovary and this is treated with laparoscopic oophorectomy. what icd-10-cm code(s) is/are reported for this procedure?
Biology
1 answer:
Gnoma [55]2 years ago
8 0

As per the American version of ICD-10-CM Diagnosis code - 2019, Code O00.202 is the representation of pregnancy of left ovary without pregnancy with in the uterus.  

The part of the code “O00” represents an ectopic pregnancy which means a pregnancy state in which the embryo is not implanted in the uterus. However it may be implanted in the fallopian tube, ovary etc.  

The code  58661 and 49321-51 represents the total or partial oophorectomy which includes process of examination of Oviduct/Ovary using Laparoscopic Procedures

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What could happen to the reproductive process of mosses during a drought? The flagellated sperm would not be able to swim to the
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The Karez well system _______.
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c. transported groundwater from a distance

Explanation:

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small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

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

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

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Answer:

D Euglena

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What are the products of cellular respiration
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