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lapo4ka [179]
3 years ago
11

The dsm-5 distinguishes between _____ forms of neurocognitive disorders.

Biology
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
<span>The DSM-5 (or DSM V) distinguishes between mild (</span>slight cognitive impairment)<span> and major (full out dementia) forms of neurocognitive disorders.

The DSM-5, (Diagnosis and Statistical-manual of Mental-disorders 5th edition), was published in 2013 by the American Psychiatric Association (APA). Mostly used by psychiatrists to classify their patients' disease.
In the chapter of DSM-5: Neurocognitive Disorders, it was added the diagnoses of mild neurocognitive disorder and major neurocognitive disorder (this is not present in the DSM 4 (1993)).
</span>
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A relationship among alleles where both alleles contribute to the phenotype of the heterozygote is
Zarrin [17]

A relationship among alleles where both alleles contribute to the phenotype of the heterozygote is codominance.

This is a phenomenon when both alleles are fully expressed. As a result, offspring will have a phenotype that is neither dominant nor recessive, it is combination of both. For example, if one allele contributes to white color of flower, another allele contributes to red color, the offspring (if we cross those two flowers) will have pink flowers (both phenotypes expressed).


7 0
3 years ago
Describe some specific failures of gene therapy trials, which have negatively affected public opinion and slowed the development
sesenic [268]

Explanation:

Gene therapy is the therapeutic delivery of the nucleic acid into the patients cells in the form of drug to treat disease. It is a complex treatment involves more potential for the treatment of the monogenic diseases and is also associated with a lot of significant risks . The children who were suffering from Wiskott- Aldrich syndrome which is a rare disease and were treated as the part of the clinical gene therapy trial which was carried out in the Germany. After treatment, health of children improved significantly. Then after one to three years following the gene therapy, seven out of ten children developed the cancer in the blood .

Public response towards gene therapy and acceptability of changing the genes varied a good deal from one scenario to the another. Many more techniques needed to be developed and more information of the diseases that need to be understood before the gene therapy which involve changes on the genetic setup of the body , it raises many unique ethical concerns. For decades researches are done to bring the gene therapy to clinic , still very less patients are given any effective gene therapy treatments . Even though gene therapy has been slow to reach patients, it's future is encouraging.

7 0
3 years ago
A species of bacteria is found deep within the Earth's crust. Which process will the bacteria use to obtain energy? A. Cellular
notka56 [123]

Answer:

CELLULAR RESPIRATION

Explanation:

7 0
3 years ago
Whats so important about the nitrogen and hydrogen in the amino acids?
hammer [34]
The acronym CHNOPS (carbon, hydrogen<span>, </span>nitrogen<span>, oxygen, phosphorus, sulfur), represents the six most </span>important<span> chemical elements whose covalent combinations make up most biological molecules on Earth. Sulfur is used in the </span>amino acids are proteins<span> cysteine and methionine.</span>
6 0
3 years ago
The trait for 'male-pattern baldness' is a recessive trait encoded for by "b". Non-balding is encoded for by a dominant allele e
wariber [46]

Answer:

<h2>Allele frequencies for B and b. </h2><h2>"b"(q) allele frequency = 0.60 </h2><h2> </h2><h2>"B"(p) allele frequency = 0.40 </h2><h2></h2><h2>Genotype frequencies; </h2><h2>BB = 0.16 </h2><h2>Bb = 0.48 </h2><h2>Bb = 0.36 </h2>

Explanation:

Given

Non-baldness (B) is dominant on baldness(b), so B is dominant over b.

Homozygous  pattern baldness male (bb) = 360,

Heterozygous  non- baldness (Bb)=  480,

Homozygous non-baldness (BB)= 160.

So, we can also denote then by genotypes  only,

BB= 160;

Bb= 480;

bb= 360;

Total= 1000

Allele frequency q² (bb) = 360/1000=0.36

allele fequency for q( b)= √o.36=0.60.

Allele frequency for p²(BB) = 160/1000=0.16

allele frequency p(B)= √0.16 = 0.4

 

Expected genotype frequencies;

BB = 160/1000  = 0.16

 

Bb = 480/1000= 0.48

 

Bb = 360/1000= 0.36

6 0
4 years ago
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