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Arlecino [84]
9 months ago
9

Haase D, Germing U, Schanz J, et al. New insights into the prognostic impact of the karyotype in MDS and correlation with subtyp

es: evidence from a core dataset of 2124 patients. Blood. 2007;110(13):4385-4395.
Biology
1 answer:
vivado [14]9 months ago
7 0

We have produced a huge, novel data set that incorporates morphologic, clinical, cytogenetic, and follow-up information from 2124 patients with myelodysplastic conditions (MDSs) at 4 organizations in Austria and 4 in Germany. Cytogenetic examinations were effectively acted in 2072 (97.6%) patients, uncovering clonal anomalies in 1084 (52.3%) patients.

Numeric and underlying chromosomal irregularities were reported for every patient and partitioned further as indicated by the quantity of extra anomalies. Along these lines, 684 unique cytogenetic classes were distinguished. The effect of the karyotype on the normal flow of the illness was concentrated on in 1286 patients treated with steady consideration as it were.

Middle endurance was 53.4 months for patients with ordinary karyotypes (n = 612) and 8.7 months for those with complex inconsistencies (n = 166). A sum of 13 uncommon irregularities were related to great (+1/+1q, t(1q), t(7q), del(9q), del(12p), chromosome 15 inconsistencies, t(17q), monosomy 21, trisomy 21, and - X), middle (del(11q), chromosome 19 peculiarities), or poor (t(5q)) prognostic effect, separately.

The prognostic significance of extra anomalies differed significantly relying upon the chromosomes impacted. For all World Health Organization (WHO) and French-American-British (FAB) order framework subtypes, the karyotype gave extra prognostic data. Our examinations offer new experiences into the prognostic meaning of uncommon chromosomal irregularities and explicit karyotypic mixes in MDS.

To learn more about prognostic impact of the karyotype in MDS from the given link

brainly.com/question/28189857

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A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density th
Volgvan

Answer:

0.153

Explanation:

We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.

So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.

ρVg = ρ'V'g + ρ"V"g

ρV = ρ'V'g + ρ"V"

Its new body volume = volume of water displaced, V = V' + V"

ρ(V' + V") = ρ'V' + ρ"V"

ρV' + ρV" = ρ'V' + ρ"V"

ρV' - ρ"V'  = ρ'V" - ρV"

(ρ - ρ")V'  = (ρ' - ρ)V"

V'/V" = (ρ - ρ")/(ρ' - ρ)

= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)

= (0.9988 g/cm³ ÷ 0.18 g/cm³)

V'/V" = 5.55

Since V = V' + V"

V' = V - V"

(V - V")/V" = 5.55

V/V" - V"/V" = 5.55

V/V" - 1 = 5.55

V/V" = 5.55 + 1

V/V" = 6.55

V"/V = 1/6.55

V"/V = 0.153

So, the fish must inflate its air sacs to 0.153 of its expanded body volume

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A researcher tests whether dosage level of some drug (low, high) causes significant differences in health. What is this type of
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Answer:

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2 years ago
Put the following steps in the correct order from 1-9
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Answer:

transcription of mRNA from DNA

small ribosomal subunit binds to mRNA

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Explanation:

The above describes the process of translation in the ribosome. After transcription of DNA to mRNA, the mRNA is taken to the ribosome to undergo translation, here the mRNA binds to the small ribosomal subuits and to other initiation factors; binding at the mRNA binding site on the small ribosomal subunit then the Large ribosomal subunits joins in.

Translation begins (codon recognition; initiating site) at the initiation codon AUG on the mRNA with the tRNA bringing its amino acid (methionine in eukaryotes and formyl methionine in prokaryotes) forming complementary base pair between its anticodon and mRNA's AUG start codon. Then translocation occurs with the ribosome moving one codon over on the mRNA thus moving the start codon tRNA from the A site to the P site, then codon recognition occurs (non-initiating site again) which includes incoming tRNA with an anticodon that is complementary to the codon exposed in the A site binds to the mRNA.

Then peptide bond formation occurs between the amino acid carried by the tRNA in the p site and the A site. When the ribosome reads a stop codon, the process stops and the polypeptide chain produced is released and the ribosomal subunits dissociates.

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Answer:

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