A group of Chinese women with PCOS who were receiving treatment saw several metabolic and hormonal imbalances improve after taking BBR. The primary consequences may be connected to the modifications in body composition in obesity and dyslipidemia. For the evaluation of the possible beneficial metabolic effects of BBR in women with PCOS, additional controlled trials are required.
PCOS:
- Objective: Insulin resistance is a common metabolic and reproductive condition called polycystic ovarian syndrome (PCOS) (IR). A quinoline derivative alkaloid known as berberine (BBR) has been utilized as an insulin sensitizer and is isolated from Chinese medicinal herbs. There may be a therapeutic benefit for PCOS from BBR. The purpose of this study was to compare the effects of BBR and metformin (MET) on the metabolic characteristics of PCOS-affected women.
- Design and methods: Ninety nine PCOS and IR patients were randomly assigned to one of three therapy groups: BBR+ compound cyproterone acetate (CPA;
), MET
CPA
, and placebo CPA 
for 
months. Before and after the treatment period, metabolic and hormonal measures as well as the clinical characteristics of the women were evaluated. - Results: Reduced waist circumference, waist-to-hip ratio (WHR;P<
), total cholesterol (TC), triglycerides (TG), and low-density lipoprotein cholesterol (LDLC; P<
), as well as increased levels of high-density lipoprotein cholesterol (HDLC) and sex hormone-binding globulin (SHBG; P<) were all observed following treatment with BBR as compared to MET. Similarly, BBR treatment decreased WHR, fasting plasma glucose, fasting insulin, homeostasis model evaluation for IR, area under the insulin curve, TC, LDLC, and TG (P<), while increasing HDLC and SHBG (P<) as compared to placebo.
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Correct question:
if a nondisjunction occurs at anaphase I of the first meiotic division, what will the proportion of abnormal gametes (for the chromosomes involved in the nondisjunction)?
Answer:
100%
Explanation:
Nondisjunction at meiosis-I means that two homologous chromosomes of at least one homologous pair fail to separate from each other during anaphase-I. This would result in the formation of one cell with one extra chromosome and the other with one less chromosome by the end of meiosis-I. Meiosis-II in these two cells would maintain this chromosome number in the daughter cells. Therefore, out of the total four gametes formed by the end of the meiosis, two would have one extra chromosome and would be denoted as "n+1". The rest of the two gametes would have one less chromosome and would be denoted as "n-1".
Answer:
Because of the fact that Snails cannot survive in nasty water.
Explanation:
