Question

Later Consider the molecular structure of a disaccharide. Identify the type of glycosidic linkage in the disaccharide. OH OBI — OH HT V/ OBA AVEL VH OH HA o1ОН H OHH OH Label the structures to show the monosaccharide units produced by hydrolysis of the disaccharide. .. Answer Bank OH CH, OH

Answer 1

Type of glycosidic linkage present in the disaccharide is

This Disaccharide molecule is referred a D-(+)-maltose. The Glycosidic linkage present in this Disaccharide is

$\alpha$(1$\rightarrow$4)

Labelling the Structure of monosaccharides produced from the hydrolysis of disaccharides

In this reaction Maltose undergoes hydrolysis to give two units of $\alpha$-glucose.
СН2ОН 1 — bH — Но 1 ОН ОН x- glucose

СН2ОН т EI OLU но бH L-glucose

42. As it is mentioned that 420 adult samples chosen is random, value of vector p will be impartial. The Behavior of the samples is autonomous, because the chosen 420 adults is lower than 10% of overall adults. In 420 adult samples, 181 of them know this company manufactures printers, so they come under the success value. The remaining 239 failures, and the success 181 are larger than 10. In case there will be a normal Sample Distribution..

Therefore H₀  : p = 0.40   

H_A : p > 0.40

0.340.360.38 0.५००.420.44.46

SD(vector p) = $\sqrt{}$0.40 *0.60 / 420 = 0.0239

vector p = 181 / 420 = 0.4310

Z = 0.4310 - 0.40 / 0.239

= 1.2971

p = normalised value of f(1.2971, 99) = 0.973

As p value greater than $\alpha$ value ($\alpha$ = 0.05), null value can be denied. Due to the Lack of Statistical proof, that the percentage is more than 40%, I will not prescribe to run the advertisements.