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katrin2010 [14]
3 years ago
11

A normal bleeding time in association with normal platelet count, and increased prothrombin time (PT) and INR, is indicative of

a. hemophilia A. b. vitamin K deficiency. c. hemophilia B. d. idiopathic thrombocytopenia.
Biology
1 answer:
Fofino [41]3 years ago
8 0

Answer:

b. vitamin K deficiency.

Explanation:

A normal bleeding time in association with normal platelet count, and increased prothrombin time (PT) and INR, is indicative of vitamin K deficiency.

This ultimately implies that, when an individual is suffering from vitamin K deficiency, he or she would experience increased bleeding.

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22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
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See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

X^2  =  \frac{(O - E)^2}{E} where O = observed frequency and E = expected frequency.

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For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

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phenotype (3) = 3/16 x 293 = 54.94

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The X^2 is calculated thus:

Phenotype            O               E                             X^2

  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

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