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1 year ago

Not all inspired air reaches the alveoli; 150 ml is stopped in the airway leading to the alveoli. this is called?

1 answer:

5 0

Not all inspired air reaches the alveoli; 150 ml is stopped in the airway leading to the alveoli. This is called dead air space.

Alveoli are the tiny air sacs present all over the lungs. These are supposed to increase the surface area of the lungs so as to increase the amount of inhaled air. The exchange of oxygen and carbon dioxide takes place through these alveoli.

Dead air space is the unventilated area where air circulation does not take place. This air is supposed to be get trapped somewhere in the respiratory pathway or it reaches an alveoli that is poorly perfused. The dead air space can be of two types: anatomical dead space and physiologic dead space.

To know more about dead air space, here

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3 years ago

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3 years ago

In the laboratory, a genetics student crossed flies with normal, long wings to flies with mutant, dumpy wings, which she believe

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Answer:

a. Chi-Square Value = 0.12

b. Degrees of freedom = 1

c. Reject or fail to reject the hypothesis = fail to reject the hypothesis

Explanation:

<u>Available data:</u>

Cross: long wings to flies with mutant, dumpy wings
F1) 100% long-winged flies
F2) 672 long-winged flies 218 dumpy-winged flies
Hypothesis: the dumpy wing is inherited as a recessive trait

Let us name the allele for Long L and the allele for dumpy l

The number of observed individuals:

Total number: 890 flies (672 + 218)
Long-winged = 672
Dump-winged = 218

In the F2 generation, if l was a recessive allele and L a dominant allele, exhibiting complete dominance, we would expect a 3:1 phenotypic ratio.

Cross: Ll x Ll

Gametes) L l L l

Punnett square) L l

L LL Ll

l lL ll

F2) 3/4 LL + Ll long-winged individuals

1/4 ll dump-winged individuals

From this ratio and the total number of individuals in the F2 generation, we can get the expected number of individuals:

4 possible genotypes ------- 890 individuals

3 LL + Ll genotype ------------X = 667.5 long-winged individuals

1 ll genotype ---------------------X = 222.5 dump-winged individuals

Chi square = ∑ (O-E)²/E

LL + Ll = (O-E)² /E

LL+ Ll = (672 - 667.5) ² / 667.5

LL + Ll = 0.03

ll= (O-E)²/E

ll= (218 - 222.5)² / 222.5

ll= 0.09

Chi square= ∑ ((O-E)²/E) = 0.03 + 0.09 = 0.12
Freedom Degrees = genotypes - alleles = 3 - 2 = 1
Significance level = 0.05

Table value/Critical value = 3.84

0.12 < 3.84 meaning that there is enough evidence to reject the null hypothesis.

Fail to reject the hypothesis. The genotypes in the population might be in equilibrium H-W. L is the dominant allele, and l is the recessive allele. The dumpy wing is inherited as a recessive trait.

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