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3 years ago

The plexus that travels through the axillary region is the

Biology

2 answers:

SCORPION-xisa [38]3 years ago

6 0

Answer:

The correct answer is: brachial plexus.

Explanation:

The brachial plexus is a net that is made of cervical nerves (more specifically, by the anterior rami of C5, C6, C7, C8 and T1) and extends from the cervicoaxillary canal to the armpit.

The nerves that form from the brachial plexus are both motor and sensory fibers that will provide innervation to the chest, the shoulder, the arm, the forearm, and the hand.

Dahasolnce [82]3 years ago

4 0

Answer:

Brachial plexus

Explanation:

Brachial plexus is a plexus of nerves that travels through the axillary region.

It is formed by ventral rami of spinal nerves C5-T1.

Five ventral rami forms roots/trunks that separate into divisions that then form cords that give rise to branches.Mojor nerves include Axillay, radial , musculocutaneous, Ulnar , median.

In the axilla it is placed lateral to the 1st part of axillary artery, it surrounds the second part of axillary artery, one cord lying medial to it, one lateral to it, and one behind it, at the lower part of axilla it gives off its terminal branches to the upper limb.

Hence, The plexus that travels through the axillary region is the brachial plexus.

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3 years ago

Read 2 more answers

In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

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3 years ago