Sand and mud that are carried and deposited by a river are____

Explain how nondisjunction can lead to aneulodies

tatuchka [14]

Answer:

nondisjunction can lead to aneulodies by results in daughter cells with abnormal chromosome numbers (aneuploidy). (if u want a more in depth answer just look it up and you'll find one, but i hope this was somewhat helpful lol)

5 0

2 years ago

Which of the following is not a piece of molecular evidence supporting the endosymbiotic theory?

Alchen [17]

A. the mitochondria and chloroplasts are much different in structure than bacteria

6 0

3 years ago

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0

3 years ago

How does the contractile vacuole in a single-celled organism function to maintain homeostasis?

dlinn [17]

Answer:

ifhdiuanv ;dj

Explanation:

7 0

2 years ago

An individual sustained a closed head injury in a rock climbing accident. In the ER the RN elevates the head of the bed at 30 de

mote1985 [20]

Answer:

Answered below

Explanation:

Head injuries may be penetrating head injury or closed head injury.

A major concern for health professionals regarding intracranial injuries resulting from closed head traumas, is the management of intracranial pressure. Elevated intracranial pressure have negative effects on the brain and immediate management of raised intracranial pressure is important to avoid cerebral edema, intracranial hematoma, brain herniation and even death.

The usual procedures in the management of head injury include elevating the head to about 30 degrees. This position promotes cerebral venous drainage and decreases intracranial venous pressure and intracranial pressure as well.

This is helpful because when the patient is left in the horizontal position, the intracranial pressure becomes significantly higher and the have fatal consequences.

8 0

2 years ago

Sand and mud that are carried and deposited by a river are_____