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3 years ago

What evidence can be examined when determining common ancestry amongst hominids?

1 answer:

6 0

We examined the biogeographic patterns implied by early hominid phylogenies and compared them to the known dispersal patterns of Plio-Pleistocene African mammals. All recent published phylogenies require between four and seven hominid dispersal events between southern Africa, eastern Africa, and the Malawi Rift, a greater number of dispersals than has previously been supposed. Most hominid species dispersed at the same time and in the same direction as other African mammals. However, depending on the ages of critical hominid specimens, many phylogenies identify at least one hominid species that dispersed in the direction opposite that of contemporaneous mammals. This suggests that those hominids may have possessed adaptations that allowed them to depart from continental patterns of mammalian dispersal.

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What happens in prophase.

Reptile [31]

During prophase, the parent cell chromosomes — which were duplicated during S phase — condense and become thousands of times more compact than they were during interphase. ... Cohesin forms rings that hold the sister chromatids together, whereas condensin forms rings that coil the chromosomes into highly compact forms.

4 0

3 years ago

Two students are given cubic boxes, measuring 10 cm on a side. Robert puts a single glass marble with a

Katyanochek1 [597]

Answer:

Both boxes have the same amount of empty space

Explanation:

Given

Cube Box

$Length = 10cm$

Robert:

1 glass marble

$Diameter = 10cm$

Susan:

1000 glass marble

$Diameter = 1cm$

Required

Whose box has more empty space?

First, we need to calculate the volume of the cubic box.

$Volume = Length^3$

Substitute 10cm for Length

$Volume = 10^3$

$Volume = 1000cm^3$

Next, calculate the volume of Robert's glass

Volume is calculated as:

$Volume = \frac{4}{3}\pi r^3$ ---- Volume of a sphere

Where

r = radius

$r = \frac{1}{2} * Diameter$

$Diameter = 10cm$ --- Given

Substitute 10 for diameter

$r = \frac{1}{2} * 10cm$

$r = 5cm$

So:

$Volume = \frac{4}{3}\pi r^3$

$Volume = \frac{4}{3} * \frac{22}{7} * 5^3$

$Volume = \frac{4}{3} * \frac{22}{7} * 125$

$Volume = 523.8cm^3$

Next, we determine the volume of Susan's 1000 glasses

Volume is calculated as:

$Volume = \frac{4}{3}\pi r^3$ ---- Volume of a sphere

Where

r = radius

$r = \frac{1}{2} * Diameter$

$Diameter = 1cm$ --- Given

Substitute 1cm for diameter

$r = \frac{1}{2} * 1cm$

$r = 0.5cm$

Substitute 0.5cm for radius in $Volume = \frac{4}{3}\pi r^3$

$Volume = \frac{4}{3} * \frac{22}{7} * 0.5^3$

$Volume = \frac{4}{3} * \frac{22}{7} * 0.125$

$Volume = 0.5238$

But there are 1000 glasses.

So, the volume of 1000 glasses is:

$Volume = 0.5238 * 1000$

$Volume = 523.8cm^3$

--------------------------------------------------------------------------------------------------------------

For Robert:

$Volume = 523.8cm^3$

For Susan:

$Volume = 523.8cm^3$

Since, they have the same volume.

Both boxes have the same amount of empty space

5 0

3 years ago

The episome in an Hfr strain is inserted near the trp (tryptophan operon) locus. This Hfr strain is grown with an F- Trp- strain

Mekhanik [1.2K]

Answer:

It is a typical Hfr mating, as trp gene is very closed to F plasmid, the Hfr bacteria donates its trp gene along with F plasmid and then process is disturbed.

Explanation:

5 0

3 years ago

Need help explaining asap please :(

vivado [14]

Answer:

Explanation:

3 0

3 years ago

The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. Given that the ste

nevsk [136]

Answer:

Approximately 6944 glucose residues are added enzymatically per second

Explanation:

Cellulose is the main structural polysaccharides in plants. It is composed of unbranched glucose monomer units linked to each other by beta 1-4 glycosidic bonds.

The cell wall and stem of plants cells are composed of cellulose fibers. They provide rigidity and support to the plant.

In the given bamboo plant, the enzymatic addition of glucose units to the growing cellulose fiber chains results in the phenomenal growth rate of the bamboo stem.

Since each glucose unit contributes ~0.5 nm to the length of a cellulose molecule, number of glucose units required for daily growth is calculated as follows:

0.5 nm = 10⁻⁹

0.3 m/0.5 x 10⁻⁹ m = 600000000 units of glucose per day

Number of seconds in a day = 24 * 60 * 60 = 86400 seconds

Number of glucose residues added per second = 600000000/86400

Number of glucose residues added per second = 6944.4 glucose molecules per second

Therefore, approximately 6944 glucose residues are added per second

8 0

3 years ago