The side of a mountain range that faces the wind often receives more what than the downwind side of the same range?

Which of these actions is most responsible for the current decline of elephant

Gennadij [26K]

Answer:

The correct answer is - option A.

Explanation:

Elephant populations are currently declining lapidary. Human influence is the main and major reason that leads to such a population decline of the elephants. The habitat loss, environmental change, and killing for human safety are the few reasons that affect the population of the elephant, but the major and most responsible cause of the killing and decrease in the elephant population is killing for the ivory of elephants,

Elephant ivory is very expensive and use for making jewelry, artifacts and many more items even illegal medicines are made by ivory.

Thus, the correct answer is option - A.

3 0

3 years ago

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What is the function of these organelles? endoplasmic reticulum: Golgi body: mitochondrion: ribosome:

liraira [26]

Answer:

Endoplasmic Reticulum: help form and store proteins.

Golgi Body: sorting and processing proteins.

Mitochondria: converts food (or protein) into energy.

Ribosome: synthesize proteins for use throughout the cell.

Hope this helps

-Amelia

7 0

3 years ago

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If a ⊥ b and b ∥ c, then _____

trasher [3.6K]

Answer:

a is perpendicular to c will be the answer.

8 0

3 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

What would be the independent variable in the following hypothesis?

Tamiku [17]

The independent variable in the hypothesis of an observed increased in sea turtle sickness once an increase in red tide can be seen would be the increase of red tide.

The reason for this is that red tide increases or decreases will happen independently of the amount of sea turtle sickness - therefore increasing sea turtle sickness will be dependent on the first variable. (independent one).

7 0

3 years ago