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3 years ago

How would results vary between observing a population directly as opposed to observing them indirectly?

Biology

2 answers:

ioda3 years ago

3 0

Answer and Explanation:

Observing population directly gives you accurate result .

Direct observation does not need any intermediary as they themselves go for survey.

So, it is cost efficient.

Whereas indirect observing may involve manipulations.

It requires someone or many to hire who helps in surveying.

It incurs cost.

Thus, direct observation is always better than indirect observation.

kondaur [170]3 years ago

3 0

Answer:

On the subjects, direct observation can be made by noticing the behaviors they possess in their natural habitat. In order to gather the information, the investigator can either hide to prevent any intrusion or can be either present physically to record the observations with the help of a camera. For example, recording or watching a chirping bird.

On the other hand, indirect observation comprises recording of the traces left behind by the subject to comprehend the cause behind the behavior demonstrated by the subject. For example, recording the distribution of the bird droppings to determine their movement.

Therefore, it can be concluded that direct observation helps in determining the behavior in a population and an indirect observation will assist in developing the cause behind the witnessed behaviors in the population.

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A student in genetics class is studying the inheritance patterns in fruit flies. He is looking at the traits for wing length, lo

Step2247 [10]

Answer:

LLgg, Llgg , two out of 16

Explanation:

Given,

Dominant - Long wings and Gray color

Recessive - Short wings and Brown color

Let the allele for long wings be "L" and the allele for short wings be "l".

Let the allele for brown color be "g" and allele for gray color be "G"

Genotype of parents

LlGg

Gametes of the parent

LG, Lg, lG, lg

Dihybrid cross is between LlGg and LlGg

The offspring produces will be as follows -

LG Lg lG lg

LG LLGG LLGg LlGG LlGg

Lg LLGg LLgg LlGg Llgg

lG LlGG LlGg llGG llGg

lg LlGg Llgg llGg llgg

Offspring with long wings and brown color – LLgg, Llgg , two out of 16

6 0

3 years ago

An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result

Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

EW EW EW EW

ew EWew EWew EWew EWew

All offspring are (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If EWew × eeww

we have; EW & ew × ew & ew

EW ew

ew EeWw eeww

offspring that

are linked in ⇒ EeWw EeWw & eeww eeww

F₂ will be

$\frac{1}{2}$ = 50% of EeWw of the total 100 offspring in the F₂ cross

$\frac{1}{2}$ = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes = 50% EeWw and 50% eeww.

For unlinked genes; If EWew × eeww

if rearrangement occurs in EWew and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

EW Ew eW ew

ew EeWw Eeww eeWw eeww

We have the following results for the unlinked genes

$\frac{1}{4}$ = EeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = Eeww 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeww 25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

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