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2 years ago

The model of exponential population growth predicts that the per capita population growth rate, r Select one:_______.

Biology

1 answer:

lawyer [7]2 years ago

5 0

<h2>Exponential Population Growth</h2>

Explanation:

The model of exponential population growth predicts that the per capita population growth rate, r :

d. does not change as the population gets larger.

A population shows exponential growth if there is no limitation on growth. All members have abundant resource availability and the population displays its intrinsic rate of increase.This is also known as geometric growth .In geometric growth, rate of increase is expressed as a constant fraction or an exponent by which a particular population is multiplied.

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2 years ago

In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eye

Korolek [52]

Explanation:

We have three genes in the same chromosome and are trying to determine their order and relative distance to each other.

The genes and possible alleles are:

claret (c/c+)
spineless (s/s+)
hairless (h/h+)

All mutations are recessive: two copies of the mutant allele are needed for the fly to show that trait.

Parental cross

claret hairless male: $\frac{s^+\ c\ h}{s^+\ c \ h}$
spineless female: $\frac{s\ c^+\ h^+}{s\ c^+ \ h^+}$

Each parent can produce 1 type of gamete only, so the F1 will be homogeneous:

F1

$\frac{\ s\ \ \ c^+\ \ h^+}{s^+\ \ c \ \ \ h}$

During meiosis, the F1 females can produce 8 types of gametes: 2 parentals and 6 recombinants (two of them, the result of a double crossing over).

If they are test crossed to homozygous recessive males (which can only produce a $\frac{s \ c\ h}{}$ gamete), the following phenotypes are obtained (I just write the alleles they inherited from the female fly, as the ones that came from the male are the same for all of them):

321 spineless (s c+ h+) ----> Parental
309 claret, hairless (s+ c h) ----> Parental
130 claret, spineless (s c h+) ----> Recombinant
140 hairless (s+ c+ h) ----> Recombinant
32 hairless, claret, spineless (s c h) ----> Recombinant
38 WT (s+ c+ h+) ----> Recombinant
18 claret (s+ c h+) ----> Double Recombinant
12 hairless, spineless (s c+ h) ----> Double Recombinant

The phenotypes observed in the highest frequency are always the parentals, and the ones in the lowest frequency are always the double recombinants.

To determine the order of the genes:

we have to write down the genotype of the F1 female three times, changing the order of the genes each time.
Then, we hypothesize what the double recombinant gametes would look like.
When the theoretical double recombinants we obtain are the same as the ones observed in the F2, we know that that is the correct order of the genes.

In this problem, only if the middle gene is h+/h the double crossing over gives us the observed double recombinant gametes, therefore hairless is the middle gene.

$\frac{s\ h^+\ c^+}{s^+\ h\ c}$

Double recombinants:

s h c+ ----> spineless hairless
s+ h+ c ----> claret

To determine the distance between the genes:

Genetic distance (m.u.) = Recombination Frequency x 100

Distance between the spineless and hairless genes:

$Distance \ [s-h]= \frac{number\ of\ recombinants \ [s-h]}{Total number of individuals} * 100\\\\\\Distance \ [s-h]= \frac{32+38+12+18}{1000} * 100\\\\Distance \ [s-h]= 10\ map\ units$

Distance between the hairless and claret genes:

$Distance \ [h-c]= \frac{130+140+12+18}{1000} * 100\\\\Distance \ [h-c]= 30\ map\ units$

<h3>The gene map for these genes is:</h3>

spineless -----------------hairless ---------------------------claret

10 m.u. 30 m.u.

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1 year ago