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tresset_1 [31]
3 years ago
9

Water conservation. What do the words bring to mind? Doing without? In reality it can be done by making very minor changes aroun

d the home. Take a look at the bar graph showing how a typical family uses water at home. Which practices described below could practically help to conserve water use at home?
Don't flush the toilet very often. Do not bathe unless you are very dirty.


Stop using the dishwasher to wash dishes. Only take a bath on the weekends.


Don’t leave your tap running while brushing your teeth, shaving, or washing the dishes. Fix leaky faucets and pipes right away.


Use environmentally friendly products whenever you can. When you must use environmentally hazardous products, be sure that you dispose of them properly.

Biology
2 answers:
ira [324]3 years ago
7 0

The correct answer is (c) Don’t leave your tap running while brushing your teeth, shaving, or washing the dishes. Fix leaky faucets and pipes right away.

As seen in the bar graph that huge amount of water is wasted in the domestic ways. By using the water in adequate amount and saving the water that is wasted while running the tap while brushing and fixing the leaks as soon as possible will save a huge amount of water. The bar graph is covering maximum wastage of water from these sections only. By reducing the water used in brushing, leaky faucets and others can save maximum water.


Alla [95]3 years ago
7 0
<span>Don’t leave your tap running while brushing your teeth, shaving, or washing the dishes. Fix leaky faucets and pipes right away.
</span>Most of the water is lost through the taps that are left running.
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The overuse of antibiotics. is a contributing reason behind emerging disease?
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What was the predator of the stickleback in loburg lake?
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The correct answer is - C. Humans.

The stickleback in Loburg Lake, Alaska, is not a native species of the lake. It is thought that this fish was introduced to this lake somewhere between 1984 and 1989. The reason for this kind of suggestion is that the stickleback in this lake had full armor like the stickleback in the seas, which is not something found among the lake populations of this species. Also, the full armor had gradually been reducing with each new generation, and in the present, the stickleback in this lake looks totally the same as the other lake populations.

The stickleback did not had any predators in Loburg Lake, thus the losing of the full armor as it was not needed, instead the stickleback was the predator. But even though it didn't had a predator in the water, the humans became its predator through the recreational fishing that takes place on this lake.

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The moon exerts a greater influence then the sun because it is closer . True or false
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In the garden pea, several different genes affect pod characteristics. A gene affecting pod color (green is dominant to yellow)
Arlecino [84]

Answer:

  1. Green/Wide/Long, G-W-L- = 186 individuals
  2. Yellow/narrow/ long, gg ww L-  = 186 individuals
  3. Green/wide/ short, G-W-ll = 186 individuals
  4. Yellow/narrow/ short, gg ww ll  = 186 individuals  
  5. Green/narrow/ long, G-wwL-  = 14 individuals
  6. Yellow/wide/ long, gg W-L- = 14 individuals
  7. Green/narrow/short, G-wwll = 14 individuals
  8. Yellow/wide/ short, ggW- ll = 14 individuals

Explanation:

<u>Available data</u>:

• One gene affecting pod color (green is dominant to yellow)  

• One gene affecting pod width (wide is dominant to narrow)

• Both genes are located on chromosome 5 and approximately 7 mu away from each other.  

• A third gene, located on chromosome 4, affects pod length (long is dominant to short).  

• Cross: A true-breeding wild-type plant (green, wide, long pods) was crossed to a plant with yellow, narrow, short pods.  

• The F1offspring were then test-crossed to plants with yellow, narrow, short pods.  

• The testcross produced 800 offspring.

<u>Cross 1: </u>

Parental) GGWWLL   x   ggwwll

F1) GgWwLl

<u>Cross 2:</u>

Parental) GgWwLl    x    ggwwll

F2) N=800

To calculate the numbers of the F2 generation, we first need to calculate the frequencies of recombination and parental of the linked genes. We know that they are 7MU apart from each other. <em>The map unit is the distance between a pair of genes for which every 100 meiotic products one results in a recombinant one.  </em>

MU = 7 means that there is 7% of recombination for genes that express color and width. This is:

MU = 7 = 7% recombination  

We have two possibilities of recombination: green/narrow and yellow/wide. Each of these recombinants has half of the possibilities of occurring, so:

7 map units = 7 % of recombination in total  

                     = % Gw + % gW  

                     = 3.5 % + 3.5 %

Gw = 3.5% = 0.035  

gW = 7%/2 = 0.035

For the parentals, we can calculate

100% - 7% = 93% of parental in total  

                  = % of GW + % gw  

                 = 46.5% + 46.5%

GW = 46.5% = 0.465

Gw = 46.5% = 0.465

The length gene is located in a different chromosome so it assorts independently. This is, of the gametes will be long, L, and the other 50% will be short, l.  

L = 50% = 0.5

L = 50% = 0.5

Now we need to figure out how to relate these frequencies. All we need to do is to multiply the frequencies of occurrence obtained previously for linked and independent genes, for each possible phenotype. This is:

  • Green/Wide, GW = 0.465
  • Yellow/narrow, gw = 0.465
  • Green/Narrow, Gw = 0.035
  • Yellow/Wide, gW = 0.035
  • Long, L = 0.5
  • Short, l = 0.5

<u>Phenotypic frequencies: </u>

  • Green/Wide/Long, G-W-L- = GW x L = 0.465 x 0.5 = 0. 2325
  • Yellow/narrow/ long, gg ww L- = gw x L = 0.465 x 0.5 = 0. 2325
  • Green/wide/ short, G-W-ll = 0.465 x 0.5 = 0. 2325
  • Yellow/narrow/ short, gg ww ll = 0.465 x 0.5 = 0. 2325  
  • Green/narrow/ long, G-wwL-  = 0.035 x 0.5 = 0.0175
  • Yellow/wide/ long, gg W-L- = 0.035 x 0.5 = 0.0175
  • Green/narrow/short, G-wwll = 0.035 x 0.5 = 0.0175
  • Yellow/wide/ short, ggW- ll= 0.035 x 0.5 = 0.0175

Finally, as we need to obtain the numbers of the individuals with those phenotypes, we need to multiply each frequency by N, which is the total number of individuals in the F2 (N = 800).

  • Green/Wide/Long, G-W-L- = 0. 2325 x 800 = 186 individuals
  • Yellow/narrow/ long, gg ww L- = 0. 2325 x 800 = 186 individuals
  • Green/wide/ short, G-W-ll = 0. 2325 x 800 = 186 individuals
  • Yellow/narrow/ short, gg ww ll = 0. 2325 x 800 = 186 individuals  
  • Green/narrow/ long, G-wwL-  = 0.0175 x 800 = 14 individuals
  • Yellow/wide/ long, gg W-L- = 0.0175 x 800 = 14 individuals  
  • Green/narrow/short, G-wwll = 0.0175 x 800 = 14 individuals
  • Yellow/wide/ short, ggW- ll = 0.0175 x 800 = 14 individuals

3 0
3 years ago
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