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Andreyy89
3 years ago
5

If waste is to remain hazardous for a long period of time, how can society protect itself from problems as occurred at Love Cana

l?
Biology
2 answers:
andrew11 [14]3 years ago
5 0
Humans can stay away from the hazards or burn it(not sure if this is what u asked for)
muminat3 years ago
4 0

In 1978, people were relocated to safe areas. The problem of Love Canal was instrumental in the passage of the the Superfund Act in 1980. This law requires companies to be responsible for hazardous chemicals that they put into the environment and to pay to clean up polluted sites, which can often cost hundreds of millions of dollars. Love Canal became a Superfund site in 1983 and as a result, several measures were taken to secure the toxic wastes. The land was capped so that water could not reach the waste, debris was cleaned from the nearby area, and contaminated soils were removed.

Sometimes it's up to the residents in an area to recognize the effects of hazardous waste and to get the government to find the responsible party and initiate cleanup. .

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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

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

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

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

<u><em>Note</em></u>: 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. <em>If individuals, whos </em><em>genes assort independently,</em><em> are test crossed, they produce a progeny with equal </em><em>phenotypic frequencies 1:1:1:1</em>. <em>If</em> we observe a <em>different distribution</em>, that is that <em>phenotypes appear in different proportions</em>, we can assume that<em> genes are linked in the double heterozygote parent</em>.  

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 <em>gro+ro+</em> and <em>gro ro</em>, while the parental gametes are <em>gro+ro</em> and <em>gro ro+.</em>

So, the genotype, in linked gene format, of the double heterozygote individual in the <u>F1</u> 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 <u>genetic distance between genes,</u> is 0.011 x 100= 1.1 MU.

<u>Genetic Linkage Map:</u>

Parental Phenotypes)  

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

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

Recombinant phenotypes)

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

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

<u>If the genes were unlinked</u> 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
Which of the following are correct when comparing aerobic respiration and anaerobic respiration
nlexa [21]

Answer:

where is the following?

Explanation:

7 0
3 years ago
Who is known as the father of taxonomy?
natima [27]
Carolus Linnaeus.

Hope this helps!
5 0
3 years ago
Which of these factors makes nuclear energy unsustainable?
Strike441 [17]

Answer:Nuclear energy generates too much toxic waste.

Explanation:

got it right on the quiz

7 0
3 years ago
1. what are the 3 types of bonds?<br> 2. describe how each is formed.
ValentinkaMS [17]
Answer: ionic, covalent, metallic bonding

Ionic - ionic bond also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Covalent - occurs when pairs of electrons are shared by atoms.

Metallic - formed between positively charged atoms in which the free electrons are shared among a lattice of cations.


Hope this helped!
7 0
2 years ago
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