Trilobites are extinct marine invertebrates. Scientists found a trilobite fossil in

Fine the area a circle with a radius of 18cm

Likurg_2 [28]

Answer: A = 1017.88

Explanation:

3 0

3 years ago

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The genetic code is carried out by the ________ molecule in most organisms.

joja [24]

The genetic code is carried out by the Deoxyribonucleus acid DNA molecule in most organisms. Most of the viruses have RNA as their genetic material which codes genetic information.

7 0

2 years ago

After playing in an old shed, two boys tell their mother that they hurt all over. She checks and finds that they both have fever

irina1246 [14]

Answer:

Considering the boys' situation, the boys live in the rural areas of one of the western United States, e.g. California.

Explanation:

Hantavirus is spread by rodents and is most common in rural areas of the western United States during the spring and summer months. Hantavirus pulmonary syndrome also occurs in South America (especially in Argentina, Chile, Uruguay, Paraguay, Brazil, and Bolivia). Hantavirus cases have also occurred in Canada and some parts of Asia, where they reportedly caused kidney disorders rather than lung problems.

5 0

3 years ago

Determine the correct steps in the activation of pka, and then place them in the correct order, starting after the adenylyl cycl

Marina CMI [18]

After the adenylyl cyclase reaction (which convert ATP into cyclic adenosine monophosphate-cAMP), cAMP level is directly increased. Next steps of PKA activation include:

• cAMP binding to regulatory subunits of PKA, as a result conformational change in the regulatory subunits of PKA is induced,

• catalytic subunits are detached and unleashed by regulatory subunits and as a result activated.

4 0

2 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

2 years ago