Why should the ureter be less robust than the vas deferens

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

3 years ago

I hate me self I’m so sad

Dmitry [639]

Answer:

be happy

Explanation:

get enough rest and eat good

8 0

2 years ago

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What did you learn about the channel on Mars from the new NASA data? After you look at the new evidence card about Mars, describ

pentagon [3]

Explanation:

The new research carried out by NASA on mars suggets that many of the Mars' ancient channel have been carved by flowing water, were actually formed by melt water beneath glacial ice sheets. The rocks looked familiar to the rocks found on Earth in these water channels. Moreover, conglomerate rock was found near the base of the channel on Mars.

6 0

2 years ago

There is a population where the frequencies of allele 1 and allele 2 are 0.7 and 0.3, respectively. Allele 1 has a selection coe

Lorico [155]

Answer:

0.8

Explanation:

There is a population where the frequencies of allele 1 and allele 2 are 0.7 and 0.3, respectively

Let's use GG to represent allele 1

Let's use gg to represent allele 2

So we can equally say that;

GG = p = 0.7

gg = q = 0.3 ( from Hardy-Weinberg Equilibrium)

So, given that the selection coefficient = 0.2

We known that the cross between GG and gg will definitely results to (GG,Gg and gg)

Then the fitness of these genes can be represented as:

1 - s, 1 and 1 - t respectively.

Thus. the allele 1's genotype fitness can be determined as

= 1 - s ( where s is the selection coefficient)

= 1 - 0.2

= 0.8

6 0

3 years ago

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How does the burning of fossil fuels and deforestation contribute to the overall amount of carbon dioxide?

Vaselesa [24]

Answer:

Forests store large amounts of carbon. When fossil fuels are burned, carbon from dead and decayed plants, animals and phytoplankton that lived hundreds of millions of years ago (before dinosaurs existed), is released into the atmosphere in the form of carbon dioxide. .

6 0

2 years ago