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3 years ago

Which of the following is a disadvantage of sexual reproduction?

2 answers:

6 0

Answer: A) Slow to repopulate

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Genetic variation is an advantage of sexual reproduction. It increases survival and the ability to adapt in a species. Species with genetic variation can respond better to environment changes, since the genetic variability allows some to be immune to disease or adapt better to a change, and live to reproduce.

Slow to repopulate is a disadvantage because the species cannot recover quickly after a large disaster. It also means they cannot adapt as quickly to an environmental change.

densk [106]3 years ago

4 0

Yeah the answer is A. because b is a advantage lol mark the other person brainly...

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NEED THIS ASAP ill mark brainliest Explain how the medical community can use computer-modeling tools, like the one shown in the

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Answer:

here u go

Explanation:

The modeling tool shows exactly how different body parts form. Doctors can use models, like the one in the video, along with ultrasound pictures of a developing baby, to determine whether the embryo is growing normally and is forming the proper body structures at the correct time. If not, the doctor could inform the parents of the problem before the baby is born. Doctors could then make a plan to treat the condition, if possible.

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2 years ago

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2 years ago

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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

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.

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Why do doctors try to remove tumors from the body?

nika2105 [10]

Doctors try to remove tumors from the body to prevent it from spreading and affecting other parts of the body.

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Answer:

Q.1: I can't help you with this, sorry :(

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Q.3: Phytoplankton is the second-order consumer because they eat first-order consumers.

Q.4: Whelks and crabs because they eat limpets, which eat producers, and they also eat seaweed.

Q.5: Gulls are carnivores because they eat the crabs, and so are crabs because they eat mullets

Q.6: Limpets and lobster would become less populated, but not yet endangered. Gulls would starve and probably disappear from this ecosystem.

Q.7: Whelks' numbers would decrease because of the number of lobsters consuming them, but then lobsters would starve because of the decline in their food. Then this would repeat, shaking the whole ecosystem.

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2 years ago

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