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4 years ago

How can i explain the answer

Biology

1 answer:

barxatty [35]4 years ago

4 0

He could have developed the trait due to random mutation of his DNA.

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Twin research studies looking at the genetic basis of personality revealed that the trait of _____ has a larger genetic componen

Colt1911 [192]

The correct answer is social potency, achievement.

Twin studies are commonly used to study the environmental or genetic effect on the development of personality or general psychological traits. Social potency is a term referring to the need for social dominance and having an impact on other people. It has been shown that the personality trait related to the need for achievement does not have a strong genetic component. This is contrary to the trait of social potency which was found to be greatly affected by the genetic background.

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4 years ago

Does pineapple eat us?

jolli1 [7]

Answer:

yea

Explanation:

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3 years ago

In the first stage of reproduction what disappears?

AnnyKZ [126]

The process of cell reproduction consists of five stages. The first stage is called prophase. This stage occurs when the chromatin coils and condenses to form the chromosomes.In this stage the <span>nuclear membrane inside the cell begins to disappear.</span>

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4 years ago

Is the following genotype homozygous dominant, homozygous recessive,or heterozygous? Gg

Romashka-Z-Leto [24]

It is heterozygous b/c the 2 letters are different, 1 being dominant and one being recessive.

4 0

4 years ago

In Drosophila, the bar mutant is characterized by eyes that are restricted to a narrow, vertical bar. When a bar female is mated

Lelechka [254]

Answer:

See the answer below

Explanation:

a) The mode of inheritance of the bar mutant is X-linked and dominant.

Assuming that barness is represented by the allele B.

Bar female = $X^BX^B$ or $X^BX^b$

Wild type male = $X^bY$

Bar female x wild type male:

$X^BX^B$ x $X^bY$

$X^BX^b$ $X^BY$ $X^BX^b$ $X^BY$

All of the F1 flies are bar because a single B allele is required for the trait to be expressed.

Bar male x wildtype female

$X^BY$ x $X^bX^b$

$X^BX^b$ $X^BX^b$ $X^bY$ $X^bY$

All the female offspring are barred while all the males are wildtype barred resulting in a 50/50 (857/905 is approximately 50/50) ratio of barred/wildtype offspring.

b) Genotypes of each of the cross above, including the parents and the F1 generations.

1. $X^BX^B$ x $X^bY$

$X^BX^b$ $X^BY$ $X^BX^b$ $X^BY$

2. $X^BY$ x $X^bX^b$

$X^BX^b$ $X^BX^b$ $X^bY$ $X^bY$

4 0

3 years ago