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ryzh
3 months ago
8

If nondisjunction occurs in meiosis ii during gametogenesis, what will be the result at the completion of meiosis?

Biology
1 answer:
enyata [2.5K]3 months ago
6 0

In the process of gametogenesis, nondisjunction during meiosis II results in at least one pair of sister chromatids failing to separate. Consequently, this leads to the formation of two cells containing the typical haploid chromosome count (n), one cell with an additional chromosome (n + 1), and a fourth cell missing a chromosome (n - 1). To summarize, the outcome is two gametes with n, one with n + 1, and one with n - 1.

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Match each scenario with the correct component of natural selection
enyata [2506]

Answer and Explanation:

Typically, sea turtles lay around 110 eggs per clutch. However, only a small fraction of the hatchlings survive to reach the ocean. Subsequently, an even fewer number are able to produce their own young. - Overpopulation

All zebras possess stripes, yet no two zebras exhibit identical stripe patterns. -Variation

A lemur species named aye-ayes feature extended, slender middle fingers that enable them to probe into tree crevices to extract insects for food.- Adaptation

6 0
3 months ago
Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where T is tall and t is short, and for
enyata [2506]

Answer:

(A) 0.0625

(B) 25%

Explanation:

A

With the following:

T indicates tall

t indicates short

R indicates red

r indicates tan.

When a short tan male goat mates with a tall red female goat of an unknown genotype;

Let's analyze separately;

A short tan male goat = ttrr

The female goat’s genotype is not specified but it is known to be tall and red

∴ We determine that she must be T_ R_;

the potential genotype probability of the female for each allele could be:

- T

T

or Tt- RR

or R

r

Nonetheless, from earlier discussion;

the chance that the female is Tt = \frac{1}{2}

the probability of the female being Rr = \frac{1}{2}

the probability of passing on the recessive allele 't' to offspring = \frac{1}{2}

the probability of passing on the recessive allele 'r' to offspring = \frac{1}{2}

∴ the likelihood that the pairing would yield short, tan offspring (ttrr) stands at;

= (\frac{1}{2})^{4}

= (\frac{1}{16})

= 0.0625

(B)

In pea plants, one is homozygous recessive for both traits, leading to the cross of Yy Rr × yy rr. Determine the green and round offspring percentage.

For Color, let:

Y = Yellow

y = Green

For shape, let;

R = Round

r = wrinkled

If a cross occurs between Yy Rr × yy rr. (yellow round and green wrinkled)

the ratio of green and round offspring will be;

If Yy Rr self-crosses; various traits (YR, Yr, yR, yr)

yy rr will produce (yr, yr)

                   YR                    Yr                    yR                    yr

yr                 YyRr                 Yyrr                 yyRr                 Yyrr

yr                 YyRr                 Yyrr                 yyRr                 Yyrr

                   Yellow              Yellow            Green              Green

                   Round               Wrinkled        Round              Wrinkled

                   Thus, the percentage of green and round offspring based on the above is:

=

%

\frac{2}{8}*100%=

%

\frac{1}{4}*100=

25%

7 0
2 months ago
If your friend was considering genetic modification to have a child and asked for your advice what would you tell her
lana [2441]

I would support her in any choices she makes and whatever path she decides to pursue.

5 0
3 months ago
Read 2 more answers
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