One polled (hornless) bull is bred to 3 cows. With Cow A, which is horned, a polled calf is produced. With Cow B, also horned, a

Response:

b

Explanation:

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 =

the probability of the female being Rr =

the probability of passing on the recessive allele 't' to offspring =

the probability of passing on the recessive allele 'r' to offspring =

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

=

=

= 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:

=

%

=

%

=

25%

Eukaryotes include options (A) and (D), while bacteria only pertains to (E), and both (B) and (C). (A) Eukaryotes are the only ones that possess a nucleus. (B) Both types duplicate their DNA prior to cell division, ensuring each cell retains a copy. (C) The structural element is known as a centromere in eukaryotes, whereas bacteria use the origin of replication for this function. (D) Centromeres are unique to eukaryotes. (E) Bacteria exclusively contain a replication origin.

Respuesta:

El orden correcto es:

1. Gran cantidad de glucosa en la sangre, enviando señales hacia el páncreas.  
2. Síntesis de hormonas por células beta.
3. Unión de hormonas con receptores en el hígado.
4. Ingesta de moléculas de glucosa de la sangre por transportadores específicos.
5. Liberación de hormonas desde los receptores

Explicación:

• El páncreas produce dos hormonas clave para la regulación de los niveles de glucosa en la sangre.
• Las dos hormonas, glucagón e insulina, aumentan y disminuyen el azúcar en la sangre respectivamente.
• La insulina reduce la glucosa en sangre al estimular las células del cuerpo para que aumenten la ingesta de glucosa.
• El cerebro detecta la alta/baja glucosa en sangre y, a continuación, envía señales al páncreas para iniciar la síntesis de insulina o glucagón por las células beta.
• Estas hormonas se unen a receptores en las células del cuerpo; en las células hepáticas, en el caso de la insulina.
• Los receptores inician la señalización celular que facilita la entrada o salida de glucosa dentro o fuera de las células del cuerpo.