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DiKsa [7]
3 years ago
15

Which type of exercise most likely causes human muscle cells to perform anaerobic respiration?

Biology
1 answer:
harina [27]3 years ago
4 0
A vigorous excersie that requires a lot of energy
You might be interested in
A population that is eaten by organisms from another population is known as
Nataly_w [17]

Answer:

B

Explanation:

A resource population is any part of the environment that can be used to meet another organism's needs. The first population acts as this because the second population is surviving by eating the first.

8 0
3 years ago
En el hombre el color negro de los ojos “A” domina sobre el color azul “a” Una pareja en la que el hombre tiene los ojos negros
umka21 [38]

Las respuestas a estas preguntas son:

a) Genotipo del padre Aa (ojos negros)

b) Cruzamiento: Aa (padre) x aa (madre)

c) Frecuencias genotípicas esperadas: 1/2 Aa; 1/2 aa  

En genética, dominancia completa se refiere al proceso  de herencia en la cual un individuo heterocigota, es decir, el individuo que presenta dos alelos diferentes para el mismo gen, presenta el mismo fenotipo que un individuo homo-cigota para el alelo dominante (el alelo dominante es aquel que enmascara la expresión del alelo recesivo en individuos heterocigotas).

En este caso, el carácter 'color de ojos' presenta un patrón de herencia monogénica, donde el alelo 'A' dominante codifica para el rasgo fenotípico ojos negros, mientras que el alelo 'a' recesivo codifica para el rasgo fenotípico ojos azules.  

En el ejemplo, la pareja tuvo progenie en la cual uno de los hijos presenta el rasgo recesivo ojos claros, con lo cual el padre debe ser heterocigota y poseer un alelo recesivo 'a'; mientras que la madre expresa el fenotipo recesivo y por lo tanto su genotipo es 'aa'. En consecuencia, el cruzamiento de un padre heterozigota 'Aa' con una madre homo-cigota recesivo 'aa' producirá una descendencia con una frecuencia genotípica esperada de 1/2 (50%) de hijos con ojos 'color negro' de genotipo Aa y 1/2 (50%) de hijos con ojos 'color celelste' de genotipo aa >>

Cruzamiento: Aa (padre) x aa (madre)

Gametos padre: 1/2 A; 1/2 a

Gametos madre: 100% a

Cuadro de Punett (combinaciones gaméticas):

                  A           a

a               Aa          aa

a               Aa          aa

En consecuencia,  este cruzamiento producirá 50% individuos ojos color negro (genotipo Aa) y 50% individuos con ojos color celelste (genotipo aa)

Podés encontrar más información sobre este tema en:

brainly.com/question/22398195?referrer=searchResults

7 0
2 years ago
If a DNA molecule has 24% adenine, what is the percentage of guanine in the same DNA molecule?
artcher [175]

Answer:

26%

Explanation:

Adenine pairs with thymine.

Guanine pairs with cytosine.

If 24% of the DNA molecule is adenine, that means that there is 24% thymine. This is because they pair together so they have to be equal. That is 48% of the DNA molecule. To find the percentage of guanine, we need to minus 48% from the total 100%.

100 - 48 = 52

Then divide by 2 because you have guanine AND cytosine.

52/2 = 26

Therefore, your final answer is 26% of guanine is in the DNA molecule.

<em>I hope this helps!!</em>

<em>- Kay :)</em>

5 0
3 years ago
1. Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently
maria [59]

Answer:

a. 1/64 AATTDD

b. 1/64 aattdd

c. 8/64=1/8 AaTtDd

d. 2/64 = 1/32 AATTDd or aattDd or AAttDd or aaTTDd

Explanation:

<u>Available data:</u>

  • Flower position: Axial (A) is dominant to Terminal (a)
  • Stem length: Tall (T) is dominant to dwarf (t)
  • Seed shape: Round (R) is dominant to wrinkled (r)

<u>Cross</u>:

Parental) AaTtDd   x   AaTtDd

Knowing that these four genes assort independently, then we can infer that the F1 will have the next genotypic proportions for each gene:

  • AA= 1/4

       Aa =2/4

       aa = 1/4

  • TT = 1/4

        Tt = 2/4

         tt = 1/4

  • DD = 1/4

        Dd = 2/4

        dd = 1/4

<em>This is the same as developing a Punnett square for each gene separately.</em>

Knowing this, to get each of the mentioned genotypes for each trait, we must multiply their respective genotypic proportions, like this:

a. homozygous for the three dominant traits, AATTDD

   AA1/4 x TT 1/4 x DD 1/4 = AATTDD 1/64

b. homozygous for the three recessive traits, aattdd

   aa 1/4 x tt 1/4 x dd 1/4 = aattdd 1/64

c. heterozygous for all three characters, AaTtDd

   Aa 2/4 x Tt 2/4 x Dd 2/4 = AaTtDd 8/16=1/8

d. homozygous for axial and tall, heterozygous for seed shape. There are four possibilities, all of them with the same result:

AATTDd --> 1/4  x 1/4 x 2/4 = 2/64=1/32

aattDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

AAttDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

aaTTDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

4 0
3 years ago
Read 2 more answers
I could use some help! 30 points!
iren2701 [21]

Answer:

not sure maybe b it sound right otherwise sorry

7 0
4 years ago
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