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vivado [14]
3 years ago
6

PLZZZ HELPPP MEEEE !!!!

Biology
1 answer:
alexandr1967 [171]3 years ago
6 0

the answer is A. changed by the reaction. when a subtrate locks with an enzyme it changes from a subtrate to a new form cuz the reaction changes it.

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What is a cancerous tumor? (in simple words pls)
Vladimir79 [104]

Answer:

Cancerous tumors are malignant, which means they can spread into, or invade, nearby tissues. In addition, as these tumors grow, some cancer cells can break off and travel to distant places in the body through the blood or the lymph system and form new tumors far from the original tumor.

7 0
2 years ago
Read 2 more answers
1. Which of the following accurately describes a typical human karyotype?
yuradex [85]
<span>Human cells contain 23 chromosomes and has a total of 46 pairs of chromosomes. The answer is letter A. Depending on the length and morphology, it has seven groups. They are arranged from largest to smallest. It has 22 pairs of homologus chromosomes known as autosomes and 1 pair of sex chromosomes, a total of 23 chromosomes. A male karyotype has a total of 46 chromosomes and gives a pair of XY while a female karyotype has a total of 46 chromosomes and gives a pair of XX. These pairing are responsible to the gender of the offspring.</span>
6 0
3 years ago
), a series of five cell divisions would produce an early embryo with how many cells?
Arada [10]
C is the correct answer
3 0
3 years ago
Party all the time was by<br><br> A. Shrek<br> B. eddie murphy<br> c. Donkey<br> D.eddie eddie
dolphi86 [110]

Answer:

C. Donkey

Explanation:

Because it was a quote from Donkey in the first or second movie of Shrek.

5 0
3 years ago
The human MN blood type is determined by two codominant alleles, L(M) and L(N). The frequency of L(M) in Eskimos on a small Arct
tatiyna

Answer:

(a) Frequency of M = 0.64

    Frequency of N = 0.04

    Frequency of MN= 0.32

(b)  Expected frequencies of M = 0.648

     Expected frequencies of MN = 0.304

     Expected frequencies of N =  0.048

Explanation:

(a) If random mating takes place in the population, then  the expected frequencies are

f(L(M)) = p = 0.8

F(L(N)) = q

q= 1 - p

= 1 - 0.8

= 0.2

Frequency of M = p^2 = ( 0.8)^2 = 0.64

Frequency of N = q^2 = (1-p)^2 = (1 - 0.8)^2 = (0.2)^2 = 0.04

Frequency of MN = 2pq = 2 * 0.8 * 0.2 = 0.32

(b)

F = inbreeding coefficient = 0.05

f(L(M)L(M)) = p^2 + Fpq = (0.8)^2 + 0.05 * 0.8 * 0.2 = 0.648

f(L(M)L(N)) = 2 pq - 2Fpq = 2 * 0.8 * 0.2 - ( 2 * 0.05 * 0.8 * 0.2) = 0.304

f(L(N)L(N)) = q^2 + Fpq = (0.2)^2 + ( 0.05 * 0.8 * 0.2) = 0.048

5 0
2 years ago
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