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Hitman42 [59]
3 years ago
12

Why are gram positive bacteria resistant to the effect of membrane attack complex (MAC)? ​

Biology
1 answer:
elena-s [515]3 years ago
8 0
Don’t click the link it’s some kind or tracking I did it don’t do it
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Which organelle packages the material used to build the cell plate in plant cells
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4 years ago
¿Porque las vitaminas, las sales minarales y el agua se consideran reguladores?
Angelina_Jolie [31]

Answer:

Vitamins and minerals boost the immune system, support normal growth and development, and help cells and organs do their jobs. For example, you've probably heard that carrots are good for your eyes. It's true! Carrots are full of substances called carotenoids (pronounced: kuh-RAH-teh-noydz) that your body converts into vitamin A, which helps prevent eye problems.

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8 0
3 years ago
At the end of the term, the professor locks the students in the classroom with plenty of food and water and leaves them there fo
levacccp [35]

Answer:

a. 0.7

b. 0.3

c. Yes

d. 90

Explanation:

As the random mating population was escaped from external evolutionary forces such as mutation, natural selection, and migration, the gene frequency will not change over time. The population will be at Hardy-Weinberg equilibrium throughout the time.

<u>As the number of sleepyheads (recessive trait) was 90 out of class of 1000 students at the start of course.</u>

<u />

q²=0.09 ∵ fraction of sleepyheads = 90/1000 = 0.09 denoted by q²

q that is the recessive gene frequency will be equal to 0.3 (by taking square root of P²

Hence, <em>allele frequency of a allele is 0.3</em>

<em></em>

As, frequency of dominant allele A is denoted by letter p and p+q=1

∴ p=0.7 As p=1-0.3

Hence,<em> allele frequency of A allele is 0.7</em>

<em></em>

Counting the probability of aa alleles (sleepyheads);

As a= 0.3

∴ aa= 0.3×0.3 = 0.09 (In terms of percentage, it is 9%)

To 0.09×1000= 90 or 9% of 1000 = 90

Hence, 90 students should be sleepyheads.

8 0
3 years ago
In horses, black color(B) dominates chestnut color(b). The trotting gait(T) dominates the pacing gait (t). A cross is made betwe
bulgar [2K]

Answer:

<h2>a.  BbTt;</h2><h2>b. BbTt X BbTt ;</h2><h2> Parent 1: BT bt Bt bT </h2><h2> Parent 2: BT bt Bt bT ;</h2><h2>c. BBTT , BbTt , BBTt , BbTT , BbTt , BbTt , BBTt , BbTT , BBTt , Bbtt , BBtt , BbTt , BbTT , bbTt , BbTt , bbTT.</h2><h2> d. 12:1:1:2</h2>

Explanation:

As given Block color is dominant over chesnut color, so B is dominant over b.

Chestnut color=bb ;

Trotting gait is dominant over  pacing gait, so T is dominant over t .

Pacing gait=tt

Now;  cross between

(a) homozygous black, pacing gait is= BBtt; and horse homozygous for chestnut color, trotting gait is= bbTT

BBtt x bbTT

gametes are ; Bt and bT,

Progeny are BbTt, BbTt,

All progeny will be BbTt  heterozygous Black color and heterozygous trotting gait

b. a cross is made between two horses, both of who are heterozygous black trotter = BbTt X BbTt

 Possible gametes of each parent are;

Parent 1: BT bt Bt bT

Parent 2: BT bt Bt bT

c.  Punnett square and  all the possible gene combinations that results are;

Cross between these gametes

BT bt Bt bT

BT bt Bt bT

possible gene combinations ; BBTT , BbTt , BBTt , BbTT , BbTt , BbTt , BBTt , BbTT , BBTt , Bbtt , BBtt , BbTt , BbTT , bbTt , BbTt , bbTT.

d. expected phenotype ration of the offspring.

Black color and trotting gait  horse 12: chestnut and trotting gait horse 1: Black color pacing gait horse  1: chestnut and trotting gait  horse 2.

so 12:1:1:2

6 0
4 years ago
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