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Morgarella [4.7K]
3 years ago
13

Can anyone help me please

Biology
1 answer:
Natalka [10]3 years ago
7 0

Answer:

hehe thanks

Explanation:

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Name the four most important ligaments of the knee. What condition is characterized by an irritation and softening of the cartil
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Answer:

1. Medial collateral ligament, lateral collateral ligament, anterior cruciate ligament and posterior cruciate ligaments.      

2. Chondromalacia patellae

Explanation:

The knee is a ginglymus that connects the thigh to the leg. It is the largest joint in the human body, that plays a vital role in the horizantal and the vertical movement of the body.

There are <u>four ligaments</u> of knee joint, that joins the femur to the tibia and also provides stability by limiting movement. They are: <u>medial collateral ligament, lateral collateral ligament, anterior cruciate ligament and posterior cruciate ligaments.</u>

Chondromalacia patellae is a medical condition that includes inflammation of the cartilage on the posterior aspect of the patella, which leads to softening of the cartilage.

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Can someone please answer number 22 , 23 , and 24 for me ? <br><br> 29 POINTS FOR THIS !
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For bacteria to continue growing rapidly when they shift from an aerobic environment to an anaerobic environment, they must
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small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

p^{2} + q^{2} + 2pq = 1\\

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

7 0
3 years ago
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