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HACTEHA [7]
2 years ago
11

Select ALL the correct answers.

Biology
2 answers:
Firdavs [7]2 years ago
5 0

Answer: Blood types O, A, and B are possible for the father.

Explanation: Since the mother has type A blood, there are two possible genotypes that she can have: AA or AO. Since her son has type O blood, he only has one genotype: OO.

For blood types,

Type A - AA or AO

Type AB - AB

Type B - BB or BO

Type O - OO

* A and B are dominant over O.

Now we have to do crosses between the mother and the father:

AA mother with AB father: child with type AA or type AB blood

AO mother with AB father: child with type AA, AO, AB, or BO blood

AA mother with OO father: child with AO blood

AO mother with OO father: child with AO or OO blood

AA mother with AA father: child with type AA blood

AA mother with AO father: child with type AA or AO blood

AO mother with AA father: child with type AA or AO blood

AO mother with AO father: child with type AA, AO, or OO blood

AA mother with BB father: child with type AB blood

AA mother with BO father: child with AB or AO blood

AO mother with BB father: child with AB or BO blood

AO mother with BO father: child with AB, AO, BO, or OO blood

The father can have blood types O, A, or B.

Leya [2.2K]2 years ago
3 0
I would say just O :) have a good day
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The answer is 44 chickens and 28 rabbits.

We have a system of two equations.
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y - the number of chickens

Both chicken and rabbit have only 1 head, so the first equation is: x + y = 72.
Since rabbits have 4 legs and chickens 2 legs, the second equation is: 4x + 2y = 200.

Now, let's solve the system:
x + y = 72
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y = 72 - x
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2x + 144 = 200
2x = 200 - 144
2x = 56
⇒ x = 56 ÷ 2 = 28
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Since
y = 72 - x = 72 - 28 = 44,
there are 44 chickens.

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