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den301095 [7]
3 years ago
11

Which is a cause for a media disruption?

Biology
1 answer:
Mkey [24]3 years ago
4 0
Outdated plugins can cause disruption
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Explanation:

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What is the ultimate source of energy for most organisms and how is this energy capture so it is useful in a food chain ?
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The table shows the number of cells at each stage of the cell cycle in a root tip meristem. One complete cell cycle takes 24 hou
topjm [15]

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im completing a challenge

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3 0
3 years ago
Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of eight offsp
Aleks [24]

Full question from alternative source:

Albinism in humans is inherited as a simple recessive trait. for the following families, determine the most likely genotypes of the parents

Part A

Two normal parents have five children, four normal and one albino. What are the most likely genotypes of the parents?

Part B

A normal male and an albino female have six children, all normal. What is the most likely genotypes of the parents?

Part C

A normal male and an albino female have six children, three normal and three albino. What are the most likely genotypes of the parents?

Part D

Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of 8 offspring. What would be the expected number of normal and albino offspring?

Answer:

Part A - both heterozygous for the trait (Aa)

Part B - female aa, male likely AA

Part C - female aa, male Aa

Part D - 4 normal, 4 albino (50:50)

Explanation:

Part A

We are told that albinism is a recessive trait. Lets call it a and the normal allele A. Since neither of the parents have it, that must mean they each possess at least one copy of the dominant allele (either AA or Aa). Most of their children are normal, but 1 is albino. The albino child must have inherited two copies of the a allele, one from each parent. This means each parent must possess an a allele and must both be Aa.

Part B

Since the trait is recessive, we already know that the albino female must have the genotype aa. If her husband in homozygous normal (AA), all their children will be heterozygous (Aa), and therefore normal, which is the case in this cross. Therefore aa and AA are the most likely genotypes. However, it is still possible that the male has the genotype Aa. This would mean they had a 50:50 chance of producing an albino offspring. We know they have six normal children, so this is unlikely.

Part C

Since the trait is recessive, we know the female has the genotype aa. Unlike the last family, this family has some albino children. This means that the husband must be a carrier of the trait, but he is unaffected, so must be Aa.

As described in part B, if he is Aa and she is aa, they will have 50:50 normal:albino children. This is the ratio in this family.

Part D

The normal children from Part B are all heterozygous for the trait (Aa), inheriting one normal allele from their father, and one albino allele from their mother. The albino children from part C are all aa. Therefore, the cross is Aa x aa. We can draw a punnett square:

                  Aa

             A         a

       a   Aa       aa

aa   a   Aa       aa

Half the children have the genotype Aa, and half have the genotype aa. That means half will be normal and half will be albino. Therefore, 4 normal, 4 albino.

4 0
3 years ago
Can someone please help me answer questions 16-20 please. Thank you!!
torisob [31]

Answer:

16) 6 electrons

17) 6 protons

18) 6 neutrons

19) 4 electrons

20) 6

Explanation:

There are 6 electrons on the rings in total and 6 protons inside the nucleus. If you count the black cir les, there are 6 of them so there are 6 neutrons. Knowing that the inner ring can only hold 2 electrons the most, 6-2= 4. So there are 4 electrons on the outer ring. Knowing that you can get the atomic number based on the number of protons an atom have, the at pool mic number would be 6.

6 0
2 years ago
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