Which is an example of potential rather than kinetic energy?

When elongated, tube –shaped cells from the lining of the intestine are treated with a certain chemical in a dish, the cells sag and becomes round blobs. the internal structures disrupted by this chemical are probably microtubules

4 0

4 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