Which of the following statements about abnormal curves is CORRECT?

What information must be entered for the onpatient access to be enabled for

leva [86]

Answer:

Home Number Cell number insurance date of brith

Explanation:

I think

6 0

2 years ago

Can a genetic disease such as NF1 be diagnosed with a karyotype? And Why?

lara31 [8.8K]

Answer:

No

Explanation:

A genetic disease cannot be diagnosed with a karyotype. That's because you can't see the bases for the disease on a karyotype.

6 0

3 years ago

Two parents who are heterozygous for type A blood and have sickle cell trait have children.

GuDViN [60]

This question is incomplete, but here is the complete question below.

In humans , blood type is a result of multiple alleles: $I^{A}$ , $I^{B}$ , $i^{o}$ .

A few simple rules of blood type genetics are that:

$I^{A}$ is dominant over $i^{o}$

$I^{B}$ is dominant over $i^{o}$ , and

$I^{A}I^{B}$ are codominant

Two plants who are heterozygous for type A blood and have sickle cell trait have children. Answer the following questions:

a. What is the genotype of the parents?

b. What are the genetic make ups of all the possible gametes they can produce.

c. Complete the dihybrid Punnet square to determine the frequency of the different phenotypes in the offspring.

(NOTE: Consider blood type and normal versus mutant hemoglobin in the various phenotypes.)

Answer:

a) $I^{A}i^{o}AS$

b) $I^{A}A,I^{A}S,i^{o}A,i^{o}S$

c) The dihybrid table is shown in the explanation below and the frequency of the different phenotypes of the offspring are in the ratio of 3:6:3:1:2:1

Explanation:

a)

From the question the genotype of the parents can be determined:

Given that Two parents are heterozygous for type A and have sickle cell trait.

∴ heterozygous for type A will be = $I^{A}i^{o}$ &

heterozygous for having sickle cell traits = AS

Therefore, the traits can be determined the genotype of the parents can be determined as: $I^{A}i^{o}AS$

b)

the genetic make ups of all the possible gametes they can produce can be determined if the parent are self crossed.

If $I^{A}i^{o}AS$ self crossed, we have:

$I^{A}$ $i^{o}$

A $I^{A}A$ $i^{o}A$

S $I^{A}S$ $i^{o}S$

∴ The genetic makeups of all the possible gametes they can produce are:

$I^{A}A$ , $I^{A}S$ , $i^{o}A$ , $i^{o}S$

c)

Below shows the dihybrid Punnet square and the frequency of the different phenotypes in the offspring

I^{A}A I^{A}S i^{o}A i^{o}S

I^{A}A I^{A}I^{A}AA I^{A}I^{A}AS I^{A}i^{o}AA I^{A}i^{o}AS

I^{A}S I^{A}I^{A}AS I^{A}I^{A}SS I^{A}i^{o}AS I^{A}i^{o}SS

i^{o}A I^{A}i^{o}AA I^{A}i^{o}AS i^{o}i^{o}AA i^{o}i^{o}AS

i^{o}S I^{A}i^{o}AS I^{A}i^{o}SS i^{o}i^{o}AS i^{o}i^{o}SS

$\frac{3}{16}$ = 18.75% : Blood type A, normal hemoglobin (normal RBCs) = I^{A}I^{A}AA, I^{A}i^{o}AA, I^{A}i^{o}AA

$\frac{6}{16}$ = 37.5% : Blood type A, normal and mutant hemoglobin (sickle cell trait) = I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}i^{o}AS, I^{A}i^{o}AS