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irinina [24]
3 years ago
15

A litter of four kittens includes one with black hair and blue eyes, one with black hair and orange eyes, one with brown tabby h

air and blue eyes, and one with brown tabby hair and orange eyes. Which of Mendel's Laws is well illustrated by this diversity of kitten types?
Biology
1 answer:
Hoochie [10]3 years ago
5 0

Answer:

Law of independent assortment

Explanation:

According to Gregor Mendel who he performed a performed a cross involving two different genes i.e. a dihybrid cross, he stated that the allele of one gene will get sorted into gametes independently of the alleles of the other gene. He called this the LAW OF INDEPENDENT ASSORTMENT.

He obtained a 9:3:3:1 ratio when he performed this cross, which was only expected if each gametes contained the two genes in a combined state i.e. the dominant allele for one gene is equally likely to contain a dominant or recessive allele for the other gene in a gamete.

In this case, it appears the black hair and blue eyes are dominant alleles respectively while the tabby hair and orange eyes are recessive alleles respectively. If these alleles assort independently and combine in different ways, it will give rise to diverse genotypes and phenotypes in the kitten as illustated in the different phenotypic expression of eye color and hair color in the kitten.

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Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X-linked recessive allele in humans. A woman whose father
prohojiy [21]

Answer:

(a) 1/2; (b) no

Explanation:

Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.

In contrast to this, her husband is diseased so his genotype will be XᵇY.

The Punnett square diagram related to the cross is attached.

(a) Proportion of their sons expected to be G6PD is 1/2:

They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.

(b) If the husband were G6PD deficient, the answer will not change.

The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.  

7 0
3 years ago
CAN ANYONE PLS ANSWER DIS!!!! THE OTHER ONES ARE THE NOTES!!
777dan777 [17]
  1. Prokaryotic cells have a cell wall surrounding their cell membrane .The Prokaryotic cells don't have a well defined nucleus .The genetic material of the cell is basically naked and not enveloped by the Nuclear membrane .
  2. Eukaryotic cells don't have a cell wall surrounding their cell membrane. The Eukaryotic cells have a well organised nucleus .The genetic material of the cell is surrounded by nuclear membrane or nuclear envelope.
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7 0
2 years ago
How many trophic levels are there in the following food web?
zhenek [66]

Answer: c.4

Explanation:

6 0
3 years ago
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
What is 25 m rounded to one significant figure?
Yuliya22 [10]

Answer:

20

Explanation:

4 0
3 years ago
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