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3 years ago

Which statement describes why zebra fish experience similar genetic diseases as humans?

Biology

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

Zebra fish have an omnivorous diet similar to that of humans. Zebra fish have nucleotide sequences similar to those of humans.

Explanation:

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Explain why science is a continuous progression of study

Sergio [31]

Well I won't go to tooooooo much of that detail but the thing is FACT BY ME that everything around u is science and when u learn about ur surroundings it is science...!!!!

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3 years ago

Anxiety disorders are characterized by all of these symptoms except ____.anxiety disorders are characterized by all of these sym

lisov135 [29]

A person can be anxious when making an important decision when faced with a problem at work. Types of disorders are;
1. Generalized anxiety disorder symptoms include
a) irritability
b) muscle tension
c)Being easily fatigued.
2. Panic disorder symptoms include
a)feeling of being out of control during a panic attack.
b)Repeated and sudden attack of intense fear.
3. Social anxiety symptoms
a)Having a very hard time making friends.
b)feelings of self-conscious in front of other people
c)Staying away from where other people are.

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3 years ago

In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the freq

k0ka [10]

Answer:

C) 64% AA, 32% Aa, 4% aa

Explanation:

The frequency of dominant allele A is = 80% = 0.8

Since the population is in Hardy-Weinberg equilibrium, p+ q = 1

Here, p= frequency of dominant allele

q = frequency of recessive allele

Given, p= 0.8

So, q = 1-p = 1-0.8= 0.2

Frequency of genotype AA = p2 = 0.8 x 0.8 = 0.64 = 64%

Frequency of genotype aa = q2 = 0.2 x 0.2 = 0.04 = 4%

Frequency of genotype Aa = 2pq = 2 x 0.8 x 0.2 = 0.32 = 32%

5 0

3 years ago

Create an mRNA copy from a DNA strand.

aleksley [76]

Answer:

RNA polymerase

Explanation:

8 0

3 years ago

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The gene

NikAS [45]

Answer:

A) 9:3:3:1

B) All bitter fruit, yellow spotted offsprings

C) Phenotypes are bitter yellow spotted (4), bitter no spot (4), sweet yellow spot (4), and sweet no spot (4). 1:1:1:1

Explanation:

This is a typical dihybrid cross involving two genes, one coding for fruit taste and the other for spot color. The allele for bitter taste (B) and yellow spot (S) is dominant over the allele for sweet taste (b) and no spot (s) respectively.

Hence, a heterozygous F1 resulting from a cross between an homozygous dominant (bitter fruit, yellow spot) and homozygous recessive (sweet fruit, no spot) will have a BbSs genotype. The heterozygous F1 offsprings are self-crossed and produce gametes BS, Bs, bS, bs. (See punnet square). The F2 offsprings will have the following phenotypes: Bitter fruit, yellow spot (9)

Bitter fruit, no spot (3)

Sweet fruit, yellow spot (3)

Sweet fruit, no spot (1)

Back cross between a F1 offspring (BbSs) and homozygous dominant parent (BBSS) will produce all bitter fruit, yellow spot offsprings (see attached image). BBSS (4), BBSs (4), BbSS (4), and BbSs (4) are the offsprings' genotypes.

For the back cross between a F1 offspring (BbSs) and a homozygous recessive (bbss) parent, the Phenotypes with their proportions are as follows:

Bitter fruit, yellow spot (BbSs, 4)

Bitter fruit, no spot (Bbss, 4)

Sweet fruit, yellow spot (bbSs, 4)

Sweet fruit, no spot (bbss, 4).

4 0

3 years ago