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3 years ago

What kinds of characters can you use to create a data matrix for estimating phylogenetic trees?

Biology

1 answer:

fomenos3 years ago

8 0

Answer: The correct option is A

any heritable trait

Explanation:

This is because phylogenetic tree is a diagram that show or indicate the evolutionary relationships between several organisms from a descent or that have common anscetors which is based on the differences or similarities of there phenotypic traits or physically expressed inheritable traits.

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Rna primers must be present on which strand during dna synthesis?

Tanya [424]

Answer:Both on the leading strand and on each okazaki fragments of the lagging strand.

Explanation:

A primer is a short single-stranded RNA nucleotide that initiate DNA synthesis at the replication fork. Because DNA polymerase can not initiate replication on it own short strand of RNA primer synthesized by an enzyme RNA primase is needed to initiate DNA polynucleotide synthesis.

RNA primer is present on both leading and lagging strand.

RNA primer of about 10 nucleotides long are present on the leading strand and on each okazaki fragments on the lagging strand to initiate replication. The RNA primer is removed and replaced by DNA nucleotides by DNA polymerase.

6 0

3 years ago

2. Dominant trait: cleft chin (C) Mother’s gametes: Cc

andre [41]

.2. Offspring Genotypes will be Cc or cc.

Offspring phenotypes : Cleft chin or no cleft chin.

% chance child will have cleft chin: 50%

3. % chance child will have arched feet: 25%

4. % chance child will have blonde hair: 50%

5. % chance child will have normal vision: 25%

Explanation:

CASE 1 :

Dominant trait: cleft chin (C)

Recessive trait: lacks cleft chin (c)

Father’s gametes: cc

Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents= $\frac{1}{2}C + \frac{1}{2} c$ ........(i)

Gametes of cc parents = $\frac{1}{2}c + \frac{1}{2}c$ .........(ii)

Combining (i) and (ii) we get,

$\frac{1}{2} Cc + \frac{1}{2} cc$

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin = $\frac{0.5}{1}$ ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

A from mother, a from father =Aa

a from mother, A from Father = Aa

a from mother, a from father = aa

Gametes of Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

Gametes of other Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

$\frac{1}{4} AA + \frac{1}{4} Aa$

+ $\frac{1}{4} Aa$ + $\frac{1}{4} aa$

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = $\frac{0.25}{1}$ × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive (bb)

Father’s gametes: Heterozygous (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in half Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous (Ff)

Father’s gametes: Heterozygous (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF then phenotype farsightedness

Genotype Ff then phenotype farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

3 0

3 years ago

Which of the following terms signals the reader that

viva [34]

A.. hopefully i’m right

4 0

3 years ago

HELP ASAP PLZ!!! Read the scenario: You are walking a trail in the woods. You notice something out of the corner of your eye and

sergejj [24]

The warmth and redness of the surrounding area of the cut is due to the immune system.

Blood carries white blood cells, which attacks pathogens (Cause of infection)

The surface become red and warm as that is a conjugating area of blood.

3 0

3 years ago

Read 2 more answers

Li is interested in ways to maximize the vitamin content in his fresh fruits and vegetables. Which of the following practices sh

OlgaM077 [116]

Answer:

Use cooking methods that minimize/eliminate contact with water.

Explanation:

Vitamins are important nutrients found in fruits and vegetables. Vitamins can be easily leached out and it is therefore recommended that the vegetables should be consumed immediately after purchase or stored in an airtight container.

The use of cooking method that eliminate contact with water such as freezing, keeping in a close container will help retain the vitamins, reduce degradation rate and prevent it form been lost.

Cutting will allow easy definition of vitamins and result in its lost.

4 0

3 years ago