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kotegsom [21]
3 years ago
11

In what way do two alleles for the same trait differ?

Biology
1 answer:
kvasek [131]3 years ago
8 0
I think it’s C explanation : i learned this a while ago
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A individual with Down syndrome (trisomy 21) has how many chromosomes
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The individual will have 42 chromosome
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Complete each statement to identify the
m_a_m_a [10]

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1.vegetative propagation 2. mitosis

Explanation:

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True or False: As ATP is used in the myofilaments, mitochondrion, and cytoplasm of muscle cells, heat is released. The circulato
Tpy6a [65]

Answer:

- As ATP is used in the myofilaments, mitochondrion, and cytoplasm of muscle cells, heat is released. True

- The circulatory system takes heat to the skin’s surface to dissipate to the environment. True

- People sweat to help cool down. True

Explanation:

Muscle contraction occurs when myosin heads bind to actin to pull it inwards. The energy for muscle contraction is provided by the hydrolysis of ATP molecules. ATP hydrolysis also provides energy for cellular processes in the mitochondria and cytoplasm of muscle cells. During ATP hydrolysis, heat is released, thereby the increase in ATP hydrolysis increases heat production (approx. 60% of the energy released from ATP hydrolysis produces metabolic heat). Moreover, on hot days, the cardiovascular system composed of the heart and blood vessels dissipates heat through the blood flowing in veins and arteries, which moves closer to the skin surface (and therefore heat is lost). Finally, sweating refers to the process of exocrine secretion of the sweat glands that release salty water through the pores of the skin. As sweat evaporates, it helps to cool down the body and regulate its temperature.

7 0
3 years ago
A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with allele
Sophie [7]

Answer:

The recombination frequency between genes E and B is 0.1 = 10%

Explanation:

<u>Available data</u>:

Cross: BbEeHh    x    bbeehh

F1) 1000 individuals in total

  • 48 bb ee Hh
  • 36 Bb ee hh
  • 400 Bb ee Hh
  • 4 Bb Ee Hh
  • 426 bb Ee hh
  • 46 Bb Ee hh
  • 38 bb Ee Hh
  • 2 bb ee hh

To know if two genes are linked, we must observe the progeny distribution. In a tri-hybrid cross, If individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1:1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the heterozygote parent.

In the present example, the phenotypic ratio shows different proportions than the expected ones if they were not linked.

We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. And simple recombinant gametes produced by the cross, which frequencies are intermediate.

Parentals:

  • 400 Bb ee Hh
  • 426 bb Ee hh

Double recombinants:

  • 4 Bb Ee Hh
  • 2 bb ee hh

Simple recombinant

  • 48 bb ee Hh
  • 36 Bb ee hh
  • 46 Bb Ee hh
  • 38 bb Ee Hh

Comparing the parental and the double recombinant we will realize that they only change in the position of the alleles E/e. This suggests that the position of the gene E is in the middle of the other two genes, B and H, because in a double recombinant only the central gene changes position in the chromatid.

B------E------H

Now we will call Region I to the area between B and E and Region II to the area between E and H.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between B and E genes, and P2 to the recombination frequency between E and H.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

So:

• P1 = (R + DR) / N = 48 + 46 + 4 + 2 / 1000 = 100/1000 = 0.1

• P2 = (R + DR)/ N  = 36 + 38 + 4 + 2 / 1000 = 80/1000= 0.08  

Now, to calculate the recombination frequency between the two extreme genes, B and H, we can just perform addition or a sum:

P1 + P2 = Pt

0.1 + 0.08 = Pt

0.18 = Pt  

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).  One centiMorgan (cM) equals one map unit (MU). The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

• P1 = 0.1 x 100 = 10%

• P2 = 0.08 x 100 = 8%

• Pt = 0.18 x 100 = 18 %

5 0
2 years ago
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