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Natasha2012 [34]
3 years ago
14

___________, the stiffening of the body after death, is the result of chemical changes within muscle tissue.

Biology
2 answers:
11111nata11111 [884]3 years ago
8 0
Rigor mortis is the answer
scZoUnD [109]3 years ago
3 0
Its called rigor mortis the stiffening of the body after death is the result of chemical changes within muscle tissue
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Which reactions of aerobic respiration occur in the inner mitochondrial membrane.
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Answer:

Mitochondria are organelles whose membranes are specialized for aerobic respiration. The matrix of the mitochondria is the site of Krebs Cycle reactions. The electron transport chain and most ATP synthesis rely on the compartments created by the inner membrane of the mitochondria.

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2 years ago
For this week's CRQ, you are to design your own genetic cross between a female and a male. Be sure to include each of the follow
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Answer and Explanation:

Punnett squares are illustrations that display the product of hybrid trait crosses by genetic breeding. Established parent genotypes and phenotypes are used and are usually compatible with Mendelian ratios and inheritance. The phnomenon influencing this behaviour is the

  • Law of Segregation,
  • Law of Independent assortment
  • and Dominance

For this cross...

Assuming Widow's peak= dominant

<u> 1. What are the dominant and recessive traits in cross?</u>

Parent generation or P generation:

gametes: Ww → Widow's peak dominant (W), No Widow's peak recessive (w)

<u>2. Is the male homozygous dominant, homozygous recessive, or heterozygous?</u>

Heterozygousity describes the allelic makeup; these traits are made of two different forms of the same gene.

Male: Ww  →  with the W.peak;  

(Thus, with both W and w this is a heterozygous trait)

<u>3. Is the female homozygous dominant, homozygous recessive, or heterozygous?</u>

However, homozygous traits are made up of the same forms of the gene called alleles.

Female: ww → without the W.peak  

(Thus, with both w ad w this is a homozygous trait)

<u>1. genotypes and phenotypes for parent and offspring...</u>

<u></u>

Assuming heterozygousity of the male , for the cross

P generation:                          Ww× ww

                            Widow's peak  ×No Widow's peak  

F1 generation:

  • genotype:                       Ww × ww       ...(from punnet square)
  • phenotype:          2 with the W.peak; 2 without

<em>∴ 50% are genotypically heterozygous while 50% is homozygous , 2 with the W.peak; 2 without </em>

6 0
3 years ago
How do the structures of prokaryotic and eukaryotic cells influence their functions?
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Contain circular pieces of DNA concentrated within a nucleoid. Some contain plasmids, which can<span> be shared with other </span>prokaryotes<span>. </span>Eukaryotes<span> contain </span>their<span> genetic material in chromosomes within a nucleus. ... </span>Prokaryotes<span> are therefore able to lack membrane-bound organelles and have less complex internal </span>structure<span>.</span>
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3 years ago
In a certain group of African people, 4% are born with sickle-cell disease, an autosomal recessive disorder. Heterozygous indivi
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Answer:

the correct answer is C) 32%

Explanation:

Sickle-cell anaemia is an autosomal recessive genetic disorder. Individuals with the homozygous recessive have sickle-shaped blood cells. Whereas, individuals with heterozygous are only carrier of sickle cell trait. The carrier individuals are resistant to malarial parasite and do not have malaria.  

As per the question, 4% of an African population is born with sickle-cell disease, then the percentage of the population is heterozygous and resistant to malaria will be:  

Hardy-Weinberg formula is equilibrium,

        p² + 2pq + q² = 1  

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p2 = percentage of homozygous dominant individuals

q2 = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Given, homozygous recessive for this gene (q2) is 4%  which is 0.04, the square root (q) is 0.2 (20%) then p should be 1-0.2 = 0.8 (20%).  

Thus, the frequency of heterozygous individuals = 2pq.  

2 (0.8 x 0.2) = 0.32 (32%).  

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3 years ago
in which of the following cells would the cycle be the shortest of cell an adult cell,an embroyonic cell,a teenagers kidney cell
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<span>an unfertilised egg cell</span>
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