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podryga [215]
3 years ago
15

What are the characteristics of living things?? ​

Biology
1 answer:
ioda3 years ago
7 0

Answer:

Cellular organization.

Reproduction.

Metabolism.

Homeostasis.

Heredity.

Response to stimuli.

Growth and development.

Adaptation through evolution.

Explanation:

Hope this helps.

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A woman who is a carrier for XLA marries a man who does not have the disorder. They have four sons. How many could be expected t
ss7ja [257]

Answer:

Their sons have 50% chance of having XLA.

Explanation:

X-linked agammaglobulinemia (XLA) is a genetic disorder that affects person immune system or ability of the body to fight against Infections. A person with XLA disease have few B cells which are specialized white blood cells for fighting Infections. If a female have XLA disease she has 50% chance of transmitting disease to their sons. A male has 50% chance of having XLA and female 25% chance.

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Why are viruses not able to make their own proteins
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Viruses don't make their own proteins because they are not cells. A virus is a nucleic acid enclosed in a protein coat, and does not contain organelles.
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3 years ago
Read 2 more answers
Why is it difficult to grow anaerobes in vitro?
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Answer:

 Anaerobes are not able to have growth in the presence of much of oxygen environment as they do not have catalse .Catalse is the enzyme for clearance of oxygen.There are also anaerobes protects their own self from oxygen by production of an enzyme known as super oxide dismutase. Therefore, it is not easy task for creation an oxygen-absent environment for anaerobes in the vitro so they cannot be grown in vitro.

5 0
3 years ago
Which of the following statements correctly describes how the jet stream effects local weather?
icang [17]

Answer:

Jet streams are like rivers of wind high above in the atmosphere. These slim strips of strong winds have a huge influence on climate, as they can push air masses around and affect weather patterns. ... Temperature also influences the velocity of the jet stream

Explanation:

7 0
3 years ago
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
marusya05 [52]

Answer:

a is the middle gene.

Distance [b-a]= 24.7 mu

Distance [a-c]= 15.8 mu

Distance [b-a} = 40.5 mu

Explanation:

A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). <u>The order of the genes here is arbitrary. </u>

The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).

The F2 shows the following phenotypic ratios:

  • 320 a⁺b⁺c⁺
  • 308 a b c
  • 102 a⁺ b c⁺
  • 112 a b⁺ c
  • 66  a⁺ b⁺ c
  • 59 a b c⁺
  • 18 a⁺ b c
  • 15 a b⁺ c⁺

Total = 1000

The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.

Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals:  a⁺b⁺c⁺ and abc.

The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.

<h3><u>1st. Determine the gene order</u></h3>

Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.

<h3><u>2nd Determine the single crossover gametes</u></h3>

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

  • The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
  • The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.
<h3><u>3) Calculate the recombination frequencies between genes </u></h3>

Recombination frequency (RF) = #Recombinants/Total progeny

  • RF [b-a]= (102+112+18+15)/1000= 0.247
  • RF [f-br]= (66+59+18+15)/1000= 0.158
<h3><u /></h3><h3><u>4) Calculate the distance in map units </u></h3>

Distance (mu) = RF x 100

Distance [b-a]= 0.247 × 100 = 24.7 mu

Distance [a-c]= 0.158 × 100 = 15.8 mu

Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu

<h3><u>The gene map therefore looks like: </u></h3>

b------------24.7 mu--------------------------a---------15.8 mu-----------c

4 0
3 years ago
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