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Amiraneli [1.4K]
3 years ago
11

Are flamingos an invasive species anywhere on Earth?

Biology
1 answer:
AVprozaik [17]3 years ago
6 0
No flamingos dont normally migrate unless there are changes in their habitat conditions
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20pt brainlyyyyyyyyyy​
nekit [7.7K]

Answer:

c

Explanation:

controls judgement and expressions

5 0
3 years ago
The almost pure cellulose obtained from the seed threads of Gossypium (cotton) is tough, fibrous, and completely insoluble in wa
olga55 [171]

Answer:

Cellulose is held by beta 1,4-glycosidic bonds making it linear while Glycogen is held by an alpha 1,4-glycosidic bond making it highly branched.

Explanation:

Cellulose and Glycogen are both carbohydrate polysaccharides formed from glucose monomers. According to the question, cellulose is a tough, fibrous, and insoluble (in water) polymer found to play a structural role in plants' cell wall while Glycogen is another polymer obtained from muscle or liver and disperses readily in hot water to make a turbid solution.

Although these two polysaccharides (cellulose and glycogen) are linked by (1, 4)-glycosidic bonds but the glucose monomers in CELLULOSE are linked by a beta 1,4-glycosidic, hence, making it a straight or linear polymer

GLYCOGEN, on the other hand, is linked by an alpha 1,4-glycosidic bond making it an highly branched polymer. This structure is responsible for the different physical properties of the two molecules.

6 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

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3 years ago
What percent of all the genes in <br> e.coli are active all the time? 10, 20, 60, or 80%?
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1. True
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