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ValentinkaMS [17]
3 years ago
15

A scientist who is researching marshes observes that marsh grass, Spartina bakeri, grows to different heights at different locat

ions in the marsh. If the scientist wants to find the reason for this difference, what is the correct first step? Conduct an experiment and change only one variable at a time. Determine which technology would assist most in answering the question. Form a research question based on knowledge of the marsh. Establish a control group for comparison.
Biology
2 answers:
9966 [12]3 years ago
7 0
I belive "form a research question based on knowledge of the marsh," in other words, your hypothesis which is the 1st step in the scientific method after observation (which your scientist has already done)
MAXImum [283]3 years ago
7 0

The correct first step would be to conduct an experiment and change only one variable at a time.

There are generally two types of variables in all the valid research, that is, independent and dependent variables. These variables permit a researcher to find the precise association among the variables.  

Thus, in the given study, if the scientists are aiming to determine the cause of length distinction in the marsh grass, then he has to perform an experiment with the application of independent and dependent variables.  


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A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
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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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The size of the exons in the genomes follows a log-normal distribution, with an average length of about 150 nucleotides, knowing that in eukaryotes, each gene contains several exons and introns (an average of 8) so the size is 8*150 = 1200 bp.

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Introns are fragment of a gene located between two exons. Introns are present in immature mRNA and absent in mature mRNA. "Non-coding" fragment of the gene.

The introns average in a gene is 3365 bp including 3'UTR and 5'UTR and intermediate introns.

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