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Pavlova-9 [17]
1 year ago
12

For carbon dioxide (CO2) to move from the outside air into the leaf through the stomata, the following condition must be true:A.

concentration of CO2 in air must be less that the concentration of CO2 in the leaf B. concentration of CO2 in leaf must be less that the concentration of CO2 in the air C. concentrations in the air and leaf must be equal D. none of the above
Biology
1 answer:
Jlenok [28]1 year ago
3 0

Condition B. concentration of carbon dioxide in the leaf must be less than the concentration of carbon dioxide in the air must be true for carbon dioxide to move from the outside air into the leaf through the stomata.

<h3>What are the stomata plants?</h3>

The stomata plants are structures in the leaves which allow them to produce the exchange of gases (i.e. carbon dioxide and oxygen) which are required and or produced during photosynthesis.

Therefore, with this data, we can see that the stomata plants are used by plants to exchange carbon dioxide with the surrounding environment.

Complete question:

For carbon dioxide (CO2) to move from the outside air into the leaf through the stomata, the following condition must be true:

A. concentration of CO2 in air must be less that the concentration of CO2 in the leaf

B. concentration of CO2 in leaf must be less that the concentration of CO2 in the air

C. concentrations in the air and leaf must be equal

D. none of the above

Learn more about the stomata plants here:

brainly.com/question/19393475

#SPJ1

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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
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⁺
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  • 112 a b⁺ c
  • 66  a⁺ b⁺ c
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  • 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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