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avanturin [10]
3 years ago
7

Imagine that four genes on a single chromosome in a mutant phenotype of Drosophila cross over at half the normal rate as in the

wild-type phenotype. How, if at all, would genetic maps of that chromosome differ between this mutant and the wild-type fly
Biology
1 answer:
lesantik [10]3 years ago
6 0

We know that frequency of recombination is proportional with the distance between the genes on the chromosome. Therefore when the recombination rate is higher that means the distance between the genes on the chromosome is bigger. If the recombination rate is lower that means the genes are closer to each other on the chromosome. In this case the cross over rate is half the normal rate in the wild-type. That means that on the genetic map the distance between the two genes on the wild type will be twice bigger than the genes of the mutated Drosophila.

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Write the reason why starch test shows blue-black color.​
Ronch [10]

Answer: Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine.

Explanation :The iodine molecule slips inside of the amylose coil.n This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black color.

6 0
2 years ago
In a diploid individual, one chromosome carries A and B genes, and the homologous chromosome carries different forms (alleles) o
Helen [10]

Answer:

D

Explanation:

This involves a dihybrid inheritance I.e. two genes are being passed on. During meiosis, specifically, the Prophase stage, homologous chromosomes (similar but non-identical chromosomes received from each parent) line side by side. According to the question, one chromosome contains A and B alleles and its homologue, received by the other parent carries a and b alleles. This means that the diploid individual has a genotype AaBb for that gene.

According to Mendel's law of independent assortment, the alleles separate independently of one another into gametes. I.e. allele A and a separates into the gametes without affecting alleles B and b of the other gene.

Crossing-over, which is the exchange of chromosomal segment occurs between the two homologues. Hence, the exchange of chromosomal segments containing alleles in the individual will possibly produce four gametes with the genotypes: AB, Ab, aB, ab.

4 0
3 years ago
In the Calvin cycle the conversion of energy poor carbon dioxide into energy rich glucose
rodikova [14]

Answer:

D)NADPH is made

A)ATP is used

Explanation:

Sorry to ask the two questions, I asked why the two options are in the process:Reduction. In the second stage, ATP and NADPH are used to convert the 3-PGA molecules into three-carbon sugar molecules, glyceraldehyde-3-phosphate ( G3P ). At this stage, it gets its name because NADPH donates, or reduces , electrons to a three-carbon intermediate to form G3P.[Ocultar detalhes]

The reduction stage of the Calvin cycle, which needs ATP and NADPH, converts 3-PGA (produced in the fixation stage) into a three-carbon sugar. This process takes place in two main stages:

Simplified diagram of the reduction step of the Calvin cycle showing the carbon atoms, but not the complete molecular structures. A 3-PGA molecule first receives a second phosphate group from ATP (generating ADP). Then, the doubly phosphorylated molecule receives electrons from NADPH and is reduced to form glyceraldehyde-3-phosphate. This reaction generates NADP + and also releases an inorganic phosphate.

Simplified diagram of the reduction step of the Calvin cycle showing the carbon atoms, but not the complete molecular structures. A 3-PGA molecule first receives a second phosphate group from ATP (generating ADP). Then, the doubly phosphorylated molecule receives electrons from NADPH and is reduced to form glyceraldehyde-3-phosphate. This reaction generates NADP + and also releases an inorganic phosphate.

First, each 3-PGA molecule receives a phosphate group from ATP, becoming a doubly phosphorylated molecule called 1,3-bisphosphoglyceride (and leaving an ADP as a by-product).

Second, 1,3-bisphosphoglycerate molecules are reduced (gain electrons). Each molecule receives two electrons from NADPH and loses one of its phosphate groups, becoming a three-carbon sugar called glyceraldehyde-3-phosphate (G3P) . This step produces NADP^+

+

start superscript, plus, end superscript and phosphate (\text P_iP

i

start text, P, end text, start subscript, i, end subscript) as by-products.

The chemical structures and real reactions are:

Reactions of the Calvin cycle reduction step, showing the molecular structures of the molecules involved.

Reactions of the Calvin cycle reduction step, showing the molecular structures of the molecules involved.

The ATP and NADPH used in these steps are products of the photo-dependent reactions (first stage of photosynthesis). That is, the chemical energy of ATP and the reducing potential of NADPH, both produced with the use of light energy, keep the Calvin cycle running. Conversely, the Calvin cycle regenerates ADP and NADP^+

+

start superscript, plus, end superscript, providing the necessary substrates for photo-dependent reactions.

7 0
2 years ago
Read 2 more answers
Suppose that meiosis occurred in the mother and father whose chromosomes you labeled in question 1. During meiosis,
Flura [38]

Explanation:

you will end up with 24 chromosomes instead of the standard 23 and have down syndrome

7 0
3 years ago
I will gave 20 point's!!!!!!!!!!!
bija089 [108]

there is mostly billions of amount of water from raindrops each day i think.

If You want a EXACT number i'll say, 1,100.

4 0
3 years ago
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