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Crazy boy [7]
3 years ago
7

You perform the following dihybrid cross: A blaBx Ab/aB. The A and B genes control different traits. What phenotypic ratio would

you expect to see in the progeny if the two genes are unlinked? What if they are linked?
a. Unlinked: 1:1: 1:1/Linked: 1: 2: 2:1
b. Unlinked: 1:1:11/Linked: 2: 1 1:2
c. Unlinked: 1:1:11/Linked: >9: 3:<3:1
d. Unlinked: 9: 33: 1/Linked: <9:3:3:
e. Unlinked: 9: 3: 3:1 Linked: >9: <3:<3:>1
f. Unlinked: 1:1:1:1Linked: <9:>3:3:<1
Biology
1 answer:
Luda [366]3 years ago
7 0

Answer:

Option-D

Explanation:

In the given question the genotype of parental allele is- AbaB X AbaB. The alleles formed will be- AB, Ab,aB, ab in which the crossover allele will be AB and ab.

1. Unlinked genes

Now of these alleles are unlinked then the alleles will assort independently of each other therefore performing the punnett square we will get dihybrid ratio as 9:3:3:1 in which  

1. A_B_ are 9

2. aaB_ are 3

3. A_bb are-3

4. aabb- 1

Since the crossover allele are AB and ab therefore A_B_ are 9 and aabb- 1 represent recombinant genotype.

2. Linked genes

When the genes will be linked then chances of crossing over will be reduced as a result of which only two type of allele will be formed which are-  Ab and aB and the recombinant frequency will become less.

Crossing these allele will  form Aabb , AaBb , AaBb  and aaBB  genotypes. The recombinant ratio will decrease as A_B_ is 2 and aabb is not present.

Thus, option-4 which states that the unlinked ration is  9:3:3:1  and linked ratio is  <9: >3: >3: <1 is correct.

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A strand of DNA that is 36 nucleotides long codes for how many amino acids?
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Answer:

See answer below

Explanation:

Hi there,

Assuming this DNA strand is fully capable of being mature mRNA (5' m7G cap and PolyA tail), nucleotides lead to an mRNA codon, which is 3 nucleotides per codon. In turn, 1 codon leads to 1 amino acid. However, as a single strand, it must be capable of terminating translation, which always requires a stop codon, and thus 3 nucleotides. Hence, we must subtract this from the total amount of codons <em>first</em>.

36 \ ncltd \ \frac{1 \ codon}{3 \ ncltd} = 12 \ codon\\12 \ codon - 1 \ codon \  = 11 \ codon\\11 \ codons\ \frac{1 A.A.}{1 \ codon}  = 11 \ A. A.

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thanks,

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GENERAL RECOMBINATION OR HOMOLOGIST

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           Suppose we have an exogenote and an endogenote, both consisting of double helices. In recombination models, the exogenote is usually referred to as donor DNA, and the endogenote as recipient DNA.

1) Start of recombination: Homologous recombination begins with an endonucleotide incision in one of the donor double helix chains. Responsible for this process is the nuclease RecBCD (= nuclease V), which acts as follows: it is randomly attached to the donor's DNA, and moves along the double helix until it finds a characteristic sequence called c

Once the sequence is recognized, the RecBCD nuclease cuts to 4-6 bases to the right (3 'side) of the upper chain (as we have written above). Then, this same protein, acting now as a helicase, unrolls the cut chain, causing a zone of single-stranded DNA (c.s. DNA) to move with its 3 ’free end

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4) Assimilation or synapse: This is the key moment of action of RecA. Somehow, the DNA-bound RecA c.s. The donor facilitates the encounter of the latter with the complementary double helix part of the recipient, so that in principle a triple helix is formed. Then, with the hydrolysis of ATP, RecA facilitates that the donor chain moves to the homologous chain of the receptor, and therefore matches the complementary one of that receptor. In this process, the chain portion of the donor's homologous receptor is displaced, causing the so-called "D-structure".

It is important to highlight that this process promoted by RecA depends on the donor and the recipient having great sequence homology (from 100 to 95%), and that these homology segments are more than 100 bases in length.

Note that this synapse involves the formation of a portion of heteroduplex in the double receptor helix: there is an area where each chain comes from a DNA c.d. different parental (donor and recipient).

5) It is assumed that the newly displaced chain of the recipient DNA (D-structure) is digested by nucleases.

6) Covalent union of the ends originating in the two homologous chains. This results in a simple cross-linking whereby the two double helices are "tied." The resulting global structure is called the Holliday structure or joint.

7) Migration of the branches: a complex formed by the RuvA and RuvB proteins is attached to the crossing point of the Holliday structure, which with ATP hydrolysis achieve the displacement of the Hollyday crossing point: in this way the portion of heteroduplex in both double helices.

8) Isomerization: to easily visualize it, imagine that we rotate the two segments of one of the DNA c.d. 180o with respect to the cross-linking point, to generate a flat structure that is isomeric from the previous one ("X structure").

9) Resolution of this structure: this step is catalyzed by the RuvC protein, which cuts and splices two of the chains cross-linked at the Hollyday junction. The result of the resolution may vary depending on whether the chains that were not previously involved in the cross-linking are cut and spliced, or that they are again involved in this second cutting and sealing operation:

a) If the cuts and splices affect the DNA chains that were not previously involved in the cross-linking, the result will be two reciprocal recombinant molecules, where each of the 4 chains are recombinant (there has been an exchange of markers between donor and recipient)

b) If the cuts and splices affect the same chains that had already participated in the first cross-linking, the result will consist of two double helices that present only two portions of heteroduplex DNA.

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