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Anvisha [2.4K]
3 years ago
6

What would be the probability of obtaining 6 offspring, 4 of which were dominant and 2 recessive if it was a test cross (Aa x aa

) instead of a cross between two heterozygotes? Give your answer to 4 decimal places.
Biology
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:0.0156

Explanation:

If the test cross is between one heterozygous (Aa) and one homologous recessive gene (aa). The possible genotype for their offspring will be

Aa aa Aa aa

The probability of dominant gene (Aa) is 2/4

The probability of recessive gene (aa) is 2/4

The probability that in six offspring 4 is dominant and two recessive is

= 2/4*2/4*2/4*2/4*2/4*2/4

=64/4096

=0.015625

=0.0156 to 4 decimal places

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a. Reduces represor binding.

b. Constitutive expression of the operon.

c. Stronger promoter if it is more similar to the consensus sequence. Weaker promoter if less similar.

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The lac operon is an operon required for lactose transport and metabolism in enteric bacteria such as <em>Escherichia coli</em>. <u>It is regulated by glucose and lactose availability</u> and consists of the following structural genes:

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  • Lac a gene: encodes the enzyme thiogalactoside transferase, which catalyzes the transfer of the acetyl group of acetyl coenzyme A to 6-OH of a thiogalactoside acceptor. This gene is not related to lactose metabolism.
  • Promoter: region of DNA recognized by RNA polymerase for transcription.
  • Operator: region of DNA located between the promoter and the beginning of the structural genes, which is recognized by the repressor protein Lac I.
  • Repressor gene (lac I): encodes the Lac I repressor protein, which recognizes the operator region, where it binds. It prevents the transcription of genes under the control of this promoter but stimulates the binding of RNA polymerase. When the repressor is absent (in the presence of inducer which in this case will be lactose or IPTG), RNA polymerase will begin transcription.

The lac operon is under a type of negative regulation, where genes can always be transcribed, except when the Lac I repressor protein is bound to the operon region, for which it has a high affinity. In this case, the promoter of the lac I gene is constitutive, so the Lac I protein is permanently expressed and remains bound in tetramer form to the operon region, preventing the transcription of structural genes.

Since lactose is the inducer of the operon, it is able to bind to the Lac I repressor protein and generate a conformational change that decreases its affinity for the operon region. Thus, the operon region is left free, <u>RNA polymerase can freely transcribe the structural genes and β-galactosidase can degrade lactose to glucose plus galactose</u>. <u>In the absence of lactose, the Lac I repressor protein maintains its high affinity for the operator region, preventing RNA polymerase from transcribing the structural genes</u>. In this way, the system remains closed with consequent energy savings for the bacterium.

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c. Since the promoter us a region of DNA recognized by RNA polymerase for transcription, if there is a mutation here it will generate a stronger promoter in the case that this part is similar to a consensus sequence (the most commonly encountered nucleotides found at a certain location). In the case that it is less similar, then it will generate a weaker promoter.

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