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marysya [2.9K]
3 years ago
8

High-throughput sequencing reveals that 30 new mutations have occurred in the coding regions of genes in an individual lizard. I

f the coding regions of this lizard comprise 150 million nucleotides per haploid genome, what is the mutation rate per nucleotide?
Biology
1 answer:
dybincka [34]3 years ago
4 0

Answer:

100 per million

Explanation:

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some mutations always occur from generation to generation. But most mutations do not persist over time in the gene pool. Which m
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3 years ago
2. Why is the kingdom Protista not valid under evolutionary classification? (1 point)
HACTEHA [7]
The best and most correct answer among the choices provided by your question is The best and most correct answer among the choices provided by your question is the last choice.
<span>
Kingdom Protista is not valid under evolutionary classification because they evolved before any other kingdom, and as such share a most recent common ancestor with each other, as opposed to any other group.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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2 years ago
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Which processes beings the formation of sedimentary rock?​
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Answer:

clastic sedimentary rock are made up of pieces of pre-existing rocks. pieces of rock are loosened by weathering, then transported to some basin or depression where sediment is trapped. if the sediment is very deeply, it becomes compacted and cemented, forming sedimentary rock.

Explanation:

Hope this helped Mark BRAINLEST!!!!

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Admitting that the "a" is a capital A for normal pigmentation and "d" is a capital D for dimpled chin, meaning that these are the dominant traits, the fraction expected to be albino with a non-dimpled chin is of 1/16.
When two heterozygous are crossed and two characteristics are being analysed, the offspring quantity that will possess the two recessive traits can be represented by 1/16. This is easily confirmed when a Punnett square is made. Considering that both parents were heterozygous, on both sides of the crossing in the Punnett square, you would have the following alleles' combination: AD, Ad, aD, and ad. The offspring that would be homozygous recessive (aadd) would correspond to only 1/16. 
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