Answer:
red-hot iron bar
Explanation:
An iceberg could even contain more heat energy than a cup of coffee or a red-hot iron bar. That's because its bigger and contains so many more molecules, each of which has some heat energy. The coffee and the iron bar are hotter (have a higher temperature), but the iceberg holds more heat because it's bigger.
Answer:
1802
Explanation:
(8)(45)(5)+10−8
(360)(5)+10−8
1800+10−8
1810−8
1802
anything else!!!!
The stages of replication is Attachment, Penetration and Replication
The following information should be considered:
The common steps in both cycles are given below:
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Answer:
Attachment, Penetration and Replication
Explanation:
A bacteriophage is a virus that attacks bacterial cells. The lytic and lysogenic cycles are two methods of viral replication. In the lytic cycle, the virions produced are released from the host cell whereas in the lysogenic cycle, viral nucleic material are incorporated into host nucleic material and are copied to daughter cells when the host cell reproduces. The common steps in both cycles are given below:
1 Attachment – in this step, the bacteriophage attaches itself to the surface of the host cell so as to insert its DNA into the host cell.
2. Penetration – the virus inserts its DNA into the host cell by penetrating the cell membrane of the host cell.
3. Replication – the viral nucleic material is replicated using the host cell's replication mechanism.
C. 1/4
D. 9/16
Answer:
A. 3/16
Explanation:
The plants with wrinkled, green seeds (rryy) and round, yellow seeds were crossed. Since the progeny did not show a "wrinkled" trait, the parent plant with "round" seeds was homozygous for seed shape and heterozygous for the seed color (RRYy). The genotype of the double recessive parent plant would be "rryy".
A cross between RRYy x rryy gives progeny in following ratio: 1/2 round, yellow seeds( RrYy ): 1/2 round, green seeds (Rryy)
The F2 progeny was obtained by self crossing RrYy plants. In F2 progeny, 3/16 plants (2/16 RRYy and 1/16 rryy) were genotypically identical to the parent plant.
b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria?
c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?
If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
The genotype of an individual refers to the sum total of genes that the individual received from its parents. Since Sally has a normal metabolism, Sally is Aa. Sally's mother is Aa while Sally's father and brother are aa.
If Sally parents have another child, using the Punnet square method, there is a 50% chance that the child will have alkaptonuria. If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
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a) Genotypes: : Aa (carrier of the alkaptonuria allele) ,Mother: Aa (carrier of the alkaptonuria allele)Father: aa (has alkaptonuria)Brother: aa (has alkaptonuria)
b) If Sally's parents have another child:The probability of the child having alkaptonuria (aa genotype) is 25%.The probability of the child being a carrier (Aa genotype) is 50%.The probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):The probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).The probability of the child being a carrier (Aa genotype) is 50%.
The probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
Sally: Aa (normal metabolism carrier)
Mother: Aa (normal metabolism carrier)
Father: aa (alkaptonuria)
Brother: aa (alkaptonuria)
b) If Sally's parents have another child:
Probability of the child having alkaptonuria (aa genotype) is 25%.
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):
Probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
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Bacterial Population
Time
Which of these most likely explains why the bacterial population stopped growing?
The bacteria mutated from photosynthetic to carnivorous organisms and consumed one another until all the individuals were dead.
The bacteria reproduced in the system but eventually died due to limited resources
The bacteria became diseased and were unable to survive in the closed system
The bacteria competed for nutrients and died after all the nutrients were consumed
The statement that most likely explains the stop of bacterial populations to grow is the bacteria competed for nutrients and died after all the nutrients were consumed.
Bacterial growth may be defined as the proliferation of bacterium into two daughter cells, in a process called binary fission.
When there are enough nutrients in the petri dish, a rapid increase is shown in the graph, but during the course of time when nutrients become limited, it affects the growth of the bacterial population.
Therefore, the statement that most likely explains the stop of bacterial populations to grow is the bacteria competed for nutrients and died after all the nutrients were consumed.
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B - The bacteria competed for nutrients and died after all the nutrients were consumed