<h2>DNA </h2>
Explanation:
1) Experiment done by Griffith:
- Griffith used two related strains of bacteria, known as R and S
- R bacteria were nonvirulent, meaning that they did not cause sickness when injected into a mouse whereas mice injected with live S bacteria developed pneumonia and died
- Griffith tried injecting mice with heat-killed S bacteria (that is, S bacteria that had been heated to high temperatures, causing the cells to die), the heat-killed S bacteria did not cause disease in mice
- When harmless R bacteria were combined with harmless heat-killed S bacteria and injected into a mouse, not only did the mouse developed disease and died, but when Griffith took a blood sample from the dead mouse, he found that it contained living S bacteria
- Griffith concluded that the R-strain bacteria must have taken up what he called a transforming principle from the heat-killed S bacteria, which allowed them to transform into smooth-coated bacteria and become virulent
2) Experiment done by Avery:
- Avery, McCarty and MacLeod set out to identify Griffith's transforming principle
- They began with large cultures of heat-killed S cells and, through a long series of biochemical steps progressively purified the transforming principle by washing away, separating out, or enzymatically destroying the other cellular components
- These results all pointed to DNA as the likely transforming principle but Avery was cautious in interpreting his results
- He realized that it was still possible that some contaminating substance present in small amounts, not DNA, was the actual transforming principle
3) Experiment done by Hershey and Chase:
- Hershey and Chase studied bacteriophage, or viruses that attack bacteria
- The phages they used were simple particles composed of protein and DNA, with the outer structures made of protein and the inner core consisting of DNA
- Hershey and Chase concluded that DNA, not protein, was injected into host cells and made up the genetic material of the phage
Answer:
linkage with approximately 33 map units between the two gene loci
Explanation:
If two genes are not linked, number of recombinants and parental offspring will be equal. Here it is clearly visible that recombinants are less than parental offspring hence the genes are linked. Given, the offspring are in following numbers:
AaBb = 106 = Parental
aabb = 94 = Parental
Aabb = 48 = Recombinant
aaBb = 52 = Recombinant
Recombination frequency = (Number of recombinants/ Total progeny) * 100 = (100/300) * 100 = 33.33 %
1% recombination frequency= 1 map unit of distance between the two gene loci. So here the distance between the two gene loci is approximately 33 map units.
Hence, these results are consistent with linkage with approximately 33 map units between the two gene loci
- there are two subspecies of this animal
- they can be found in india and sri lanka
- threats of these cats are habitat loss, hunting for their meat, and the crossbreading with other domestic cats
- they are listed as vulnerable because they are down to less than 10,000 cats
- it is one of the smallest cat species (two times smaller than a domestic cat)
- these animals can climb trees to avoid predators (making them very agile) but spend most time on ground
- they are nocturnal
- they are carnivores and eat lizards,birds, frogs and rodents.
- when captive in zoos, they are very friendly towards zoo keepers and there are 56 kept in zoos worldwide.
- in captivity, these animals can live to around 18 years.
hope this helps and please mark brainliest!
The answer is true
becouse variegated leaves have areas with no epidermis
i hope it help...... have a good one
