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Answer: The combination produced is RRtt.
The combination rrTt does not appear in this punnett square.
Explanation:
As the problem does not indicate more information, I assume that punnett square is made between two RRtt genotypes. It can produce gametes that have the alleles Rt. A gamete is a sexual cell (egg or sperm) that has only one allele of each gene. And, each gamete codes for a different gene. Since here we have two different types of alleles (R and t), it is a dihybrid cross.
<u>A Punnett square is a diagram used to predict the genotypes of a cross or breeding experiment.</u> It is used to determine the genotypes and phenotypes of the offspring. To do it, you have to label the rows with one parent's gametes and label the columns with the other parent's genotype. Then, have each box inherit letters from its row and column, and interpret the results.
Then, the punnett square (shown in the picture) will be between RRtt and RRtt whose gametes can only be Rt, so Rt x Rt will be made.
The result in an offspring which will be 100% RRtt, and there is no rrTt combination here because non of the parents have an r allele, then none of the children can inherit it.
To have an offspring where rrTt genotype is shown, a different genotype must be used, in which both parents must have at least one r allele and one t allele. For example, it could be Rrtt x rrTt.
Answer:
Explanation:
We are asked to find how much of a 40 gram sample remains after 12 years.
Iron-55 has a half-life of 3 years. Therefore, after 12 years, 4 half-lives have been completed.
- 12 years/3 years = 4 half-lives
Every time a half-life is completed, half of the sample's mass decays. Remember we start with a 40 gram sample.
- 1 half- life: 40 g / 2 = 20 g
- 2 half-lives: 20 g / 2= 10 g
- 3 half-lives: 10 g / 2 = 5 g
- 4 half-lives: 5 g / 2 = 2.5 g
There is also a formula that can be used to solve this problem.
Where A₀ is the initial amount, t is the time, and hl is the half-life.
We know 40 grams is the inital amount, 12 years is the time, and 3 years is the halflife.
- A₀= 40 g
- t= 12
- hl= 3
After 12 years, <u>2.5 grams </u> of Iron-55 will remain.