Because their jobs required them to set all day.
Answer:
Peppered moths get their color in lieu to become the fittest among all so that they can survive easily.
Well first i just wanna say that life on mars has been discovered,not little green men but Bactria found on the surface and small traces of water have been found as well. some forms of life need little water to survive (like bacteria) but they still need water so yes there is water on mars the question is will we find it.
Answer:
Choice A, BCAD
Explanation:
Recombination frequency is a number that describes the proportion of recombinant offspring produced in a cross. In other words, it tells us how often does crossing over happen between the loci of two observed genes. The bigger the RF is, the larger is the distance between two genes. That means that genes B and D (RF 14%) are the most distant ones, and all the others are located inbetween them. You have to sketch the order of genes based on how far away they are from each other.
If you want to find out more, I recommend reading this article: https://www.khanacademy.org/science/biology/classical-genetics/chromosomal-basis-of-genetics/a/linkage-mapping
hope it helps:)
Really, there are multiple questions all rolled into one.I will try to answer them patiently and systematically.
First summarize data.
A. Neither Olivia nor Marcus have freckles (recessive, ff)
B. Both are heterozygous for the hairline trait (dominant Ww)
C. Neither Marcus nor Olivia can roll their tongues (rr).
D. All four children have dimples (dominant Dx)
E. Both Olivia and Marcus are EE (unattached earlobe trait).
F. Marcus can not detect the bitter taste (pp for PTC gene)Olivia has been found to be able to detect the bitter taste, but she is heterozygous for the trait. (Pp for PTC gene)
A. Freckles, F (Dominant)
"Neither Olivia nor Marcus have freckles" =>
both have genotype ff.
None of the children have freckles (i.e. P(F)=0% for freckles in all children)
B. Widow's Peak, W (dominant)
"Both are heterozygous for the hairline trait"
So both have genotype Ww.
Punnett square
W w
W WW Ww
w Ww ww
Since W is a dominant trait, only ww (25%) will have straight hairline, 75% will inherit the widow's peak.
50% of the children will be homozygous (Ww).
C. Rolling tongues, R (dominant)
"Neither Marcus nor Olivia can roll their tongues"
means that both are homozygous recessive, with genotype rr.As in freckles, all children will have genotype rr, so none of them will roll their tongues.
None will be heterozygous. The whole family's genotype is rr.
D. Dimples, D (dominant)
"D. All four children have dimples"
implies that all children have genotype DD or Dd.
It is likely that at least one parent has genotype DD in order to have 100% of children have DD or Dd.Here are some possibilities
Case 1: DD + DD (both homoozygous dominant)
D D
D DD DD
D DD DD
Phenotype: 100% have dimples
Case 2: DD + Dd (one homoozygous dominant, and other heterozygous)
D d
D DD Dd
D DD Dd
Phenotype: 100% have dimples
Case 3: DD + dd (one homoozygous dominant, and other homozygous recessive)
D D
D DD DD
d Dd Dd
Phenotype: 100% have dimples
Case 4: Dd + Dd (both heterozygous)
D d
D DD Dd
d Dd dd
Phenotype: 75% have dimples, 25 do not.Note: all 4 children could have dimples, with probability 31.6%
Case 5: Dd + dd (Heterozygous + homozygous recessive)
D d
d Dd dd
d Dd dd
Phenotype: 50% have dimples, 50 do not.Note: All four children could have dimples, with probability 6.25%.
Case 6: dd + dd (Both homozygous recessive)
D d
d dd dd
d dd dd
Phenotype: all children have no dimples.
Conclusion:Likely genotypes of parents: DD+DD, DD+Dd, DD+dd
Possible genotypes of parents: Dd+Dd, Dd+dd
Impossible genotype of parents: dd+dd
Therefore we know with certainty that at least one of the parents has dimples.
E. Unattached Earlobe trait, E (dominant)
"Both Olivia and Marcus are EE"
(i.e. unattached earlobe trait).
This means that the whole family will have genotype EE, i.e. all are homozygous dominant, and have unattached earlobes.
F. Bitter taste, P (incomplete dominance)
"Marcus can not detect the bitter taste (pp for PTC gene)
Olivia has been found to be able to detect the bitter taste, but she is heterozygous for the trait. (Pp)"
P p
p Pp pp
p Pp pp
Probability for each single child being able to taste the ptc paper is 1/2.
Probability for all children being able to taste the ptc paper is (1/2)^4=1/16.
If Violet cannot taste the ptc paper, her genotype is pp.
We do not know for sure how many of the children can taste the ptc paper.
The most like situation is only half of them can taste, so do the parents. Therefore, half of the family can taste the ptc paper.
Finally, as to "please answer it correctly", I believe I did. :)