CREATING PUNNETT SQUARES FOR THE FISHER FAMILY There are six people in the Fisher family. Olivia and Marcus are the parents. The
y have four children: Violet, Nathan, Jonas, and Claudia. Neither Olivia nor Marcus have freckles. What is Olivia's genotype? What is Marcus' genotype? What is the probability that each child has of inheriting freckles? Both Marcus and Olivia are heterozygous for the hairline trait. What is the probability that each of their children could inherit a widow's peak? What is the probability that each of their children could inherit a straight hairline? What is the probability that each of their children could be homozygous dominant for the trait? Neither Marcus nor Olivia can roll their tongues. How many of their children could be born with the ability to roll their tongues? How many of their children could be heterozygous? What is the genotype of the whole family? All four children have dimples. What is likely the genotypes of their parents? Do we know with certainty that one of the parents has to have dimples? Both Olivia and Marcus are EE for the earlobe trait. Among their children, what is the ratio of free earlobes to attached earlobes? What percentage of the children are homozygous? Marcus can not detect the bitter taste. Olivia has been found to be able to detect the bitter taste, but she is heterozygous for the trait. What is the probability that each of their children could taste the ptc paper? If Violet can not taste the ptc paper, what is Violet's genotype? Given what you know, how many of the family can taste the ptc paper? please answer it correctly there or 16 questions
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. :)
In eukaryotic organisms, gamete mother cells are diploid (2N) and have two complete sets of chromosomes. Meiosis in male and female gamete mother cells form haploid male and female gametes (N) respectively. This occurs since meiosis reduces the number of the chromosome to half in the daughter cells. The fusion of haploid male and female gametes during fertilization restores the diploid chromosome number of the species and forms diploid zygote (2N). Repeated mitotic divisions in the diploid zygote form the diploid organism.
The veinlets form a network in the leaf lamina that is why the type of venation reticulate meaning network. So all dicots have dicotyledonous seeds, a tap root system and leaves with reticulate venation. ( with some exceptions of dicots having parallel venation in leaves like Calophyllum)
