Question

please answer all 4 questions

Question 19 3 pts This type of replication bubble: Replication origin is here. o Fits models of replication directionality in eukaryotes but not prokaryotes o Cannot tell which model it fits without more information o Fits models of replication directionality in prokaryotes but not eukaryotes o Fits models of replication directionality in both prokaryotes and eukaryotes Question 20 3 pts When considering human blood types, the Bombay phenotype is an example of: o Incomplete dominance . Epistasis o Incomplete penetrance o Codominance o Sex linkage D Question 21 3 pts A specific type of dwarfism is a genetically inherited recessive trait. Based upon this pedigree, what is the probability that Max and Natalie will have another affected daughter? Joe Curtis Anne Max Natalie Devon Ken Kimberly Ryan Donna o 1/2 o 1/4 o 1/16 o 1/8 D Question 22 3 pts Given the following cross: AabbCcDdEe X AaBb CcDdee, what is the probability of getting the recessive phenotype? o 1/26 o 1/64 01/256 oO

Answer 1

19) Replication bubble is the unwound DNA that is open and is exposed in order to initiate the DNA replication process. The replication bubble is same for eukaryotes and prokaryotes except the number of sites for origin of replication differs in them. Correct answer : fits models of replication directionality in both prokaryotes and eukaryotes.

20) Bombay Phenotype is where the individual having this blood group lack any H antigen. This means no A or B antigen is found in them also. This is a case of : epistasis. Epistasis is where the effect of mutation of a gene also affects the presence or absence of the gene related to it.

21) The disease is genetically recessive which means that it may not be present at every generation which is why the parental generation does not have it. Because Curtis is recessive and suppose his genotype is "dd", The parents Kate and Joe must be Dd and Dd. Similarly, if Max is dd then Kate and Joe must be Dd and Dd. So Natalie must be having Dd or DD. The probability of having one recessive d in her is 1/4. So the probability of having another daughter with dwarfism is : 1/4 x 1/4 = 1/16

22) By crossing the genotypes, the gametes will be about 8. So the number of offsprings will be : 0