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A mother is mildly affected with a mitochondrial disease. As she is only mildly affected, she decides to have 3 children. Her first child is also mildly affected. However, her second child is moderately affected, and her third child was severely affected and died in infancy. How can the three children display such different phenotypic variation?

What is the complement DNA strand to 5′ – GGA CGT – 3′?   

Fill in the blanks below: Heterochromatin is _________ than euchromatin and heterochromatin has a _________ density of transcriptionally active DNA compared to euchromatin.

Do individuals with monosomy X or trisomy X have any phenotypic symptoms?

During meiosis I there is a non-disjunction of chromosome 11. What would be the genetic make-up of the resulting embryo following fertilization by a normal gamete?

The pedigree below shows the inheritance of a mitochondrial disorder. What is the most likely mode of inheritance for this mitochondrial disorder? Assume complete penetrance.   

Which statement describing DNA is incorrect.

Colorblindness and adrenoleukodystrophy (ALD) are both X linked recessive phenotypes (C = normal; c = colorblind and A = normal; a = adrenoleukodystrophy). The genes are 12 m.u. apart on the X chromosome.  Adult onset cerebral ALD begins in middle age and has a rapid onset to death. Cara’s dad is color blind and has just started showing signs of ALD. Assume there are no recessive alleles on Cara’s mother’s side of the family (i.e., assume Cara is C___A c___a). Cara’s 3-yr old son Oliver has normal vision. What is the probability that Oliver will develop ALD?    

James is mildly affected with the mitochondrial disorder MERRF (myoclonic epilepsy and ragged red fiber disease), due to a mutation in his mtDNA. His partner is phenotypically normal and does not have a family history of mitochondrial disease. What is the likelihood the couple will have a child affected with MERFF?

The rhesus (Rh) blood group gene (Rh positive = R; Rh negative = r) and a gene that causes a form of anemia called elliptocytosis (E = anemia; e = no anemia) show linkage. Of 200 offspring from a cross where one parent is Rh negative with no anemia (r___e r___e) and the other parent is Rh positive with anemia (R___E r___e): 80 are R___E r___e       80 are r­___e r­­­­___e        20 are R___e  r___e       20 are r___E r___e.   Based off these 200 offspring, how many map units apart are the genes?