A wild-type chromosome has the following segments: A B C middot D E F G H I An individual is heterozygous for the following chromosome mutation: A B C middot D H I How would the mutated and wild-type chromosomes pair in prophase I of meiosis? D E F G H of the wild-type chromosome will form a deletion loop. E F G of the wild-type chromosome will form a deletion loop. D E F G H of the wild-type chromosome will form an Inversion loop. E F G of the wild- type chromosome will form an inversion loop. Red-green color blindness is a human x-linked recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color Wind. His 46, XY brother also is color blind. Both parents have normal color vision. Where did the no disjunction that gave rise to the young man with Klinefelter syndrome take place? Assume that no crossing over took place In prophase I of meiosis. Mother - meiosis I Mother - meiosis II Father - meiosis I Father - meiosis II Both parents Functional epidermal lies bull nose (JEB) Is a severe skin disorder that results in blisters over the entire body. The disorder is caused by autosomal recessive mutations at any one of three loci that help to encode lamina 5, a major component In the dermal-epidermal basement membrane. Leena Pulicine and colleagues described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et al. 1997. American Journal of Human Genetics 61:611-619); the child had healthy unrelated parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes. Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA fingerprinting demonstrated that the male assumed to be the father had, In fact, conceived the child. Assuming that no new mutations occurred in this family, what might best explain the presence of an autosomal recessive disease in the child when the mother is heterozygous and the father Is homozygous normal? The infant inherited both copies of chromosome 1 from the mother, which is an example of unit parental disomic. The infant is monatomic for chromosome 1, which he received from his mother but did not receive a chromosome 1 from his father. Epitasis masked the disease phenotype in the father, but complementation of the parental alleles gave rise to a child with functional epidermal lies bull nose. The infant has tiresome 1, where he inherited two recessive alleles from his mother. A translocation between chromosome 1 and chromosome 22 generated a recessive phenotype. Plasmid DNA, like bacterial DNA: is circular. is single stranded. comes in homologous pairs. All of the above. None of the above. Solution 10. Answer is B. EFG of the wild type will form a deletion loop. Because EFG in one chromosome has no matching pair in other chromosome. So EFG loop will be formed and it will be removed. 13. Answer is A. Plasmid DNA like bacte.

1. A wild-type chromosome has the following segments: A B C middot D E F G H I An individual
is heterozygous for the following chromosome mutation: A B C middot D H I How would the
mutated and wild-type chromosomes pair in prophase I of meiosis? D E F G H of the wild-type
chromosome will form a deletion loop. E F G of the wild-type chromosome will form a deletion
loop. D E F G H of the wild-type chromosome will form an Inversion loop. E F G of the wild-
type chromosome will form an inversion loop. Red-green color blindness is a human x-linked
recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color
Wind. His 46, XY brother also is color blind. Both parents have normal color vision. Where did
the no disjunction that gave rise to the young man with Klinefelter syndrome take place? Assume
that no crossing over took place In prophase I of meiosis. Mother - meiosis I Mother - meiosis II
Father - meiosis I Father - meiosis II Both parents Functional epidermal lies bull nose (JEB) Is a
severe skin disorder that results in blisters over the entire body. The disorder is caused by
autosomal recessive mutations at any one of three loci that help to encode lamina 5, a major
component In the dermal-epidermal basement membrane. Leena Pulicine and colleagues
described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et
al. 1997. American Journal of Human Genetics 61:611-619); the child had healthy unrelated
parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes.
Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the
LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA
fingerprinting demonstrated that the male assumed to be the father had, In fact, conceived the
child. Assuming that no new mutations occurred in this family, what might best explain the
presence of an autosomal recessive disease in the child when the mother is heterozygous and the
father Is homozygous normal? The infant inherited both copies of chromosome 1 from the
mother, which is an example of unit parental disomic. The infant is monatomic for chromosome
1, which he received from his mother but did not receive a chromosome 1 from his father.
Epitasis masked the disease phenotype in the father, but complementation of the parental alleles
gave rise to a child with functional epidermal lies bull nose. The infant has tiresome 1, where he
inherited two recessive alleles from his mother. A translocation between chromosome 1 and
chromosome 22 generated a recessive phenotype. Plasmid DNA, like bacterial DNA: is circular.
is single stranded. comes in homologous pairs. All of the above. None of the above.
Solution
10. Answer is B. EFG of the wild type will form a deletion loop. Because EFG in one
chromosome has no matching pair in other chromosome. So EFG loop will be formed and it will
be removed.

2. 13. Answer is A. Plasmid DNA like bacterial DNA is circular and double stranded. Prokaryotes
lack chromosome structure instead they have circular nucleoid structure.