Question

Read the passage and answer the following question.

The genes that code for the human leukocyte antigen(HLA) antigens, also known as the major histocompatibility complex(MHC) genes, are all located on human chromosome $6$. Two brothers and a sister participated in a bone marrow donor registration, and their HLA typing results are given.

ID	HLA-A	HLA-B	HLA-C	HLA-DR
$1$	A$1$, A$3$	B$5$, B$7$	C$3$, Cw$19$	DR$3$, DR$9$
$2$	A$1$, A$6$	B$2$, B$6$	C$2$, C$16$	DR$4$, DR$13$
$3$	A$1$, A$6$	B$2$, B$6$	C$2$, C$16$	DR$4$, DR$13$

What is your explanation for the single HLA-A result for sibling $3$?

• A. A gene-deletion event occured during crossing over
• B. Nondisjunction occurred during meiosis
• C. A laboratory error caused the lost data; it should have a second result
• D. A metabolic error prevented the synthesis of the second antigen
• E. Both the parents were heterozygous for HLA-A$1$

Answer 1

The correct option is E Both the parents were heterozygous for HLA-A$1$

HLA is inherited as a "set" of the three HLA groups, A, B, DR known as halotype. In given halotype of sibling 3, all four HLAs (HLA-A, B , C and DR) are different refecting the heterozygosity of two parents for the same. The sibling has HLA of two types: A1 and A6 i.e. one parent have HLA-A1 and other has HLA-A6. Both parents are heterozygous for HLA-A. Thus, the correct answer is option E.