According to the Hardy–Weinberg law, if a population has 1 in 1600 of its members affected by a homozygous recessive disorder, how many members will be heterozygous carriers?
D. This can be calculated using the Hardy–Weinberg equilibrium. If p and q are the allele frequencies of recessive copy ‘a’ and dominant copy ‘A’, respectively, then p2 gives the frequency of homozygous aa and q2 gives the frequency of homozygous AA. 2pq gives the frequency of the heterozygous aA. 1/1600 is the frequency of homozygous individuals, that is p2. Hence, p = 1/40. If p = 1/40, then q will be 1 − p = 39/40. The number of heterozygous carries is given by 2 × 1/40 × 39/40 = 1/20 approximately. As one can see easily from this calculation, at any given time in a population there are more heterozygous carriers than diseased individuals. So negative eugenics, that is elimination of all diseased individuals, cannot eliminate an autosomal recessive disease.
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Nucleic acids constitute the chemical base of genes.
Which of the following is not a component of nucleic acids?
E. Arachidonic acid is a fatty acid often found in membrane phospholipids. It is not a component of nucleic acids. Nucleic acids are made up of a purine or pyrimidine base, a pentose sugar moiety, and a phosphate group. Depending on whether the sugar is deoxyribose or ribose sugar, nucleic acids are either DNA or RNA.
Which of the following genetic studies compares the frequency of a marker in groups of patients versus unrelated controls?
A. Association studies simply compare the frequency of a particular marker, for example a polymorphism, in diseased and normal populations. These are usually case–control studies and are comparatively easy to carry out. Linkage studies investigate the cosegregation of a disease and a set of genetic markers. Here, the aims are to determine linkage among candidate loci and determine the genetic distance between loci in an attempt to narrow down the site of the genetic abnormality. Linkage study is possible only if at least one parent has a double heterozygote make-up, that is heterozygous at both marker and disease loci. Family study refers to a genetic study whereby cases are ascertained by interviewing all available relatives of an identified proband. Age correction must be applied in such family-based case ascertainments. Adoption study investigates shared traits or phenotypes among adoptees, adopting families, and biological families in various combinations. Ecological study is not a specific genetic study.
In genetic twin studies, pair-wise concordance differs from proband-wise concordance in that:
A. In twin studies, case ascertainment and zygosity assignment is done initially. Later, concordance or discordance is measured to determine the heritability. This could be done by counting the proportion of the total number of concordantly affected twins among all pairs studied (pair-wise concordance) or by calculating the proportion of the number of affected twins among all cotwins studies (probands-wise concordance). The latter is possible if there is a twin register maintained with systematic ascertainment. Zygosity does not influence selection of the methods. Multifactorial diseases are commonly studied using twin studies by both methods.
The ratio of clinically affected to unaffected offspring for an autosomal recessive disorder where both parents are carriers is:
B. Autosomal recessive diseases often skip generations. The usual pattern of inheritance is from two heterozygous carriers, who are often unaware of their carrier status until their child is born homozygous with the disease. The chance of having an offspring with homozygous inheritance is 1 in 4; in other words, one affected child for every three unaffected children born (1: 3). This ratio becomes 1: 1 (50%) if one of the parents is homozygous and suffering from the disease.
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