Concept 14.4: Many human traits follow Mendelian patterns of inheritance Flashcards
Unable to manipulate the mating patterns of people, geneticists instead analyze the results of matings that have already occurred. They do so by collecting information about a family’s history for a particular trait and assembling this information into a family tree describing the traits of parents and children across the generations—a family
pedigree
a three-generation pedigree that traces the occurrence of a pointed contour of the hairline on the forehead. This trait, called a ______________, is due to a dominant allele, W.
widow’s peak
Because the widow’s-peak allele is dominant, all individuals who lack a widow’s peak must be
homozygous recessive (ww).
The two grandparents with widow’s peaks must have the Ww genotype, since some of their offspring are
homozygous recessive.
The offspring in the second generation who do have widow’s peaks must also be heterozygous, because they are the products of
Ww x ww matings.
The third generation in this pedigree consists of two sisters. The one who has a widow’s peak could be either ________________________________, given what we know about the genotypes of her parents (both Ww).
homozygous (WW) or heterozygous (Ww)
figure 14.15 pedigree analysis
What is the probability that the child will have a widow’s peak? This is equivalent to a Mendelian F1 monohybrid cross (Ww × Ww), and therefore the probability that a child will inherit a dominant allele and have a widow’s peak is
¾ (¼ WW + ½ Ww).
What is the probability that the child will be unable to taste PTC? We can also treat this as a monohybrid cross (Tt × Tt), but this time we want to know the chance that the offspring will be homozygous recessive (tt). That probability is
¼.
Finally, what is the chance that the child will have a widow’s peak and be unable to taste PTC? Assuming that the genes for these two characters are on different chromosomes, the two pairs of alleles will assort independently in this dihybrid cross (WwTt × WwTt). Therefore, we can use the multiplication rule:
¾ (chance of widow’speak) × ¼ (chance of inability to taste PTC) = 3⁄16 (chance of widow’s peak and inability to taste PTC).
Thousands of genetic disorders are known to be inherited as simple
recessive traits
An allele that causes a genetic disorder (let’s call it allele a) codes for either a malfunctioning protein or
no protein at all.
In the case of disorders classified as recessive, heterozygotes (Aa) typically have the normal phenotype because one copy of the normal allele (A) produces a
sufficient amount of the specific protein.
Thus, a recessively inherited disorder shows up only in the homozygous individuals (aa) who inherit a recessive allele from each
parent
Although phenotypically normal with regard to the disorder, heterozygotes may transmit the recessive allele to their offspring and thus are called
carriers
figure 14.16 albinism: a recessive trait
A mating between two carriers corresponds to a Mendelian F1 monohybrid cross, so the predicted genotypic ratio for the offspring is
1 AA : 2 Aa : 1 aa.
Thus, each child has a 1/4 chance of inheriting a double dose of the recessive allele; in the case of albinism, such a child will have .
albinism
From the genotypic ratio, we also can see that out of three offspring with the normal phenotype (one AA plus two Aa), two are predicted to be heterozygous carriers, a
2/3 chance.
Recessive homozygotes could also result from Aa x aa and aa matings, but if the disorder is lethal before reproductive age or results in sterility (neither of which is true for albinism), no aa individuals will
reproduce
Tay-Sachs disease, which we described earlier in this chapter, is disproportionately high among
Ashkenazic Jews, Jewish people whose ancestors lived in central Europe.
The probability of passing on recessive traits increases greatly, however, if the man and woman are close
relatives