Both haemophilia and color blindness are X-linked recessive traits. A women who

Question

Both haemophilia and color blindness are X-linked recessive traits. A women
who is a carrier for haemophilia marries a colorblind man. Both the man and
woman have a normal karyotype. The couple decides to have a child. Which of
the listed children represents the phenotype least likely to occur?

A Klinefelter syndrome (XXY) son with haemophilia
A Klinefelter syndrome (XXY) son otherwise normal
A colorblind Klinefelter syndrome (XXY) son
An XYY son with haemophilia
A Turner syndrome (XO) otherwise normal daughter
A colorblind daughter with Turner syndrome

Explanation / Answer

Normal fruit flies have grayish-yellow bodies, red eyes, and wings that are long-enough to be able to fly. As Morgan and his students were breeding fruit flies, they found mutant flies with black bodies, some with stumpy, vestigial wings, and some with a brighter, orangish-red eye color that they called “cinnabar,” and by breeding flies, they were able to determine that all three of these mutations were recessive and were on the autosomes. Fruit fly researchers use a different type of symbolism to represent their genetic crosses. Those researchers use a plus sign (+) to indicate anything that is the “wild” type and a letter or two to represent a mutant allele (capital for dominant, lower case for recessive). Thus a fruit fly with a homozygous grayish-yellow body would be labeled as “++” while a black-bodied fly would be “bb” and a heterozygous fly would be “+b”. The symbol for vestigial wings is “vg” and the symbol for cinnabar eyes is ”cn”. Through breeding, Morgan and his students were able to obtain flies that had both a black body and vestigial wings (bbvgvg). They bred some of those flies with wild type flies to obtain flies that were heterozygous for both traits (+b+vg). Then they did a testcross by crossing the +b+vg heterozygous flies with bbvgvg flies, and counted a total of 2300 offspring. If the genes were on different chromosomes, the Punnett square for this cross would look like this: ++ +vg b+ bvg bvg +b+vg +bvgvg bb+vg bbvgvg so out of the 2300 offspring, they would expect to get 575 of each of the four types. If the genes were linked on the same chromosome, the Punnett square for this cross would look like this: ++ bvg bvg +b+vg bbvgvg so out of the 2300 offspring, they would expect to get 1150 of each of the two types. However, when Morgan and his students actually counted the offspring, they were surprised by the results: +b+vg +bvgvg bb+vg bbvgvg 965 185 206 944 Since most of the flies were +b+vg and bbvgvg, that suggested that the two genes were linked. However, the researchers noticed that (206 + 185) × 100 = 17% 2300 of the flies were recombinant types: they exhibited a different combination of phenotypes than either of their parents. Morgan suggested that this was due to an exchange of equal chromosome segments during synapsis in meiosis which he called crossing over. Morgan and his students subsequently determined that the gene for cinnabar eyes was also linked to, was on the same chromosome as, black body and vestigial wings. They determined that the rate of crossing over between b and cn was about 9%, and between vg and cn was about 9.5%. Alfred Sturtevant, one of Morgan’s grad students, suggested that the farther apart genes are on a chromosome, the greater the chances of crossing over. Thus, he suggested that the rates of crossing over (recombination frequency) could be used to predict the order in which the genes occur. In this case, Sturtevant said that, based on the observed recombination frequencies, cn was in between b and vg. Eventually, Sturtevant and his fellow students were able to arrange four groups of fruit fly genes. Since other researchers had found four sets of chromosomes in fruit flies, this added further evidence to support the idea that genes are on chromosomes.

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