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Chapter 15: Q15-7TYU (page 313)

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossover? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

Short Answer

Expert verified

Fifty percent of the offspring obtained from crossover will exhibit recombinant phenotypes because A and B are located far from each other.

It is inferred from the results that the genes are unlinked as the number of recombinants produced is high.

Step by step solution

01

Recombinant frequency

The offspring that result from the crossing over of DNA (deoxyribonucleic acid) during meiosis is called recombinants.The frequency of recombinants is less than non-recombinants because crossing-over is a random process that occurs during meiosis.

The frequency of recombinants depends on the distance between the linked genes.The farther the genes are on the chromosome, the higher the chances of recombination frequency.

Recombination frequency is a measure of recombinants considering the total offspring produced in a cross.

02

Percentage of recombinants produced from crossover

The genotype of animal heterozygous for bothAand B genes isAaBb. The gametes produced by this animal areAB,Ab, aB, andab. The genotype of homozygous recessive isaabb. The gamete produced by this animal isab.

When a heterozygous and homozygous animal is crossed(AaBb脳aabb), four types of offspring are produced. The genotypes of the offspring produced areAaBb,Aabb,aaBb,andaabb.

Out of the four types of progeny, AaBb and aabb are parental types, whereas Aabb and aaBb are recombinants. Thus, fifty percent of the offspring resulting from crossover are recombinants.

03

Interpretation of the result

A and B genes are the same chromosome, and they are 50 map units apart which means they have a recombination frequency of 50 percent.A recombination frequency of 50 percent means that the two genes are unlinked.

If A and B were on different chromosomes, then also the results would have been the same. This is because crossing-over between genes that are located far from each other would have occurred during meiosis.

Fifty of the results obtained are recombinants. This confirms that the genes are unlinked as the percentage of recombination is high.

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Most popular questions from this chapter

The continuity of life is based on heritable information in the form of DNA. In a short essay (100-150 words), relate the structure and behavior of chromosomes to inheritance in both asexually and sexually reproducing species.

The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum

Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found a male.

Butterflies have an X-Y sex determination system that is different from that of flies or humans. Female butterflies may be either XY or X0, while butterflies with two or more X chromosomes are males. This photograph shows a tiger swallowtail gynandromorphy, which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual looking butterfly.

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.
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