Chapter 7

look at # 29

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4. Why is the frequency of recombinant gametes always half the frequency of crossing over?

Crossing over occurs at the four-strand stage, when two homologous chromosomes, each consisting of a pair of sister chromatids, are paired. Each crossover involves just two of the four strands and generates two recombinant gametes. The remaining two strands that were not involved in the crossover generate two nonrecombinant gametes. Therefore, the frequency of recombinant gametes is always half the frequency of crossovers, because only half of the gametes from each crossing-over event are recombinant gametes.

23. In German cockroaches, bulging eyes (bu) are recessive to normal eyes (bu+) and curved wings (cv) are recessive to straight wings (cv+). Both traits are encoded by autosomal genes that are linked. A cockroach has genotype bu+bu cv+cv, and the genes are in repulsion. Which of the following sets of genes will be found in the most-common gametes produced by this cockroach? a. bu+ cv+ b. bu cv c. bu+ bu d. cv+ cv e. bu cv+

The most common gametes will have (e) bu cv . Equally common will be gametes that have bu+cv, not given among the choices. Since these genes are linked, in repulsion, the wild-type alleles are on different chromosomes. Thus the cockroach has one chromosome with bu+cv and the homologous chromosome with bu cv+. Meiosis always produces nonrecombinant gametes at higher frequencies than recombinants, so gametes bearing bu cv+ will be produced at higher frequencies than (a) or (b), which are the products of recombination. The choices (c) and (d) have two copies of one locus and no copy of the other locus. They violate Mendelian segregation: Each gamete must contain one allele of each locus.

17. In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What phenotypes and phenotypic proportions are expected in the progeny of this cross?

The recombinants should total 32.6%, so each recombinant phenotype will be 16.3% of the progeny. Because the F1 inherited a chromosome with heart-shaped leaves and numerous spines (hl ns) from one parent and a chromosome with normal leaves and few spines (Hl Ns) from the other parent, these are the nonrecombinant phenotypes, and together they total 67.4%, or 33.7% each. The two recombinant phenotypes are heart-shaped leaves with few spines (hl Ns) and normal-shaped leaves with numerous spines (Hl ns). Heart-shaped, numerous spines 33.7% Normal-shaped, few spines 33.7% Heart-shaped, few spines 16.3% Normal-shaped, numerous spines 16.3%

19. Alleles A and a are at a locus that is located on the same chromosome as is a locus with alleles B and b. Aa Bb is crossed with aa bb and the following progeny are produced: Aa Bb 5 Aa bb 45 aa Bb 45 aa bb 5 What conclusion can be made about the arrangement of the genes on the chromosome in the Aa Bb parent?

The results of this testcross reveal that Aa bb and aa Bb, with far greater numbers, are the progeny that received nonrecombinant chromatids from the Aa Bb parent. Given that all the progeny received ab from the aa bb parent, the nonrecombinant progeny received either an Ab or an aB chromatid from the Aa Bb parent. Therefore, the A and B loci are in repulsion in the Aa Bb parent. Aa Bb and aa bb are the recombinant classes, and their frequencies indicate that the genes A and B are 10 m.u. apart.

15. In silkmoths (Bombyx mori), red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild- type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are: wild-type eyes, wild-type wings 418 red eyes, wild-type wings 19 wild-type eyes, white-banded wings 16 red eyes, white-banded wings 426 a. What phenotypic proportions would be expected if the genes for red eyes and white-banded wings were located on different chromosomes? b. What is the percent recombination between the genes for red eyes and white- banded wings?

(a) 1⁄4 wild-type eyes, wild-type wings 1⁄4 red eyes, wild-type wings 1⁄4 wild-type eyes, white-banded wings 1⁄4 red eyes, white-banded wings (b) The F1 heterozygote inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+ wb+) from the other parent. These are therefore the phenotypes of the nonrecombinant progeny, present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white-banded wings. RF = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0% The distance between the genes is four map units.

2. In a testcross for two genes, what types of gametes are produced with (a) complete linkage (b) independent assortment (c) incomplete linkage?

(a) Complete linkage of two genes means that only nonrecombinant gametes will be produced; the recombination frequency is zero. (b) Independent assortment of two genes will result in 50% of the gametes being recombinant and 50% being nonrecombinant, as would be observed for genes on two different chromosomes. Independent assortment may also be observed for genes on the same chromosome if they are far enough apart that one or more crossovers occur between them in every meiosis. (c) Incomplete linkage means that greater than 50% of the gametes produced are nonrecombinant and less than 50% of the gametes are recombinant; the recombination frequency is greater than 0 and less than 50%.

14. In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over? b. What will the results of the testcross be if the loci assort independently? c. What will the results of the testcross be if the loci are linked and 20 m.u. apart?

(a) With absolute linkage, there will be no recombinant progeny. The F1 inherited banded and yellow alleles (BBCY) together on one chromosome from the banded yellow parent and unbanded and brown alleles (BOCBw) together on the homologous chromosome from the unbanded brown parent. Without recombination, all the F1 gametes will contain only these two allelic combinations, in equal proportions. Therefore, the F2 testcross progeny will be 1⁄2 banded, yellow (BBBBCYCY) and 1⁄2 unbanded, brown (BOBBCBwCY). (b) With independent assortment, the progeny will be: 1⁄4 banded, yellow (BBBBCYCY) 1⁄4 banded, brown (BBBBCBwCY) 1⁄4 unbanded, yellow (BOBBCYCY) 1⁄4 unbanded, brown (BOBBCBwCY) (c) The recombination frequency is 20%, so each of the two classes of recombinant progeny must be 10%. The recombinants are banded, brown and unbanded, yellow. The two classes of nonrecombinants are 80% of the progeny, so each class must be 40%. The nonrecombinants are banded, yellow and unbanded, brown. In summary: 40% banded, yellow (B^B B^B C^Y C^Y ) 40% unbanded, brown (B^O B^B C^Bw C^Y) 10% banded, brown (B^B B^B C^Bw C^Y) 10% unbanded, yellow (B^O B^B C^Y C^Y)