LS chapter 5

The parental class =

1 - (SCO 1 + SCO 2 + DCO).

a. One gene; A completely dominant to a.

2 (A- and aa); 3:1

b. One gene; A and a codominant.

3 (AA, Aa and aa); 1:2:1

c. One gene; A incompletely dominant to a.

3 (AA, Aa and aa); 1:2:1

d. Two unlinked genes; no gene interactions; A com- pletely dominant to a, and B completely dominant to b.

4 (A- B-, A- bb, aa B-, aa bb); 9:3:3:1

e. Two genes, 10 m.u. apart; no gene interactions; A completely dominant to a, and B completely domi- nant to b.

4 (A- B-, A- bb, aa B-, aa bb; because the genes are linked, the frequency of the four classes will be different than that seen in part d);

When the 3 genes are genetically linked, these 8 classes will be found as __.

4 reciprocal pairs

unlinked; complete dominance cross with incomplete dominance:

6:3:3:2:1:1

A parent that is heterozygous for 3 genes will give.

8 classes of gametes

Recominants __be detected if genotype is only heterozygous for one gene

CANNOT

The DCO class is the result of a simultaneous crossover in region 1 and region 2. Recombination in two separate regions of the chromosome should be independent of each other, so we can apply the product rule to calculate the expected frequency of DCOs. Thus frequency of DCO =

DCO = rf at region 1 x rf at region 2

In a testcross using males heterozygous for both of these genes, a different set of results was obtained: wild-type wings, wild-type eyes 247 dumpy wings, brown eyes 242

In this cross the heterozygous parent is the male and there is no recombination in Drosophila males!

f. Two unlinked genes; no gene interactions; A and a codominant, and B incompletely dominant to b.

Nine phenotypes in total. There are three phenotypes possible for each gene. The total number of combinations of phenotypes is (3)2 = 9 ((AA, Aa and aa) x (BB, Bb and bb)). 1:2:1 x 1:2:1

g. Two genes, 10 m.u. apart; A completely dominant to a, and B completely dominant to b; and with recessive epistasis between the genes.

Normally there are four phenotypic classes, as in parts d and e (A-B-, A-bb, aaB- and aabb).

Thus, rf at region 1=

SCO 1 + DCO

rf at region 2=

SCO 2 + DCO

2c. Suppose you mated the F1 females from the cross in part (a) to wild-type males. Why would this cross fail to inform you whether the two genes are linked?

The F1 female will still make four types of gametes in equal frequency. However the male parent in this cross can only contribute sc+ j+ gametes so all the progeny will be phenotypically wild type. Thus you could not determine the frequency of parental and recombinant gametes from the F1 female.

How to determine the arrangement of alleles in the parent ?

The arrangement of alleles in the parent is determined by looking at the phenotypes of the largest classes of progeny (the parental reciprocal pair)

How can you measure distance greater than 50 mu?

The measured recombination frequency can be no higher than 50% when looking at data involving a single pair of genes.

h. Two unlinked duplicated genes (that is, A and B perform the same function); A and B completely dominant to a and b, respectively.

Two genes means four phenotypic classes (A- B-, aa B-, A- bb and aa bb). Because gene function is duplicated the first three classes are all phenotypically equivalent in that they have function,

e___k____mb: Why combining two smaller distances gives better representation?

because single crossovers between e and mb and does not include any of the double crossovers that occurred in the e - k and k - mb regions simultaneously.

Genes may assort independently either because they are on different chromosomes or __

because they are far apart on the same chromosome.

Genes on the same chromosome are physically __

connected or linked

The least probable meiotic event is a __ which is a recombination event in the region between M and C (region 1) and simultaneously in the region between C and S (region 2).

double crossover

In other words, each meiotic event in the heterozygous parent must give 2 reciprocal products that occur in __ frequency

equal

In the case of genes on separate chromosomes there will be

four equally frequent classes of progeny.

supposed you've done frequency for F2 brothers and they were true test cross, what would be the frequency for F2 sisters if they were true test cross too?

genotypes, phenotypes and frequencies of the female progeny would be the same as their brothers

In a test cross, the gamete from the __ determines the phenotype of the progeny.

heterozygous parent

The greater the distance between linked genes, the __ the recombination frequency (rf).

higher

The recombination frequencies of pairs of genes indicate ___

how often 2 genes are transmitted together

2d. Suppose you mated females from the true-breed- ing stock with javelin bristles to males with sca- brous eyes and javelin bristles. Why would this cross fail to inform you whether the two genes are linked?

javelin ♀ (sc+sc+ jj) x scabrous javelin ♂ (scsc jj) → F1 javelin ♀ (scsc+ jj). This F1 female will make sc+ j and sc j) parental gametes. Her recombinant gametes will be the same genotypes as her parental gametes thus making it impossible to detect crossing over.

For linked genes, the frequency is

less than 50%.

This particular reciprocal pair will also be the most __event and so the most__pair of products.

likely ; frequent

For instance, a meiosis with __ will produce the parental gametes, M C S and m c s in equal frequency

no recombination

0 mu

no recombination. Two genes are linked.

what genotype from the parents are needed to detect crossing over or independent assortment?

one parent must be heterozygous for two genes or markers. and the other parent must be homozygous recessive. The cross overs detected will occur between the two genes.

In the case of genes on the same chromosome (even genes that are very far apart on the same chromosome) there will

only be two classes of progeny - the parental classes.

The __distances are the most accurate

shortest

The remaining gametes are produced by a __ in region 1 (the reciprocal pair known as single crossovers in region 1) and a single crossover in region 2 (the reciprocal pair known as single crossovers in region 2).

single crossover

Gene pairs that are close together on the same chromosome are genetically linked because__

the alleles on the same homolog are transmitted together (parental types) into gametes more often than not during meiosis

Recombination frequencies become more inaccurate as __

the distance between genes increases.

greater than or equal to 50 mu

the genes assort independently

Gene pairs that assort independently exhibit a recombination frequency of 50% because__

the number of parental types = the number of recombinant

two smaller distances Do not sum to the longer distance because

these recombination frequencies are based on two point crosses.

How to distinguish between genes on separate chromosomes and same chromosome?

use heterozygous male.