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Flashcards in Lecture 10 Deck (15):
1

Test cross involving two loci comes out with 1:1:1:1 ratio, what does that mean?

assorted independently -in this case the homozygous parent doesn’t contribute to the phenotype (always produces the same gamete) then can work out the genotype of the other parent

2

If a test cross ratio is away from 1:1:1:1 what does it mean?

-genes linked

3

Why can independent assortment occur?

-there are many more genes than there are chromosomes (only 24 chromosomes) -several genes coexist on the same chromosome -consequently independent assortment of these genes may not occur

4

On which chromosomes can you have aneuploidy and survive?

-mapping of how many genes on which chromosome- the smallest number on chromosomes 13, 18 and 21- those are the ones where you can survive having an extra one (aneuploidy)

5

What is a recombinant?

recombination of the maternal and paternal chromatids, the gametes resulting from a cross over = much less frequent than the parental non-recombinant

6

What is the frequency of recombination dependent upon?

measure of the relative distance between gene loci, the loci closer together= smaller possibility of recombination (up to 50 map units)

7

What is map distance?

-relative distance between gene loci on the same chromosome are measured by recombination frequency -that’s how gene mapping was started -1 map unit (centimorgan cM) = 1% recombination

8

How do you draw a map of loci?

draw a line with the gene names, uner it put number of cM or m u (map units)

9

How far a way must loci be from each other to assort independently?Why?

50 map units apart are said to assort independently: Loci are not linked if they are on the same chromosome but far enough apart for crossing over to occur often enough to produce equal numbers of recombinants and parental combinations. -the loci assort independently but are part of the same linkage group -can never have more than 50% recombinants- due to the sister chromatids (reason for the 50 cM limit)

10

How do you solve problems in this area? (when unequal ratios)

-unequal ratios=linkage problem -2 loci- 4 phenotypes (2parental 2 recombinants) -3 loci- 8 phenotypes (2 parental, 2 recombinants, 4 double recombinants) -double recombinants are the lowest frequency -middle locus swaps over in double recombinants that’s how you find it, the two outecr loci in a double recomb are in the same place as the parent but middle swaps -when something is wild type- heterozygous the normal alleles -when writing the attributes of offspring- the ones wrotten are the mutant all the rest assumed normal -remember that in male drosphila and lucila there is no crossing over

11

How do you write linkage?

AB

ab

12

What is cis arrangement?

AB

ab

                           ab together= remember by- „sisters“ cis

13

What is trans arrangment?

Ab

aB

14

How do you calculate map distance?

map distance = number of recombinant offspring   x 100
                                 total offspring

 

result in cM(centimorgan)

15

Do recombinants change depending on the allelic arrangement?

-yes

                 AB       -cis                                       aB       -trans

ab                                                     Ab
-recombinants are Ab and aB          -recombinants are AB an ab