Question

In: Biology

The black color cost in hamsters is due to a dominant gene (B).. A recessive allele...

The black color cost in hamsters is due to a dominant gene (B)..

A recessive allele (b) at this locus results in a brown coat when homozygous (b/b). However, neither coat color is expressed when the organism is homozygous for the allele (a) at a separate locus. The a/a genotype results in a white (albino) coat, regardless of the allele at the B locus. The wild-type allele (+) at the (a) locus allows normal coat coloration, whether the genotype is +/+ or +/a.

The following experiments were performed to better understand these relationships.

Experiment 1:

A female hamster with the genotype B/B; +/+ is crossed with a male hamster of genotype b/b; a/a.

Experiment 2:

Female offspring from the cross in Experiment 1 were backcrossed to the (b/b; a/a) parent. The distribution of coat coloration among the progeny was as follows; black (66), brown (34), and white (100).

2. A true-breeding strain of black hamsters is available. What is the genotype of this strain?

A. B/b; +/+

B. B/b; +/a

C. B/B; +/+

D. B/B; +/a

3. A strain of hamsters known to be homozygous (b/b; a/a) at both loci is available. What is the phenotype of these animals?

A. Black

B. Brown

C. White

D. A mixture of white and brown

4. What will be the phonotype(s) of the F1 animals resulting from the cross in Experiment 1?

A. All black

B. All brown

C. All white

D. Both black and brown

5. Experiment 2 suggests that the two genetic loci discussed in the passage are:

A. linked

B. unlinked.

C. recessive.

D. Not enough information to determine linkage.

6. Based on the results from Experiment 2, what is the genetic map distance (frequency of recombination) between the two loci discussed in the passage?

A. 17 centimorgans

B. 34 centimorgans

C. 68 centimorgans

D. Impossible to determine; they are unlinked.

Expert Solution

2 B. B/b; +/a becacuse Sometimes there is crossover. The number of crossover gametes will always be less than the number of non-crossover gametes. Crossing over B+/ba will give you two new gametes: Ba and b+. The crossover offspring will be Ba/ba and b+/ba.

Therefore, we've come up with 4 genotypes of offspring. Two of them are parental (non-crossover). Two of them are crossover (and less common).

Parental: B+/ba (Black) 66
Parental: ba/ba (Albino) 66
Crossover: Ba/ba (Albino) 34
Crossover: b+/ba (Brown) 34

3.A. Black

4.A.All black

5. A.linked becuse,The calculation for this data is:

(RR, Rr) Percentage of nontarget genes from donor parent = (1/2)ⁿ⁺¹
where , n = number of backcross.
However, the rate at which genes entering a cross from the donor parent (R = rust resistant) are eliminated during backcrossing will be influenced by linkage . Linkage is a term used to describe genes that do not independently assort (one of the requirements for Mendelian inheritance). Linked genes tend to stay together and unlinked genes independently assort. The physical basis for linkage is how close the genes are to each other on a chromosome. If genes are far apart or on different chromosomes, they are unlinked and independently assort. If they are near to each other on the chromosome they are more frequently passed on together than separately.

6. B.34 centimorgans because,The trick is that the two parental genotypes will have the same probability of happening with each other. The same goes for crossover. I filled in the numbers next to the categories. Notice that you can go ahead and write the numbers for Black parental and Brown crossover right away. Then, you can assume that the number of parental Albinos is 66 (same as the other parental number) and the number of crossover Albinos is 34 (same as the other crossover number). It checks out since 66 + 34 is 100.

To determine centimorgans, divide the number of crossover individuals by the total individuals.
68 / 200 = 34.
You actually could have shortcutted to the answer by realizing that brown individuals are only due to crossover. You could take 34 and double it to get the number of albino individuals, then divide by 200.

gladiator answered 2 years ago

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