Question

You are following the inheritance of three genes: A, B and C. A and B are on the same chromosome; C is on a different chromosome from A and B (for example; A and B are on chromosome 1 and C is on chromosome 4).

Which option correctly represents the genotype of an individual who is heterozygous for all three genes?

What is the recombination frequency for unlinked genes (i.e., genes that sort independently)? Frequency is given as a number from 0 to 1.

Remember, even if genes sort independently you can still calculate a recombination frequency.

Select one:

a. 0.25

b. 1

c. 0.5

d. 0

Genes W and R are on the same chromosome. In a cross between a WR/wr and wr/wr individual, of 1000 offspring, 10 are recombinants and 990 are parental.

The distance between W and R is 1 map unit (m.u.).

Select one:

True

False

Nail patella syndrome is a rare autosomal dominant condition causing misshapen nails and kneecaps. The Nail patella gene is approximately 12 map units apart from the I gene which determines ABO blood type.

• Nail patella syndrome (N) is dominant to normal

• I^A is codominant to I^B; I^A is dominant to i

• I^B is codominant to I^A ; I^B is dominant to i

• I^A I^A and I^A i = Type A blood

• I^B I^B and I^Bi = Type B blood

• I^A  I^B = Type AB blood

• ii = Type O blood

Sam has Nail patella syndrome and Type AB blood, Sam's mother had Type B blood and had normal nails and kneecaps. Tai has normal nails and kneecaps, and Type O blood. What is the likelihood Sam and Tai's child will have Nail patella syndrome, regardless of blood type (i.e., can be any blood type)?

Select one:

a. 1

b. 0

c. 2/3

d. 1/2

Nail patella syndrome is a rare autosomal dominant condition causing misshapen nails and kneecaps. The Nail patella gene is approximately 12 map units apart from the I gene which determines ABO blood type.

• Nail patella syndrome (N) is dominant to normal

• I^A is codominant to I^B; I^A is dominant to i

• I^B is codominant to I^A ; I^B is dominant to i

• I^A I^A and I^A i = Type A blood

• I^B I^B and I^Bi = Type B blood

• I^A  I^B = Type AB blood

• ii = Type O blood

Sam has Nail patella syndrome and Type AB blood, Sam's mother had Type B blood and had normal nails and kneecaps. What correctly represents Sam's genotype?

Hint: there is a reason you're told the phenotype of Sam's mother.

Select one:

a. I^BN/I^An

b. I^Bn/I^AN

c. I^BN/I^AN

d. I^AI^B/Nn

Nail patella syndrome is a rare autosomal dominant condition causing misshapen nails and kneecaps. The Nail patella gene is approximately 12 map units apart from the I gene which determines ABO blood type.

• Nail patella syndrome (N) is dominant to normal

• I^A is codominant to I^B; I^A is dominant to i

• I^B is codominant to I^A ; I^B is dominant to i

• I^A I^A and I^A i = Type A blood

• I^B I^B and I^Bi = Type B blood

• I^A  I^B = Type AB blood

• ii = Type O blood

Sam has Nail patella syndrome and Type AB blood, Sam's genotype is I^AN/I^Bn

Tai has Type O blood and normal nails and kneecaps.

What is the chance Sam and Tai's child will have Type B blood and Nail patella syndrome? Answer is given in percent

Select one:

a. 6%

b. 25%

c. 0%

d. 50%

e. 12%

There are three genes - A, B, C - on the same chromosome. Each gene has two alleles in a dominant/recessive relationship: A dominant to a; B dominant to b; C dominant to c.

Which genotype option(s) could possibly represent an individual heterozygous at all three genes? You may choose more than one.

Select one or more:

a. ACb/acB

b. AaBbCc

c. ABC/abc

d. Cab/cAB

e. A/a; B/b; C/c

There are three genes - A, B, C - on the same chromosome. You have been studying gene A and know that the distance between A and B is 5 m.u. (map units), and the distance between A and C is 3 m.u.

Based on the information you have, above, what are potential arrangements of all three genes on the chromosome? You can choose more than one.

Note: answers give the genes and map units between them, in order. For example if we had: S--10mu--N--3mu--T that means the genes are in the order SNT, and there are 10 map units between S and N, and 3 map units between N and T.

Select one or more:

a. A--3mu--C--2mu--B

b. B--5mu--A--3mu--C

c. A--3mu--C--5mu--B

d. A--2mu--B--3mu--C

Answer 1

Answer 1:

Given that there are three genes, A, B and C. We need to determine the genotype of an individual who is heterozygous for all the genes.

A heterozygous genotype means that the two alleles for a gene are different. One allele is the dominant one and the other allele is recessive.

Thus, genotype of heterozygous individual will be AaBbCc.

Answer 2:

Recombinant frequency represents the proportion of non-parental types in offspring. When the genes are unlinked, i.e. are inherited independently, then 50% progeny are parental types and 50% are recombinants. When genes are linked, the frequency of recombination becomes less than 50% , as linked genes tend to get inherited together.

Hence, the recombination frequency for unlinked genes is 50% , or 0.5.

The correct choice is

(C) 0. 5.

​​​​​​Answer 3:

Given that out of 1000 offspring, 10 are recombinants.

Recombination frequency can be calculated as,

= Recombinants/total × 100

= (10/1000) × 100

= 1%

Map distance is equal to the recombination frequency.

Thus map distance= 1 map units

The given statement is TRUE.