In labrador retrievers, black fur (B) is dominant to brown fur (b). Normal visio

Question

In labrador retrievers, black fur (B) is dominant to brown fur (b). Normal vision (N) is dominant to retinal atrophy (n) which causes blindness. Cross BbNn X BbNn. What is the phenotypic ratio of the offspring?

4 black, normal: 4 black, blind: 4 brown, normal: 4 brown blind

3 black, normal : 1 brown blind

100% black and normal

9 black, normal: 3 black, blind: 3 brown, normal : 1 brown blind

Savannah was 19 when she died of complications from her cystic fibrosis. Neither of her parents had the disease, but they were carriers. Cystic fibrosis is a recessive condition. What were the chances that her parents would produce a child with cystic fibrosis?

75%

0%

100%

25%

Alexandra is a carrier for hemophilia (which is an X-linked recessive condition). She marries Bob who neither has, nor is a carrier for hemophilia. What are the chances that this couple produce a child with hemophilia (the actual disease)?

100%

0%

25%

50%

Conner Hopf was diagnosed with Tay-Sachs disease before his first birthday. Neither of his parents showed symptoms of the disease, but this disease is genetic. This means that Tay-Sachs is a _____ disorder.

ATA GCA GTA CCT

UAU CGU CAU GGA

TAT CGT CAT GGA

AUA GCA GUA CCU

25%

50%

Not enough information

100%

Using the rules of complementary base pairing, replicate this strand of DNA:

     ATA GCA GTA CCT

environmental

sex-linked

dominant

The mom from Little People Big World has a form of dwarfism called achrondroplasia which is a dominant disorder. The father has a completely unrelated form of dwarfism. Each time this couple have a child, what are the chances that they produce a child with achondroplasia?

recessive

dominant

environmental

spontaneous mutation

Red-green color blindness is a ______ disorder.

4 black, normal: 4 black, blind: 4 brown, normal: 4 brown blind

3 black, normal : 1 brown blind

100% black and normal

9 black, normal: 3 black, blind: 3 brown, normal : 1 brown blind

Explanation / Answer

1. Parents              Black, normal (BbNn)                             x               Black, normal (BbNn)

Gametes                BN, Bn, bN, bn                                                         BN, Bn, bN, bn

Progeny                  

BBNN

Black, normal

BBNn

Black, normal

BbNN

Black, normal

BbNn

Black, normal

BBNn

Black, normal

BBnn

Black, blind

BbNn

Black, normal

BBnn

Black, blind

BbNN

Black, normal

BbNn

Black, normal

bbNN

Brown, normal

bbNn

Brown, normal

BbNn

Black, normal

Bbnn

Black, blind

bbNn

Brown, normal

bbnn

brown, blind

So, the phenotypic ratio is 9 Black, normal: 3 Black, blind: 3 Brown, normal : 1 Brown blind

2. Cystic fibrosis is caused by a recessive allele. Let's assume C is the normal allele and c is diseased. Parents are carriers which means they will have genotypes Cc

So,      Parents                            Cc                                x                            Cc

Gametes                                      C, c                                                     C, c

Progeny            

So, 1 progeny out of four may be diseased. Hence the probability of producing a diseased child (i.e. with cystic fibrosis is 1/4 i.e. 25%.

3. Alexandra is a carrier of hemophilia which means she has one normal X chromosome and another Xh. Bob does not have hemophilia, so his genotype will be XY.

So,      Parents                                   XXh                     x                       XY

Gametes                                          X, Xh                                              X, Y

Progeny       

XX

Normal

XXh

Carrier

XY

Normal

XhY

Diseased

So, 1 progeny out of four may be diseased. Hence the probability of producing a diseased child (i.e. with hemophilia is 1/4 i.e. 25%.

4. Connor........................

Here, we observe that the parents do not show the disease, but the child does. Also, it is known that the disease is genetic. So, the disease appears to be recessive. Both the parents might be carriers having one normal allele which is dominant over the disease causing allele. Hence, the parents appear normal. But the child must have inherited the diseased allele from both the parents and hence was found to have Tay-Sachs disease.

5. In DNA, A pairs with T and C pairs with G.

Template DNA                                              ATA GCA GTA CCT

Complementary DNA                                   TAT CGT CAT GGA

6. Achondroplasia is a dominant disorder. In the present case mother has achondroplasia but the father does not have the disease. Let's denote disease allele with A and the normal allele with a. Now the mother may have AA or Aa i.e she may be homozygous or heterozygous for the condition. However, the homozygous achondroplasia generally results in death. So, the mother must be heterozygous (Aa). Father does not have the disease, so he must have the genotype aa.

Parents                                Aa   (Achondroplasia )                             x                   aa (No Achondroplasia )

Gametes                    A, a                                                                                      a

Progeny                           

Aa

(Achondroplasia )

aa

(No Achondroplasia )

So, out of 2 types of progeny one can have Achondroplasia. This means that the probability of producing a diseased child (i.e. with Achondroplasia) is 1/2 i.e. 50%.

7. Red green color blindness is a___________

recessive sex-linked disorder. This is actually an X-linked recessive disorder.

Gametes BN Bn bN bn BN

BBNN

Black, normal

BBNn

Black, normal

BbNN

Black, normal

BbNn

Black, normal

Bn

BBNn

Black, normal

BBnn

Black, blind

BbNn

Black, normal

BBnn

Black, blind

bN

BbNN

Black, normal

BbNn

Black, normal

bbNN

Brown, normal

bbNn

Brown, normal

bn

BbNn

Black, normal

Bbnn

Black, blind

bbNn

Brown, normal

bbnn

brown, blind

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