Genetics Answers

The gene se. The gene sequence of a chromosome is HIJKL.MNOPQ .
What chromosome change resulted in the following sequences of the
chromosome?
(a) HIJKNM.LOPQ
(b) HIJKL.MNONOPQ
(c) HIJKLLK.MNOPQ
(d) HIJKL.MNCDEFG

The frequency of recessive allele in a large randomly mating population is 0.2. What is the frequency of : (a) the dominant allele? (b) hetrozygote?

The gene sequence of chromosome is HIJKLMNOPQwhat chromosome change resulted in the following sequence of the chromosome?(a) HIJKNMLOPQ(b)HIJKLMNONOPQ(c)HIJKLLKMNOPQ(d)HIJKLMNCDEFG

The gene sequence of a chromosome is HIJKL.MNOPQ.
What chromosome change resulted in the following sequences of the chromosome?
(a) HIJKNM.LOPQ
(b) HIJKL.MNONOPQ
(c) HIJKLLK.MNOPQ
(d) HIJKL.MNCDEFG

With the help of clear and labelled diagrams describe how aneuploids are formed.

In peas Tall (T) is dominant over dwarf (t), Yellow is dominant over green (y), and smooth (S) is dominant over wrinkled (s).What fraction of the offspring in the following cross would be homologous recessive for all gene pairs in the cross: YyTtss X YyttSs ?

Two parents with cleft chins, both heterozygous for cleft chin (Cc), have three children with cleft chins. The parents are sure that their fourth child will not have a cleft chin. Draw a Punnett Square to see if this is possible.
Determine the probability that the fourth child will have a cleft chin.


Using Models: What information does a Punnett square give? Does the Punnett square support the parents' prediction? Explain your answer.
Predicting: If these parents have a fifth child, what are the chances that the child will have a cleft chin? Explain your answer.

Describe how aneuploids are formed

A gene is found on the X chromosome of humans. A recessive allele leads to loss of the gene function. Using a notation that distinguishes a normal X chromosome and a mutated X chromosome, indicate the genotype of an affected male, a normal male, an affected female, a normal female who carries the mutation, and normal female who can have all normal sons. Try the same exercise for a mutated allele that is dominant over the normall allele.