Acouple plans to have a child. the mother is homozygous for an autosomal recessive mutation. the father is heterozygous for the same mutation. with respect to this mutation, what are the possible genotypes of any children they have? calculate the probability of each genotype's occurrence. show how you got your answer.

Answer:

The possible genotypes of the children are Hh and hh.The probability of genotype Hh = 1/2The probability of genotype hh = 1/2

Explanation:

Autosomal genes are the genes present on any of the chromosomes except the sex chromosomes.An autosomal recessive mutation is the mutation in an autosomal gene which only expresses itself in the phenotype in homozygous condition and affects both males and female equally. For the given question, let the autosomal gene be represented by H (dominant) and h (recessive)According to the question,

Since the mother is homozygous her genotype would be hh and the father is heterozygous so his genotype would be Hh.

The following cross is made to know the genotypes of the children :

hh X Hh--------> Hh Hh hh hh

So the possible genotypes of the children are Hh and hh as shown in the cross.

Note : An image is attached representing the cross in a Punnett square.

The probability of an event = No. of favourable outcomes/total no. of outcomes

The probability of occurrence of genotype Hh would be = 2/4 = 1/2

Similarly, the probability of occurrence of genotype hh would be 2/4 = 1/2

Answer:

The possible genotypes of the children are Hh and hh.The probability of genotype Hh = 1/2The probability of genotype hh = 1/2

Explanation:

Autosomal genes are the genes present on any of the chromosomes except the sex chromosomes.An autosomal recessive mutation is the mutation in an autosomal gene which only expresses itself in the phenotype in homozygous condition and affects both males and female equally. For the given question, let the autosomal gene be represented by H (dominant) and h (recessive)According to the question,

Since the mother is homozygous her genotype would be hh and the father is heterozygous so his genotype would be Hh.

The following cross is made to know the genotypes of the children :

hh X Hh--------> Hh Hh hh hh

So the possible genotypes of the children are Hh and hh as shown in the cross.

Note : An image is attached representing the cross in a Punnett square.

The probability of an event = No. of favourable outcomes/total no. of outcomes

The probability of occurrence of genotype Hh would be = 2/4 = 1/2

Similarly, the probability of occurrence of genotype hh would be 2/4 = 1/2

Hunting in the darksome fish rely on senses beyond sight. in the darkness of the marianas trench, a fish called the rattail has an extremely sensitive sense of smell. the fish uses scent, rather than sight, to steer toward food. once it's near food, the rattail uses a special organ on the bottom of its chin to guide itself by feel.marine snowone staple of the deep sea diet is "marine snow," flakes of organic, edible substances that sink down from the upper ocean. animals at the ocean's bottom find sustenance in everything from the bodies of dead creatures to fecal matter. marine snow is diverse enough to support numerous species. for instance, several species of snails and worms feed exclusively on deceased whales.microorganismsthe deep sea is an intricate ecosystem, and even the ocean floor itself plays a part. hydrothermal vents are cracks in the sea floor where erupting magma mixes with ocean water to produce plumes of hot, sulfur-rich chemicals. some microbes can use these chemicals for energy, sustaining their existence and that of the animals who depend on their roles in the deep sea food chain.