In humans, the absence of small molars and six-fingered teeth are dominant in relation to the norm.

In humans, the absence of small molars and six-fingered teeth are dominant in relation to the norm. A man with six-fingered fingers and the absence of small molars, heterozygous for both of the above characteristics, marries a normal woman for these characteristics. A) How many types of gametes does a woman have? B) What is the probability (in%) of the birth in the family of a child who will inherit both anomalies from the father? Q) How many different phenotypes are there among children in this family? D) What is the probability (in%) of the birth of a healthy child in the family? E) How many different genotypes can there be among children?

Let’s designate the gene that causes the absence of small molars in a person as B, then the gene that causes the development of a normal dental formula will be b.

Let’s designate the six-fingered gene as C, then the gene for the normal structure of the hand will be c.

Heterozygous six-fingered male without small molars – BbCc. It produces four types of spermatozoa – bc, BC, bc and bC.

A woman with a normal structure of the hand and the composition of the teeth – VVSS, because the norm refers to recessive traits and manifests itself only in the case of homozygosity of the genotype. It produces oocytes of the same type – bc.

The possible offspring of this married couple will be represented by the following options:

children without anomalies of the hand and teeth (bbcc) – 25%;

children with six fingers and without small molars (BbCc) – 25%;

children without small molars with a normal hand (Bbcc) – 25%;

six-toed children with normal teeth (bbCc) – 25%.

A) 1 type of gametes;

B) 25%;

C) 4 phenotypes;

D) 25%;

E) 4 genotypes.